10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB). Visualize:
Solve:
For the bullet,
KB =
1 1 mB vB2 = (0.01 kg)(500 m /s) 2 = 1250 J 2 2
For the bowling ball,
K BB =
1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2 2
Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 1000 times smaller than the mass of the bowling ball, its speed is 50 times larger.
10.2. Model: Model the hiker as a particle. Visualize:
The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley is y0 = −8.5 m. Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is
∆U = U gf − U gi = mgyf − mgyi = mg( yf − yi ) = (65 kg)(9.8 m /s 2 )[ 4420 m − ( −85 m )] = 2.87 × 10 6 J Assess: Note that ∆U is independent of the origin of the coordinate system.
10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize:
Solve:
For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,
KC = KT ⇒
1 1 mC vC2 = mT vT2 ⇒ vC = 2 2
mT vT = mC
20, 000 kg (25 km / hr ) = 112 km / hr 1000 kg
Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
10.4. Model: Model the oxygen and the helium atoms as particles. Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is four times heavier than the helium atom, so mO = 4 mHe . Solve: The energy conservation equation K O = K He is
v 1 1 2 2 mO vO2 = mHe vHe ⇒ ( 4 mHe )vO2 = mHe vHe ⇒ He = 2.0 vO 2 2 Assess: The result vHe = 2 vO , combined with the fact that mHe = 14 mO , is a consequence of the way kinetic energy is defined: it is directly proportional to the mass and to the square of the speed.
10.5. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize:
Solve:
(a) The kinetic energy of the car is
KC =
1 1 mC vC2 = (1500 kg)(30 m /s) 2 = 6.75 × 10 5 J 2 2
(b) Let us relabel KC as Kf and place our coordinate system at yf = 0 m so that the car’s potential energy Ugf is zero, its velocity is vf, and its kinetic energy is Kf. At position yi, vi = 0 m /s or K i = 0 J, and the only energy the car has is Ugi = mgyi . Since the sum K + Ug is unchanged by motion, K f + U gf = K i + U gi . This means
K f + mgyf = K i + mgyi ⇒ K f + 0 = K i + mgyi ⇒ yi =
( Kf − Ki ) (6.75 × 10 5 J - 0 J ) = = 45.9 m (1500 kg)(9.8 m / s 2 ) mg
(c) From part (b),
1 2 1 2 mv − mv (vf2 − vi2 ) ( Kf − Ki ) 2 f 2 i yi = = = 2g mg mg Free fall does not depend upon the mass.
10.6. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Visualize:
The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c). Solve: The quantity K + Ug is the same during free fall: K f + U gf = K i + U gi . We have 1 2 1 (a) mv1 + mgy1 = mv02 + mgy0 2 2
⇒ y1 = (v02 − v12 ) 2 g = [(10 m / s) 2 − (0 m / s) 2 ]/(2 × 9.8 m / s 2 ) = 5.10 m
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. 1 2 1 (b) mv2 + mgy2 = mv02 + mgy0 2 2 Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m / s. 1 2 1 (c) mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g( y0 − y3 ) 2 2 ⇒ v32 = (10 m / s) 2 + 2(9.8 m / s 2 )[0 m − ( −20 m )] = 492 m 2 / s 2 This means the magnitude of v3 is equal to 22.2 m/s. Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which it was tossed up, but in the opposite direction.
10.7. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball, considered as a particle, does not change during its motion. Visualize:
The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: The quantity K + Ug is the same during free fall. Thus, K f + U gf = K i + U gi . 1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g( y1 − y0 ) (a) 2 2 2 2 ⇒ v0 = (0 m / s) + 2(9.8 m / s 2 )(10 m − 1.5 m ) = 166.6 m 2 / s 2 ⇒ v0 = 12.9 m / s 1 2 1 mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g( y0 − y2 ) (b) 2 2 ⇒ v22 = 166.6 m 2 / s 2 + 2(9.8 m / s 2 )(1.5 m − 0 m ) ⇒ v2 = 14.0 m / s Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable. Also, note that mass does not appear in the calculations that involve free fall.
10.8. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential energy does not change during its sliding motion. Visualize:
Solve: The quantity K + Ug is the same at the top of the ramp as it was at the bottom. The energy conservation equation K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g( yf − yi ) 2 2 2 2 2 ⇒ vi = (0 m / s) + 2(9.8 m / s )(1.026 m − 0 m ) = 20.11 m 2 / s 2 ⇒ vi = 4.48 m / s Assess: An initial push with a speed of 4.48 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems reasonable.
10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion. Visualize:
The vertical displacement of the skateboarder is equal to the radius of the track. Solve: The quantity K + Ug is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g( yf − yi ) 2 2 2 2 vi = (0 m / s) + 2(9.8 m / s 2 )(3.0 m − 0 m ) = 58.8 m / s ⇒ vi = 7.67 m / s Assess: Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.
10.10. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of the ball’s kinetic and gravitational potential energy, therefore, does not change during the rolling motion. Visualize:
Solve: Since the quantity K + Ug does not change during rolling motion, the energy conservation equations apply. For the linear segment the energy conservation equation K 0 + U g 0 = K1 + U g1 is
1 1 1 1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ m v12 + mg(0 m ) = m(0 m / s) 2 + mgy0 ⇒ mv12 = mgy0 2 2 2 2 2 For the parabolic part of the track, K1 + U g1 = K 2 + U g 2 is
1 2 1 1 1 1 mv1 + mgy1 = mv22 + mgy2 ⇒ mv12 + mg(0 m ) = m(0 m / s) 2 + mgy2 ⇒ mv12 = mgy2 2 2 2 2 2 Since from the linear segment we have 12 m v12 = mgy0 , we get mgy0 = mgy2 or y2 = y0 = 1.0 m. Thus, the ball rolls up to exactly the same height as it started from. Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is no rolling friction. This result is an important consequence of energy conservation.
10.11.
Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other. Visualize:
The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K + Ug is the same at the lowest point of the trajectory as it was at the highest point. Thus, K1 + U g1 = K 0 + U g 0 means
1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v12 + 2 gy1 = v02 + 2 gy0 2 2 ⇒ v12 + 2 g(0 m ) = (0 m / s) 2 + 2 gy0 ⇒ v1 = 2 gy0 From the pictorial representation, we find that y0 = L − L cos 30°. Thus,
v1 = 2 gL(1 − cos 30°) = 2(9.8 m /s 2 )(0.75 m )(1 − cos 30°) = 1.403 m /s (b) Since the quantity K + Ug does not change, K 2 + U g 2 = K1 + U g1 . We have
1 2 1 mv2 + mgy2 = mv12 + mgy1 ⇒ y2 = (v12 − v22 ) 2 g 2 2 2 ⇒ y2 = [(1.403 m /s) − (0 m /s) 2 ]/(2 × 9.8 m /s 2 ) = 0.100 m Since y2 = L − L cosθ , we obtain
cosθ =
L − y2 (0.75 m ) − (0.10 m ) = = 0.8667 ⇒ θ = cos −1 (0.8667) = 30° (0.75 m ) L
That is, the pendulum swings to the other side by 30° . Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar chart in the figure.
10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the swing’s motion. Visualize:
Solve: The quantity K + Ug is the same at the highest point of the swing as it is at the lowest point. That is, K 0 + U g 0 = K1 + U g1 . It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing and child will also be maximum. The above equation is
1 2 1 mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g( y0 − y1 ) 2 2 ⇒ v12 = (0 m / s) 2 + 2 g( y0 − 0 m ) ⇒ v1 = 2 gy0 We see from the pictorial representation that
y0 = L − L cos 45° = (3.0 m ) − (3.0 m )cos 45° = 0.879 m ⇒ v1 = 2 gy0 = 2(9.8 m / s 2 )(0.879 m ) = 4.15 m / s Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.15 m/s is reasonable.
10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitational potential energy, therefore, does not change during the car’s motion. Visualize:
Solve:
(a) The initial energy of the car is
K0 + Ug 0 =
1 2 1 mv0 + mgy0 = (1500 kg)(10.0 m /s) 2 = 7.50 × 10 4 J 2 2
The energy of the car at the top of the hill is
K1 + U g1 = K1 + mgy1 = K1 + (1500 kg)(9.8 m / s 2 )(5 m ) = K1 + 7.35 × 10 4 J If the car just wants to make it to the top, then K1 = 0. In other words, an energy of 7.35 × 104 J is needed to get to the top. Since this energy is less than the available energy of 7.50 × 104 J, the car can make it to the top. (b) The conservation of energy equation K 0 + U g 0 = K 2 + U g 2 is
7.50 × 10 4 J =
1 2 1 mv2 + mgy2 ⇒ 7.50 × 10 4 J = (1500 kg)v22 + (1500 kg)(9.8 m / s 2 )( −5.0 m ) 2 2 ⇒ v2 = 14.1 m / s
Assess: A higher speed on the other side of the hill is reasonable because the car has increased its kinetic energy and lowered its potential energy compared to its starting values.
10.14. Model: Assume an ideal spring that obeys Hooke’s law. Visualize:
Solve: (a) The spring force on the 2.0 kg mass is Fsp = − k∆y. Notice that ∆y is negative, so Fsp is positive. This force is equal to mg, because the 2.0 kg mass is at rest. We have − k∆y = mg. Solving for k:
k = −( mg / ∆y) = −(2.0 kg)(9.8 m / s 2 )/( −0.15 m − ( −0.10 m )) = 392 N / m (b) Again using − k∆y = mg:
∆y = − mg / k = −(3.0 kg)(9.8 m / s 2 )/(392 N / m ) y ′ − ye = −0.075 m ⇒ y ′ = ye − 0.075 m = −0.10 m − 0.075 m = −0.175 m = −17.5 cm The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number.
10.15. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a particle. Visualize:
Solve:
We will use the subscript s for the scale and sp for the spring.
(a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives
∑(F
) = Fs on m − w = 0 ⇒ Fs on m = w = mg = (5.0 kg)(9.8 m / s 2 ) = 49 N
on m y
(b) In this case, the force is
∑(F
) = Fs on m + Fsp − w = 0 ⇒ 20 N + k∆y − mg = 0
on m y
⇒ k = ( mg − 20 N )/ ∆y = ( 49 N − 20 N )/ 0.02 m = 1450 N / m (c) In this case, the force is
∑(F
) = Fsp − w = 0 ⇒ k∆y − mg = 0
on m y
⇒ ∆y = mg / k = ( 49 N )/(1450 N / m ) = 0.0338 m = 3.4 cm
10.16. Model: Assume that the spring is ideal and obeysr Hooke’s law. Also model the sled as a particle. Visualize: The only horizontal force acting on the sled is Fsp .
Solve:
Applying Newton’s second law to the sled gives
∑(F
) = Fsp = ma x ⇒ k∆y = ma x
on sled x
⇒ ax = k∆ x / m = (150 N / m )(0.20 m )/ 20 kg = 1.50 m /s2
10.17. Model: Assume that the spring is ideal and obeys Hooke’s law. Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring is given as Fsp = k∆ x, where ∆x is the change in length of the spring. If the mass m is at rest, then Fsp is also equal to the weight w = mg. Solve: We have Fsp = k∆ x = mg. We want a 0.10 kg mass to give ∆x = 0.010 m. This means k = mg / ∆x = (0.10 kg)(9.8 N / m )/(0.010 m ) = 98 N / m
10.18. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law. Visualize:
Solve:
According to Newton’s second law the force on the student is
∑(F
) = Fspring on S − w = ma y
on S y
⇒ Fspring on S = w + ma y = mg + ma y = (60 kg)(9.8 m / s 2 + 3.0 m / s 2 ) = 768 N Since Fspring on S = FS on spring = k∆y, k∆y = 768 N. This means ∆y = (768 N )/(2500 N / m ) = 0.307 m.
10.19. Model: Assume an ideal spring that obeys Hooke’s law. Solve: The elastic potential energy of a spring is defined as Us = 12 k ( ∆s) 2 , where ∆s is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. We have ∆s = 20 cm = 0.20 m and k = 500 N / m . This means
Us =
1 1 k ( ∆s) 2 = (500 N / m )(0.20 m ) = 10 J 2 2
Assess: Since ∆s is squared, Us is positive for a spring that is either compressed or stretched. Us is zero when the spring is in its equilibrium position.
10.20. Model: Assume an ideal spring that obeys Hooke’s law.
Solve: The elastic potential energy of a spring is defined as Us = 12 k ( ∆s) 2 , where ∆ s is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. ∆Us = 0 when the spring is at its equilibrium length and ∆s = 0. We have Us = 200 J and k = 1000 N / m. Solving for ∆s:
∆s =
2Us / k = 2(200 J ) / 1000 N/ m = 0.632 m
10.21. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy
K + Us is conserved. Also model the book as a particle. Visualize:
The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely transformed into the kinetic energy of the book. Solve: The conservation of energy equation K 2 + Us 2 = K1 + Us1 is
1 2 1 1 1 mv2 + k ( x 2 − x e ) 2 = mv12 + k ( x1 − x e ) 2 2 2 2 2 Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to
1 2 1 mv2 = k ( x1 − 0 m ) 2 ⇒ v2 = 2 2
kx12 = m
(1250 N / m )(0.040 m ) 2 = 2.00 m /s (0.500 kg)
Assess: This problem can not be solved using constant-acceleration kinematic equations. The acceleration is not a constant in this problem, since the spring force, given as Fs = − k∆x, is directly proportional to ∆x or | x − x e |.
10.22. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanical energy K + Us is conserved. Also, model the block as a particle. Visualize:
The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the equilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm. Solve: The conservation of energy equation K 2 + Us 2 = K1 + Us1 is
1 2 1 1 1 mv2 + k ( x 2 − x e ) 2 = mv12 + k ( x1 − x e ) 2 2 2 2 2 Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.02 m, we get
1 1 k ( x 2 − x e ) 2 = mv12 ⇒ ∆x = ( x 2 − x e ) = 2 2
m v1 k
That is, the compression is directly proportional to the velocity v1 . When the block collides with the spring with twice the earlier velocity (2v1), the compression will also be doubled to 2(x2 − xe) = 2(2.0 cm) = 4.0 cm. Assess: This problem shows the power of using energy conservation over using Newton’s laws in solving problems involving nonconstant acceleration.
10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved. Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the spring. This give x1 = xe = 0 and (x2 − xe) = 60 cm. Solve: The conservation of energy equation K 2 + Us 2 = K1 + Us1 is
1 2 1 1 1 mv2 + k ( x 2 − x e ) 2 = mv12 + k ( x1 − x e ) 2 2 2 2 2 Using v2 = 0 m/s, (x2 − xe) = 0.60 m, and x1 = xe = 0 m gives:
1 1 k ( x 2 − x e ) 2 = m v12 ⇒ v1 = 2 2
k ( x2 − xe ) = m
250 N / m (0.60 m ) = 3.00 m / s 10 kg
10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved. Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the plane, x 2 − x e = 30 m. Solve: The conservation of energy equation K 2 + Us 2 = K1 + Us1 for the spring-jet plane system is
1 2 1 1 1 mv2 + k ( x 2 − x e ) 2 = mv12 + k ( x1 − x e ) 2 2 2 2 2 Using v2 = 0 m /s, x1 = xe = 0 m, and x2 − xe = 30 m yields
1 1 k ( x 2 − x e ) 2 = mv12 ⇒ v1 = 2 2
k ( x 2 − x1 ) = m
Assess: A landing speed of 60 m/s or ≈ 120 mph is reasonable.
60, 000 N / m (30 m ) = 60 m /s 15, 000 kg
10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and mechanical energy. Visualize:
Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities (vfx )1 and (vfx ) 2 are unknowns and can not be determined from one equation. Solve: Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2 These two equations can be solved for (vfx )1 and (vfx ) 2 , as shown by Equations 10.39 through 10.43, to give Energy conservation:
(vfx )1 =
m1 − m2 50 g − 20 g (vix )1 = (2.0 m / s) = 0.857 m / s m1 + m2 50 g + 20 g
(vfx ) 2 =
2 m1 2(50 g) (vix )1 = (2.0 m / s) = 2.86 m / s m1 + m2 50 g + 20 g
Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move along the +x direction.
10.26. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force acts on the system (clay + block). We also represent our system as a particle. Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
( m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 Using (vix )1 = v0 and (vix ) 2 = 0, we get
vfx =
m1 0.05 kg (vix )1 = (vix )1 = 0.0476(vix )1 = 0.0476 v0 (1 kg + 0.05 kg) m1 + m2
(b) The initial and final kinetic energies are given by
1 1 1 1 2 m1 (vix )12 + m2 (vix ) 22 = (0.050 kg)v02 + (1 kg)(0 m /s) = (0.025 kg)v02 2 2 2 2 1 1 K f = ( m1 + m2 )vf2x = (1 kg + 0.050 kg)(0.0476) 2 v02 = 0.00119 v02 2 2 Ki =
K − Kf 0.00119 × 100% = 95.2% The percent of energy lost = i × 100% = 1 − 0.025 Ki
10.27. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. Visualize:
Solve: In the case of a perfectly elastic collision, the two velocities (vfx )1 and (vfx ) 2 can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity vfx and can be determined from just the momentum conservation equation. (a) Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2 These two equations can be solved as shown in Equations 10.39 through 10.43: m − m2 (100 g) − (300 g) (vfx )1 = 1 (vix )1 = (10 m / s) = −5.0 m / s (100 g) + (300 g) m1 + m2 Energy conservation:
2 m1 2(100 g) (vix )1 = (10 m / s) = +5.0 m / s (100 g) + (300 g) m1 + m2 (b) For the inelastic collision, both balls travel with the same final speed vfx . The momentum conservation equation pfx = pix is (vfx ) 2 =
( m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 100 g ⇒ vfx = (10 m /s) + 0 m /s = 2.5 m /s 100 g + 300 g Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.
10.28. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision obeys the momentum as well as the energy conservation law. Visualize:
Solve:
Momentum conservation: mP (vix ) P + mC (vix ) C = mP (vfx ) P + mC (vfx ) C
1 1 1 1 mP (vix ) 2P + mC (vix ) 2C = mP (vfx ) 2P + mC (vfx ) 2C 2 2 2 2 These two equations can be solved, as described in the text through Equations 10.39 to 10.43: Energy conservation:
(vfx ) P =
m − 12 mP m P − mC 7 7 mP + mC (vix ) P = P (2.0 × 10 m /s) = −1.69 × 10 m /s m P + mC mP + 12 mP
(vfx ) C =
2 mP 2 mP 6 7 (vix ) P = (2.0 × 10 m /s) = 3.08 × 10 m /s m P + mC mP + 12 mP
After the elastic collision the proton rebounds at 1.69 × 10 7 m /s and the carbon atom moves forward at 3.08 × 10 6 m /s.
10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by E = K + U, we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE) and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: (a) We have E = 5.0 J and this energy is a constant. For x < 1.0, U > 5.0 J and, therefore, K must be negative to keep E the same (note that K = E − U or K = 5.0 J − U). Since negative kinetic energy is unphysical, the particle can not move to the left. That is, the particle will move to the right of x = 1.0 m. (b) The expression for the kinetic energy is E − U. This means the particle has maximum speed or maximum kinetic energy when U is minimum. This happens at x = 2.0 m. Thus,
K max = E − U min = (5.0 J ) − (1.0 J ) = 4.0 J
1 2 mvmax = 4.0 J ⇒ vmax = 2
2( 4.0 J ) = m
8.0 J = 20 m / s 0.020 kg
The particle possesses this speed at x = 2.0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are the turning points of the motion.
10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U, we can draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: The kinetic energy is given by E − U, so we have
1 2 mv = E − U ⇒ v = 2( E − U )/ m 2 Using U B = 2.0 J, UC = 3.0 J, and U D = 0 J, we get
vB = 2(5.0 J − 2.0 J )/ 0.5 kg = 3.46 m/s vD = 2(5.0 J − 0 J )/ 0.5 kg = 4.47 m/s
vC = 2(5.0 J − 3.0 J )/ 0.5 kg = 2.83 m/s
10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from the potential energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, but the sum K + U does not change. Solve: The equation for total energy E = U + K means K = E − U, so that K is maximum when U is minimum. We have
K max =
1 2 mvmax = 5.0 J − U min 2
⇒ vmax = 2(5.0 J − U min )/ m = 2(5.0 J − 1.0 J )/ 0.002 kg = 63.2 m /s
10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as it reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just makes it to the top with zero kinetic energy. Solve: (a) The energy equation K A + U A = K top + U top is
1 2 mvA + U A = 0 J + U top 2 ⇒ vA = 2(U top − U A )/ m = 2(5 J − 2 J )/ 0.100 kg = 7.75 m /s (b) To go from point B to point A, K B + U B = K top + U top is
1 2 mvB + U B = 0 J + U top 2 ⇒ vB = 2(U top − U B )/ m = 2(5 J − 0 J )/ 0.100 kg = 10.0 m /s Assess: The particle requires a higher kinetic energy to reach A from B than to reach B from A.
10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and gravitational potential energy does not change as the vehicle coasts down the hill. Visualize:
The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that 35 km / hr = (35000 m / 3600 s) = 9.722 m / s Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2 1 mvi + mgyi = mvf2 + mgyf â&#x2021;&#x2019; vi2 + 2 gyi = vf2 2 2 â&#x2021;&#x2019; vf = vi2 + 2 gyi = (9.722 m / s) 2 + 2(9.8 m / s)(15 m ) = 19.7 m / s = 71 km / hr You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 19.7 m/s or 71 km/hr at the bottom of the hill, when your speed at the top of the hill was 35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the final kinetic energy.
10.34. Model: This is case of free-fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls. Visualize:
The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2 1 mvi + mgyi = mvf2 + mgyf â&#x2021;&#x2019; vi2 + 2 gyi = vf2 2 2 vf = vi2 + 2 gyi = (80 m /s) 2 + 2(9.8 m /s 2 )(10 m ) = 81.2 m /s Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of the velocity vector.
10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the rock is thrown. Model the Frisbee and the rock as particles. Visualize:
The coordinate system is put on the ground for this system, so that yf = 16 m . The rock’s final velocity vf must be at least 5.0 m/s to dislodge the Frisbee. Solve: (a) The energy consrevation equation for the rock K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi 2 2 This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodge the Frisbee. This equation is true for all angles that will take the rock to the Frisbee. (b) Using the above equation we get
vf2 + 2 gyf = vi2 + 2 gyi
vi = vf2 + 2 g( yf − yi ) = (5.0 m / s) 2 + 2(9.8 m / s 2 )(16 m − 2 m ) = 17.3 m / s
Assess: Kinetic energy is defined as K = 12 mv 2 and is a scalar quantity. Scalar quantities do not have directional properties.
10.36. Model: For the marble spinning around the inside of a smooth pipe, the sum of the kinetic and gravitational potential energy does not change. Visualize:
We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is 10 cm, and therefore ytop = y2 = 2R = 0.20 m. At an arbitrary angle θ, measured from the bottom of the circle, y = R − R cosθ. Solve: (a) The energy conservation equation K 2 + U g 2 = K1 + U g1 is
⇒
1 2 1 mv2 + mgy2 = mv12 + mgy1 2 2
⇒ v2 = v12 + 2 g( y1 − y2 ) = (3.0 m / s) 2 + 2(9.8 m / s 2 )(0 m − 0.20 m ) = 2.25 m / s (b) Expressing the energy conservation equation as a function of θ : 1 1 K (θ ) + U g (θ ) = K1 + U g1 ⇒ mv 2 (θ ) + mgy(θ ) = mv12 + 0 J 2 2
⇒ v(θ ) = v12 − 2 gy(θ ) = v12 − 2 gR(1 − cosθ ) Using v1 = 3.0 m/s, g = 9.8 m/s2, and R = 0.10 m, we get v(θ ) = 9 − 1.96(1 − cosθ ) (m/s) (c) The accompanying figure shows a graph of v for a complete revolution (0° ≤ θ ≤ 360°).
Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kinetic energy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since some of the kinetic energy has been transformed into the marble’s potential energy.
10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Visualize:
Please refer to Figure P10.37. Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band results from a negative displacement of the rock. That is, ( Fsp ) x = − kx, where x is the rock’s displacement from the equilibrium position due to the spring force Fsp. (b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as Fsp = − kx, the rubber band obeys Hooke’s law. (c) From the graph, | ∆Fsp | = 20 N for | ∆x | = 10 cm. Thus,
| ∆Fsp |
20 N = = 200 N / m 0.10 m | ∆x | (d) The conservation of energy equation K f + Usf = K i + Usi for the rock is k=
1 2 1 2 1 2 1 2 1 1 1 1 mvf + kx f = mvi + kx i ⇒ mvf2 + k (0 m ) 2 = m(0 m /s) 2 + kx i2 2 2 2 2 2 2 2 2 vf =
k xi = m
200 N / m (0.30 m ) = 19.0 m /s 0.05 kg
Assess: Note that xi is ∆x, which is the displacement relative to the equilibrium position, and xf is the equilibrium position of the rubber band, which is equal to zero.
10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + Us is conserved. Visualize:
Note that xf = xe and xi − xe = ∆x. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation K f + Usf = K i + Usi for the single spring is 1 2 1 1 1 mvf + k ( x f − x e ) 2 = mvi2 + k ( x i − x e ) 2 2 2 2 2 Using the value for vf given in the problem, we get 1 2 1 1 1 mv0 + 0 J = 0 J + k ( ∆x ) 2 ⇒ mv02 = k ( ∆x ) 2 2 2 2 2 Conservation of energy for the two-spring case: 1 1 1 1 mVf2 + 0 J = 0 J + k ( x i − x e ) 2 + k ( x i − x e ) 2 mVf2 = k ( ∆x ) 2 2 2 2 2 Using the result of the single-spring case, 1 mVf2 = mv02 ⇒ Vf = 2 v 0 2 Assess: The block separates from the spring at the equilibrium position of the spring.
10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + Us is conserved. Visualize:
The springs in both cases have the same compression ∆x. We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected springs for the two-spring case (b). Solve: The conservation of energy for the single-spring case: 1 1 1 1 K f + Usf = K i + Usi ⇒ mvf2 + k ( x f − x e ) 2 = mvi2 + k ( x i − x e ) 2 2 2 2 2 Using xf = xe = 0 m, vi = 0 m/s, and vf = v0, this equation simplifies to 1 2 1 mv0 = k ( ∆x )2 2 2 Conservation of energy in the case of the two springs in series, where each spring compresses by ∆x /2, is
K f + Usf = K i + Usi ⇒
1 1 1 1 mVf2 + 0 = mvi2 + k ( ∆x / 2) 2 + k ( ∆x / 2) 2 2 2 2 2
Using xf = x e′ = 0 m and vi = 0 m/s, we get
1 1 1 mVf2 = k ( ∆x )2 2 2 2 Comparing the two results we see that Vf = v0 / 2 . Assess: The block pushes on the spring until the spring has returned to its equilibrium length.
10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conserves both momentum and mechanical energy. Visualize:
The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from a height of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can then apply the conservation laws to find earth’s recoil velocity. Solve: (a) To get the speed of the ball at collision:
(v1 ) 2B = (v0 ) 2B − 2 g( y1 − y0 ) B = (0 m/s) − 2(9.8 m/s 2 )( −10.0 m ) ⇒ (v1 ) B = 2(9.8 m/s 2 )(10.0 m ) = 14.0 m/s In an elastic collision, the velocity of the object that is struck is 2 mB 2(0.50 kg) (14.0 m / s) = 2.33 × 10 −24 m / s ( v2 ) E = (v1 ) B = mE + mB 5.98 × 10 24 kg (b) The time it would take earth to move 1.0 mm at this speed is given by
speed =
distance 10 −3 m = = 4.29 × 10 20 s = 1.36 × 1013 years velocity 2.33 × 10 −24 m /s
10.41. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also assume zero rolling friction between the marble and the incline. Visualize:
This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble’s speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble and the steel ball. Solve: The conservation of energy equation K1 + U g1 = K 0 + U g 0 gives us:
1 1 mM (v1 ) 2M + mM gy1 = mM (v0 ) 2M + mM gy0 2 2 1 (v1 ) 2M = gy0 ⇒ (v1 ) M = 2 gy0 . When the marble collides with the 2 steel ball, the elastic collision gives the ball velocity 2 mM 2 mM 2 gy0 ( v2 ) S = (v1 ) M = m M + mS m M + mS Using (v0 ) M = 0 m / s and y1 = 0 m, we get
Solving for y0 gives 2
y0 =
1 m M + mS (v2 ) S = 0.258 m = 25.8 cm 2 g 2 mM
10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in an elastic collision. Visualize:
Solve:
For a package with mass m the conservation of energy equation is
K1 + U g1 = K 0 + U g 0 ⇒
1 1 m(v1 ) 2m + mgy1 = m(v0 ) 2m + mgy0 2 2
Using (v0 )m = 0 m / s and y1 = 0 m,
1 m(v1 ) 2m = mgy0 ⇒ (v1 ) m = 2 gy0 = 2(9.8 m / s 2 )(3.0 m ) = 7.668 m / s 2 (a) For the perfectly inelastic collision the conservation of momentum equation is pfx = pix ⇒ ( m + 2 m)(v2 ) 3 m = m(v1 ) m + (2 m)(v1 ) 2 m Using (v1 )2 m = 0 m / s, we get (v2 ) 3 m = (v1 ) m / 3 = 2.56 m / s (b) For the elastic collision, the mass m package rebounds with velocity m − 2m 1 ( v3 ) m = (v1 ) m = − (7.668 m / s) = −2.56 m / s m + 2m 3 The negative sign with (v3)m shows that the package with mass m rebounds and goes to the position y4 . We can determine y4 by applying the conservation of energy equation as follows. For a package of mass m: 1 1 K f + U gf = K i + U gi ⇒ m(v4 ) 2m + mgy4 = m(v3 ) 2m + mgy3 2 2 Using (v3 ) m = −2.55 m / s, y3 = 0 m, and (v4 ) m = 0 m /s, we get
mgy4 =
1 m( −2.56 m / s) 2 ⇒ y4 = 33.3 cm 2
10.43. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the mechanical energy K + Us + Ug is conserved. Visualize:
The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate system at the position where the ice cube has compressed the spring 10.0 cm. That is, y0 = 0. Solve: The energy conservation equation K 2 + Us 2 + U g 2 = K 0 + Us 0 + U g 0 is
1 2 1 1 1 mv2 + k ( x e − x e ) 2 + mgy0 = mv02 + k ( x − x e ) 2 + mgy0 2 2 2 2 Using v2 = 0 m / s, y0 = 0 m, and v0 = 0 m / s,
k( x − xe )2 1 (25 N / m )(0.10 m ) 2 k ( x − x e ) 2 ⇒ y2 = = = 25.5 cm 2 2 mg 2(0.050 kg)(9.8 m / s 2 ) The distance traveled along the incline is y2 / sin 30° = 51.0 cm. Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve the problem. mgy2 =
10.44. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free fall problem, the mechanical energy K + U g + Us is conserved. Also, model the safe as a particle.
Visualize:
We have chosen to place the origin of our coordinate system at the free end of the spring which is neither stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so energy is conserved throughout the process. This means that the initial energy—as the safe is released—equals the final energy—when the safe is at rest and the spring is fully compressed. Solve: The conservation of energy equation K1 + U g1 + Us1 = K 0 + U g 0 + Us 0 is
1 2 1 1 1 mv1 + mg( y1 − ye ) + k ( y1 − ye ) 2 = mv02 + mg( y0 − ye ) + k ( ye − ye ) 2 2 2 2 2 Using v1 = v0 = 0 m/s and ye = 0 m, the above equation simplifies to
mgy1 + ⇒k=
1 2 ky1 = mgy0 2
2 mg( y0 − y1 ) 2(1000 kg)(9.8 m /s 2 )(2.0 m − ( −0.5 m )) = = 1.96 × 10 5 N / m. ( −0.5 m 2 ) y12
Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring.
10.45. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + Us + Ug is conserved during the launch of the ball. Visualize:
This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves the spring. This will yield the ball’s kinetic energy as it leaves the spring. Solve: Using the equations of kinematics,
1 a x (t3 − t2 ) 2 ⇒ 5.0 m = 0 m + (v2 cos 30°)(t3 − 0 s) + 0 m 2 (v2 cos 30°)t3 = 5.0 m ⇒ t3 = (5.0 m / v2 cos 30°)
x 3 = x 2 + v2 x (t 3 − t 2 ) +
1 a y (t 3 − t 2 ) 2 2 1 −1.5 m = 0 + (v2 sin 30°)(t3 − 0 s) + (9.8 m / s 2 )(t3 − 0 s) 2 2 y3 = y 2 + v 2 y ( t 3 − t 2 ) +
5.0 m 5.0 m 2 Substituting the value for t3, ( −1.5 m ) = (v2 sin 30°) − 4.9 m / s v2 cos 30° v2 cos 30°
(
⇒ ( −1.5 m ) = +(2.887 m ) −
)
2
163.33 ⇒ v2 = 6.102 m /s v22
The conservation of energy equation K 2 + Us 2 + U g 2 = K1 + Us1 +U g1 is
1 2 1 1 1 mv2 + k (0 m ) 2 + mgy2 = mv12 + k ( ∆s) 2 + mgy1 2 2 2 2 Using y2 = 0 m, v1 = 0 m/s, ∆s = 0.2 m, and y1 = −(∆s)sin 30º, we get
1 2 1 mv2 + 0 J + 0 J = 0 J + k ( ∆s) 2 − mg( ∆s)sin 30° ( ∆s) 2 k = mv22 + 2 mg( ∆s)sin 30° 2 2 (0.2 m ) 2 k = (0.02 kg)(6.102 m / s) 2 + 2(0.02 kg)(9.8 m / s 2 )(0.2)(0.5) ⇒ k = 19.6 N / m Assess: Note that y1 = −( ∆s)sin 30° is with a minus sign and hence the gravitational potential energy mgy1 is − mg( ∆s)sin 30°.
10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the mechanical energy K + Us + Ug is conserved. Visualize:
Solve:
(a) When releasing the block suddenly, K 2 + Us 2 + U g 2 = K1 + Us1 + U g1
1 2 1 1 1 mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2 Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye, we get
0J+
1 ( 490 N / m )( y2 − y1 ) 2 + mgy2 = 0 J + 0 J + mgy1 ⇒ (245 N / m )( y2 − y1 ) 2 = mg( y1 − y2 ) 2 ⇒ (245 N / m )( y1 − y2 ) 2 = (5.0 kg)(9.8 m/s 2 )( y1 − y2 ) ⇒ ( y1 − y2 ) = 0.20 m
(b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium.
Fnet = k∆y − mg = 0 ⇒ ∆y =
mg (5.0 kg)(9.8 m / s 2 ) = = 0.10 m k 490 N / m
(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring.
10.47. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical energy K + Us + Ug is conserved. Visualize:
We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the spring with velocity v2 as the spring reaches its equilibrium position. Solve: (a) The conservation of mechanical energy equation is
K 2 + Us 2 + U g 2 = K1 + Us1 + U g1
1 1 1 1 mv22 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2
Using y2 = ye, y1 = 0 m, and v1 = 0 m/s, this simplifies to
1 2 1 mv2 + 0 J + mgy2 = 0 J + k ( y1 − ye ) 2 + 0 2 2 1 1 (0.4 kg)v22 + (0.4 kg)(9.8 m /s 2 )(0.30 m ) = (1000 N / m )(0.30 m ) 2 ⇒ v2 = 14.8 m /s 2 2 (b) Let us use the conservation of mechanical energy equation once again to find the highest position (y3 ) of the rock where its speed (v3) is zero: 1 1 K 3 + U g 3 = K 2 + U g 2 ⇒ mv32 + mgy3 = mv22 + mgy2 2 2 v22 1 2 (14.8 m /s) 2 ⇒ 0 + g ( y3 − y 2 ) = v 2 ⇒ ( y3 − y 2 ) = = = 11.2 m 2 2 g 2(9.8 m /s 2 ) If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m. This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit.
10.48. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy
K + Ug + Us is conserved. Visualize:
Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m. Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by using this to find the spring constant.
Fsp = w ⇒ − k ( y1 − ye ) = − ky1 = mg ⇒ k = −
mg (0.10 kg)(9.8 m / s 2 ) =− = 9.8 N / m y1 −0.10 m
Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end, the spring has maximum stretch and the clay is instantaneously at rest. Thus
K 2 + (U g ) 2 + (Us ) 2 = K 0 + (U g ) 0 + (Us ) 0 ⇒ 12 mv22 + mgy2 + 12 ky22 = 12 mv02 + mgy0 + 0 J Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes
2 mgy0 2 mg =0 y2 − k k y22 + 0.20 y2 − 0.10 = 0
mgy2 + 12 ky22 = mgy0 ⇒ y22 +
The numerical values were found using known values of m, g, k, and y0 . The two solutions to this quadratic equation are y2 = 0.231 m and y2 = –0.432 m. The point we’re looking for is below the origin, so we need the negative root. The distance of the pan from the ceiling is L = |y2| + 50 cm = 93.2 cm
10.49. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy K + Us + Ug is conserved. Visualize:
We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches the roller coaster car. Solve: (a) The energy conservation equation for the car going to the top of the hill is
K 2 + U g 2 + Us 2 = K 0 + U g 0 + Us 0
1 2 1 1 1 mv2 + mgy2 + k ( x e − x e ) 2 = mv02 + mgy0 + k ( x 0 − x e ) 2 2 2 2 2
Noting that y0 = 0 m, v2 = 0 m /s, v1 = 0 m /s, and | x 0 − x e | = 2.0 m , we obtain 1 0 J + mgy2 + 0 J = 0 J + 0 J + k (2.0 m ) 2 2 2 2 mgy2 2( 400 kg)(9.8 m /s )(10 m ) ⇒k= = = 19, 600 N / m (2.0 m ) 2 (2.0 m ) 2 We now increase this value for k by 10% for safety, giving a value of 21,600 N/m. (b) The energy conservation equation K 3 + U g 3 + Us 3 = K 0 + U g 0 + Us 0 is
1 2 1 1 1 mv3 + mgy3 + k ( x e − x e ) 2 = mv02 + mgy0 + k ( x 0 − x e ) 2 2 2 2 2 Using y3 = −5.0 m, v0 = 0 m / s, y0 = 0 m, and | x0 − xe | = 2.0 m, we get
1 2 1 mv3 + mg( −5.0 m ) + 0 J = 0 J + 0 J + k ( x 0 − x e ) 2 2 2 1 1 ( 400 kg)v32 − ( 400 kg)(9.8 m /s 2 )(5.0 m ) = (2.156 × 10 4 N / m )(2.0 m ) 2 2 2 ⇒ v3 = 17.7 m /s
10.50. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change. Model Julie as a particle. Visualize:
We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From geometry, Julie launches off the end of the ramp at a 30° angle. 1 1 Solve: Energy conservation: K1 + U g1 = K 0 + U g 0 ⇒ mv12 + mgy1 = mv02 + mgy0 2 2 Using v0 = 0 m /s, y0 = 25 m, and y1 = 3 m, the above equation simplifies to
1 2 mv1 + mgy1 = mgy0 ⇒ v1 = 2 g( y0 − y1 ) = 2(9.8 m /s 2 )(25 m − 3 m ) = 20.77 m /s 2 We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll consider the vertical motion: 1 1 y2 = y1 + v1y (t2 − t1 ) + a y (t2 − t1 ) 2 0 m = 3 m + (v1 sin 30°)(t2 − t1 ) + ( −9.8 m / s 2 )(t2 − t1 ) 2 2 2 (20.77 m /s)sin 30° (3 m ) (t2 − t1 ) 2 − (t2 − t1 ) − =0 ( 4.9 m /s 2 ) ( 4.9 m /s 2 )
(t2 − t1 ) 2 − (2.119 s)(t2 − t1 ) − (0.6122 s 2 ) = 0 ⇒ (t2 − t1 ) = 2.377 s For the horizontal motion:
1 a x (t2 − t1 ) 2 2 x 2 − x1 = (v1 cos 30°)(t2 − t1 ) + 0 m = (20.77 m /s)(cos 30°)(2.377 s) = 42.7 m x 2 = x1 + v1 x (t2 − t1 ) +
Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries Julie through a 90° turn.
10.51. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Visualize:
Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-and-after pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the ball’s rebound velocity is
( vf ) b =
m b − mB −80 g ( vi ) b = (5.0 m /s) = −3.33 m /s 120 g m b + mB
The ball’s speed is 3.33 m/s. (b) An elastic collision gives the block speed
( vf ) B =
2 mB 40 g ( vi ) b = (5.0 m /s) = 1.667 m /s 120 g m b + mB
To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the block + spring system. That is K1 + Us1 = K 0 + Us 0 :
1 1 1 1 mB (vf′ ) 2B + k ( x1 − x 0 ) 2 = mB (vf ) 2B + k ( x 0 − x 0 ) 2 2 2 2 2 (x1 − x 0 ) =
0 + k ( x1 − x 0 ) 2 = mB (vf )2B + 0
(0.100 kg )(1.667 m /s )2 /(20 N/ m ) = 11.8 cm
(c) Momentum conservation pf = pi for the perfectly inelastic collision means
( m B + m B )v f = m b ( v i ) b + m B ( v i ) B (0.100 kg + 0.020 kg)vf = (0.020 kg)(5.0 m/s) + 0 m/ v ⇒ vf = 0.833 m/s The maximum compression in this case can now be obtained using the conservation of energy equation K1 + Us1 = K 0 + Us 0 :
0 J + (1 / 2)k ( ∆x )2 = (1 / 2)( mB + mb )vf 2 + 0 J ⇒ ∆x =
mB + mb vf = k
0.120 kg (0.833 m /s) = 0.0645 m = 6.45 cm 20 N / m
10.52. Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the mechanical energy of the system (bullet + block + spring) is conserved. Visualize:
We place the origin of our coordinate system at the end of the spring that is not anchored to the wall. Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means
m ( m + M )vf = m(vi ) m + M (vi ) M ⇒ ( m + M )vf = mvB + 0 kg m /s ⇒ vf = v m + M B where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation K1 + Us1 = K e + Use as the bullet embedded block compresses the spring is:
1 1 1 1 m(v ′f ) 2 + k ( x1 − x e ) 2 = ( m + M )(vf ) 2 + k ( x e − x e ) 2 2 2 2 2 2
0J+
1 2 1 m 2 kd = ( m + M ) v + 0 J ⇒ vB = m + M B 2 2
( m + M )kd 2 m2
(b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,
vB = (0.005 kg + 2.0 kg)(50 N / m )(0.10 m ) 2 / (0.005) 2 = 200 m /s (c) The fraction of energy lost is 1 2 1 2 2 mvB − ( m + M )vf2 m + M vf m+M m 2 2 = 1− = 1− 1 2 m vB m m + M mvB 2 m 0.005 kg = 1− = 1− = 0.9975 (0.005 kg + 2.0 kg) m+M That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated inside the block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is more accurate to say, “The energy is transformed to thermal energy.”
10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy does not change during the car’s motion. Visualize:
We place the origin of the coordinate system at the ground level directly below the car’s starting position. This is a two-part problem. If we first find the maximum speed at the top of the hill (point C), we can use energy conservation to find the maximum initial height. Solve: Because the its motion is circular, at the top of the hill the car has a downward-pointing centripetal accelr eration ac = −( mv 2 / r ) jˆ. Newton’s second law at the top of the hill is
( Fnet ) y = n y + w y = n − mg = m( ac ) y = −
mv 2 mv 2 v2 ⇒ n = mg − = m g − R R R
If v = 0 m / s, n = mg as expected in static equilibrium. As v increases, n gets smaller—the apparent weight of the car and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the maximum speed vmax occurs when n → 0 . We see that vmax = gR . Now we can use energy conservation to relate point C to the starting height: 1 1 1 K C + UC = K i + U i ⇒ mvC2 + mgyC = mv12 + mgyi ⇒ mgR + mgR = 0 J + mghmax , 2 2 2 where we used vC = vmax and yC = R. Solving for hmax gives hmax = 23 R. (b) If R = 10 m, then hmax = 15 m.
10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle θ. Visualize:
We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newton’s second law at this point is
w+T =
mv 2 r
where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. In this case,
mg =
mvC2 ⇒ vC2 = gr = gL 3 r
The ball should have kinetic energy at least equal to
1 2 1 L mvc = mg 2 2 3 for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle θ. The equation for the conservation of energy is
K f + U gf = K i + U gi ⇒
1 2 1 mvf + mgyf = mvi2 + mgyi 2 2
Using vf = vc , yf = 13 L, vi = 0, and the above value for vc2 , we get
1 L L L mg + mg = mgyi ⇒ yi = 2 3 3 2 That is, the ball is a vertical distance
1 2
L above the peg’s location or a distance of 2L − L = L 3 2 6
below the point of suspension of the pendulum, as shown in the figure on the right. Thus,
cosθ =
L/6 1 = ⇒ θ = 80.4° L 6
10.55. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to go over the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational potential energy is conserved during the block’s motion. We will use this conservation equation in the second part to find the minimum height the block must start from to make it around the loop. Visualize:
We place the origin of our coordinate system directly below the block’s starting position on the frictionless track. Solve: The free-body diagram on the block implies
w+n=
mvc2 R
For the block to just stay on track, n = 0 . Thus the critical velocity vc is
mvc2 ⇒ vc2 = gR R The block needs kinetic energy 12 mvc2 = 12 mgR to go over the top of the loop. We can now use the conservation of mechanical energy equation to find the minimum height h. w = mg =
K f + U gf = K i + U gi ⇒
1 2 1 mvf + mgyf = mvi2 + mgyi 2 2
Using vf = vc = gR , yf = 2 R, vi = 0 m /s, and yi = h, we obtain
1 gR + g(2 R) = 0 + gh ⇒ h = 2.5 R 2
10.56. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless, so
that the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision, as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume the spring to be an ideal one that obeys Hooke’s law. Visualize:
We place the origin of our coordinate system directly below the bobsled’s initial position. Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p1 = p0 , or
( mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ ( mL + mB )v1 = mL (v0 ) L + 0 mL 40 kg ⇒ v1 = (12 m / s) = 8.0 m / s ( v0 ) L = + + 40 kg 20 kg m m L B The energy conservation equation: K 2 + Us 2 + U g 2 = K1 + Us1 + U g1 is
1 1 1 1 ( mL + mB )v22 + k ( x 2 − x e ) 2 + ( mL + mB )gy2 = ( mL + mB )v12 + k ( x e − x e ) 2 + ( mL + mB )gy1 2 2 2 2 Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m) sin 20° = 17.1 m, v1 = 8.0 m/s, and (mL + mB) = 60 kg, we get
0J+
1 1 (2000 N / m )( x 2 − x e ) 2 + 0 J = (60 kg)(8.0 m / s) 2 + 0 J + (60 kg)(9.8 m / s 2 )(17.1 m ) 2 2
Solving this equation yields ( x 2 − x e ) = 3.46 m. (b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical energy that the speed at the bottom depends only on the vertical descent ∆y. Only the ramp’s height h is important, not its shape or angle.
10.57. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum. Visualize:
The figure shows three distinct moments of time: the time before the collision, the time after the collision but before the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate system on the 100 g ball when it is hanging motionless. Solve: For a perfectly elastic collision, the ball moves forward with speed
(v1 ) M =
2 mm 1 ( v0 ) m = ( v0 ) m 3 mm + m M
In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up after the collision. That is, K 2 + U g 2 = K1 + U g1 . We have
1 1 M (v2 ) 2M + Mgy2 = M (v1 ) 2M + Mgy1 2 2 Using (v2 ) M = 0 J, (v1 ) M =
( v0 ) m , y1 = 0 m, and y2 = L − L cosθ , the energy equation simplifies to 3 g( L − L cosθ ) =
1 (v0 ) 2m 2 9
⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.8 m /s 2 )(1.0 m)(1 − cos 50°) = 7.94 m /s
10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S′, find the final velocities of the balls in S′, and then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 200-g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initial velocities in the S frame are (vix )1 = 4 m /s and (vix ) 2 = −3 m /s.
Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collision situation as seen in frame S′. The after-collision velocities in S′ are shown in figure (c), and figure (d) indicates velocities in S after they have been transformed to frame S from S′. Solve: (a) In the S frame, (vix )1 = 4 m /s and (vix )2 = −3 m /s . S′ is the reference frame of the 200-g ball, so V = –3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v – V. So, (vix )1′ = (vix )1 − V = 4 m /s − ( −3 m /s) = 7 m /s (vix )′2 = (vix ) 2 − V = −3 m /s − ( −3 m /s) = 0 m /s Figure (b) shows the “before” situation, where ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′:
(vfx )1′ =
m1 − m2 100 g − 200 g 7 (vix )1′ = (7 m /s) = − m /s 100 g + 200 g 3 m1 + m2
(vfx )′2 =
2 m1 2(100) g 14 (vix )1′ = (7 m /s) = m /s 100 g + 200 g 3 m1 + m2
Finally, we need to apply the reverse Galilean transformation v = v′ + V, with the same V, to transform the aftercollision velocities back to the lab frame S:
7 (vfx )1 = (vfx )1′ + V = − m /s − 3 m /s = −5.33 m /s 3 14 (vfx ) 2 = (vfx )′2 + V = m /s − 3 m /s = 1.67 m /s 3 Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.33 m/s; the 200 g ball is moving right at 1.67 m/s.
(b) The momentum conservation equation pfx = pix for a perfectly inelastic collision is
( m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 (0.100 kg + 0.200 kg)vfx = (0.100 kg)( 4.0 m/s) + (0.200 kg)( −3.0 m/s) ⇒ vfx = −0.667 m/s Both balls are moving left at 0.667 m/s.
10.59. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S′. After finding the final velocities of the balls in S′, we can then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 400-g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initial velocities in the S frame are (vix )1 = + 4.0 m /s and (vix )2 = +1.0 m /s.
Figures (a) and (b) show the before-collision situations in frames S and S′, respectively. The after-collision velocities in S′ are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S from S′. Solve: In frame S, (vix )1 = 4.0 m /s and (vix )2 = 1.0 m /s. Because S′ is the reference frame of the 400-g ball, V = 1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v – V. So, (vix )1′ = (vix )1 − V = 4.0 m /s − 1.0 m /s = 3.0 m /s (vix )′2 = (vix ) 2 − V = 1.0 m /s − 1.0 m /s = 0 m /s Figure (b) shows the “before” situation in frame S′ where the ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′:
(vfx )1′ =
m1 − m2 100 g − 400 g (vix )1′ = (3.0 m /s) = −1.80 m /s 100 g + 400 g m1 + m2
(vfx )′2 =
2 m1 2(100 g) (vix )1′ = (3.0 m /s) = 1.20 m /s 100 g + 400 g m1 + m2
Finally, we need to apply the reverse Galilean transformation v = v′ + V, with the same V, to transform the aftercollision velocities back to the lab frame S.
(vfx )1 = (vfx )1′ + V = −1.80 m/s + 1.0 m/s = −0.80 m/s (vfx ) 2 = (vfx )′2 + V = 1.20 m/s + 1.0 m/s = 2.20 m/s Figure (d) shows the “after” situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.20 m/s. Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities.
10.60. Model: Use the model of the conservation of mechanical energy. Visualize:
Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E = 12 J, this occurs at the points x = 1 m and x = 7 m. (b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and total energy line. U = 8 J at x = 2 m, so K = 12 J − 8 J = 4 J. The speed corresponding to this kinetic energy is
v=
2K = m
2( 4 J ) = 4.0 m /s 0.5 kg
(c) Maximum speed occurs for minimum U. This occurs at x = 1 m and x = 4 m, where U = 0 J and K = 12 J. The speed at these two points is
v=
2K = m
2(12 J ) = 6.93 m /s 0.5 kg
(d) The particle leaves x = 1 m with v = 6.93 m/s. It gradually slows down, reaching x = 2 m with a speed of 4.00 m/s. After x = 2 m, it speeds up again, returning to a speed of 6.93 m/s as it crosses x = 4 m. Then it slows again, coming instantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and move back to the left. (e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either oscillate back and forth over the range 1.0 m ≤ x ≤ 1.5 m or over the range 3 m ≤ x ≤ 5 m.
10.61. Model: We will use the conservation of mechanical energy. Visualize:
The potential energy (U) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two minima correspond to the nitrogen atom’s position on both sides of the plane containing the three hydrogens. Solve: (a) At room temperature, the total energy line is below the “hill” in the center of the potential energy curve. That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There’s a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axis—that is, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B. (b) The total energy line is now well above the “hill,” and the turning points of the nitrogen atom’s motion (where the total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side of the H3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has sufficient energy to get through.
10.62. Model: The potential energy of two nucleons interacting via the strong force is U = U 0 [1 − e − x / x0 ] where x is the distance between the centers of the two nucleons, x 0 = 2.0 × 10 −15 m, and U 0 = 6.0 × 10 −11 J. Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the strong force. This force exists between nucleons at very small separations. Solve: (a)
(b) For x = 5.0 × 10 −15 m, U = 55.1 × 10 −12 J. This energy is represented by a total energy line. (c) Due to conservation of total energy, the potential energy when x = 5.0 × 10 −15 J is transformed into kinetic energy until x = twice the radius = 1.0 × 10 −15 m. At this separation, u = 15.7 × 10 −12 J. Thus, 1 2 1 2 mv + mv + 15.7 × 10 −12 J = 55.1 × 10 −12 J ⇒ v = 1.53 × 10 8 m /s 2 2 Assess: A speed of 1.53 × 108 m/s is approximately 0.5 c where c is the speed of light. This speed is understandable for the present model.
10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by how much is it compressed?
10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial speed must the auto have had at the bottom of the hill? (b)
We place the origin of our coordinate system at the bottom of the hill. (c) The solution of the equation is vi = 14.9 m/s. This is approximately 30 mph and is a reasonable value for the speed at the bottom of the hill.
10.65. (a) A spring gun is compressed 15 cm to launch a 200-g ball on a horizontal, frictionless surface. The ball has a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun. (b)
We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because the surface is frictionless, the mechanical energy for the system (ball + spring) is conserved. (c) The conservation of energy equation is
K f + Usf = K i + Usi 1 2 1 1 1 mv1 + k (0 m ) 2 = m(0 m /s) 2 + k ( â&#x2C6;&#x2019;0.15 m ) 2 2 2 2 2 2 2 (0.200 kg)(2.0 m /s) = k ( â&#x2C6;&#x2019;0.15 m ) k = 35.6 N / m
10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the spring is firmly anchored to a fixed post on the surface. How far will the spring compress? (b)
(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the conservation of energy equation yields â&#x2C6;&#x2020;x 2 = 31.6 cm.
10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30째 incline. A 500-g block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with which the block is launched up the incline? (b) The origin is placed at the end of the uncompressed spring. This is point from which the block is launched as the spring expands.
(c) Solving the energy conservation equation, we get vf = 2.65 m/s.
10.68. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, so it is an isolated system. The interaction forces within the system are the spring force of the bungee cord and the gravitational force. These are both conservative forces, so mechanical energy is conserved. Visualize:
We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the lowest point. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need to note that the end of the unstretched cord is at y0 = y1 − 30 m = 70 m, so U2 s = 12 k ( y2 − y0 ) 2 . Also note that U1s = 0, since the cord is not stretched. The energy conservation equation is 1 K 2 + U2 g + U2 s = K1 + U1g + U1s ⇒ 0 J + mgy2 + k ( y2 − y0 ) 2 = 0 J + mgy1 + 0 J 2 Multiply out the square of the binomial and rearrange: 1 1 mgy2 + ky22 − ky0 y2 + ky02 = mgy1 2 2 mgy 2 mg 2 1 ⇒ y22 + − 2 y0 y2 + y02 − = y22 − 100.8 y2 + 980 = 0 k k This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a height above the point where the cord started to stretch. So we find that your distance from the water when the bungee cord stops stretching is 10.9 m.
10.69. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + Ug + Us is conserved. Visualize:
We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use lengths along the ramp with the variable s rather than x. Solve: (a) The conservation of energy equation K 2 + U g 2 + Us 2 = K1 + U g1 + Us1 is
1 2 1 1 mv2 + mgy2 + k ( ∆s) 2 = mv12 + mgy1 + k (0 m ) 2 2 2 2 1 1 1 2 2 m(0 m /s) + mg(0 m ) + k ( ∆s) = m(0 m /s) 2 + mg( 4.0 m + ∆s)sin 30° + 0 J 2 2 2 1 1 (250 N / m )( ∆s) 2 = (10 kg)(9.8 m /s 2 )( 4.0 m + ∆s) 2 2 This gives the quadratic equation:
(125 N/ m )( ∆s) 2 − ( 49 kg ⋅ m/s 2 )∆s − 196 kg ⋅ m 2 /s 2 = 0 ⇒ ∆s = 1.46 m and − 1.07 m (unphysical) The maximum compression is 1.46 m. (b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the block’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring is neither compressed nor stretched.
1 2 1 1 1 mv + mgy + k ( ∆s) 2 = mv12 + mgy1 + k (0 m ) 2 2 2 2 2 1 2 1 mv + mg( − ∆s sin 30°) + k ( ∆s) 2 = 0 J + mg( 4.0 m sin 30°) + 0 J 2 2 1 1 2 k ( ∆s) − ( mg sin 30°)∆s + mv 2 − mg sin 30°( 4.0 m ) = 0 2 2 To find the compression where v is maximum, take the derivative of this equation with respect to ∆s:
1 1 dv k 2( ∆s) − ( mg sin 30°) + m 2v −0=0 2 2 d∆s Since
dv = 0 at the maximum, we have d∆s ∆s = ( mg sin 30°)/ k = (10 kg)(9.8 m /s 2 )(0.5)/(250 N / m ) = 19.6 cm
10.70. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon (C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion. Visualize:
The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the spring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the cannon. Solve: The momentum conservation equation pfx = pix is
mB (vfx ) B + mC (vfx ) C = mB (vix ) B + mC (vix ) C Since the initial momentum is zero,
(vfx ) B = −
200 kg mC (vfx ) C = − (vfx ) C = −20(vfx ) C mB 10 kg
The mechanical energy conservation equation for the cannon + spring K f + Usf = K i + Usi is 1 1 1 1 1 m(vf ) 2C + k ( ∆x ) 2 = m(vi ) 2C + 0 J ⇒ 0 J + k ( ∆x ) 2 = m(vfx ) 2C 2 2 2 2 2
⇒ (vfx ) C = ±
k (20, 000 N / m ) (0.5 m ) = ±5.0 m /s ∆x = ± m 200 kg
To make this velocity physically correct, we retain the minus sign with (vfx)C. Substituting into the momentum conservation equation yields: (vfx ) B = −20( −5.0 m /s) = 100 m /s
10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + Ug + Us is conserved. Visualize:
We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m and the block’s mass is M. Solve: (a) The energy conservation equation K 2 + Us 2 + U g 2 = K1 + Us1 + U g1 becomes
1 1 1 1 ( m + M )v22 + k ( y2 − ye )2 + ( m + M )g( y2 − ye ) = ( m + M )v12 + k ( y1 − ye )2 − ( m + M )g( y1 − ye ) 2 2 2 2 Noting v2 = 0 m/s, we can rewrite the above equation as k ( ∆y2 ) 2 + 2( m + M )g( ∆y2 + ∆y1 ) = ( m + M )v12 + k ( ∆y1 ) 2 Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi which is ( m + M )v1 = mvB + Mvblock . Since vblock = 0 m/s, we have
m v1 = v m + M B We can also find the magnitude of y1 from the equilibrium condition k ( y1 − ye ) = Mg.
Mg k With these substitutions for v1 and ∆y1, the energy conservation equation simplifies to ∆y1 =
k ( ∆y2 ) 2 + 2( m + M )g( ∆y1 + ∆y2 ) = ⇒ vB2 = 2
m 2 vB2 Mg + k k (m + M )
2
m + M (m + M ) M 2 g 2 m + M g( ∆y1 + ∆y2 ) − + k ( ∆y2 ) 2 m2 m m2 k 2
We still need to include the spring’s maximum compression (d) into this equation. We assume that d = ∆y1 + ∆y2, that is, maximum compression is measured from the initial position (y1) of the block. Thus, using ∆y2 = d − ∆y1 = ( d − Mg / k ) , we have
m + M2 m + M M 2 g2 m + M v B = 2 gd − + k ( d − Mg / k ) 2 2 2 m k m m (b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m, 2
12
2
2.010 kg 2.0 kg 2 2 2 v = 2 (9.8 m /s )(0.45 m ) − (2.010 kg) (9.8 m /s ) /(50 N / m ) 0.010 kg 0.010 kg 2 B
+ (50 N / m )(2.010 kg)
1 [0.45 m − (2.0 kg) × (9.8 m /s 2 )/ 50 N / m ]2 (0.010 kg) 2
⇒ vB = 453 m /s Assess: This is a reasonable speed for the bullet.
10.72. Model: This is a collision between two objects, and momentum is conserved in the collision. In addition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved. Visualize:
Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at maximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2. This is an important observation. Solve: First relate part 1 to part 2. Conservation of energy is 1 1 1 1 1 1 mA vA2 1 = mA vA2 2 + mB vB2 2 + k ( ∆x max ) 2 = ( mA + mB )v22 + k ( ∆x max ) 2 2 2 2 2 2 2 where ∆xmax is the spring’s compression. Ug is not in the equation because there are no elevation changes. Also note that K 2 is the sum of the kinetic energies of all moving objects. Both v2 and ∆xmax are unknowns. Now add the conservation of momentum: m A v A1 m A v A1 = m A v A 2 + m B v B 2 = ( m A + m B )v 2 ⇒ v 2 = = 2.667 m /s mA + mB Substitute this result for v2 into the energy equation to find: 1 1 1 k ( ∆x max ) 2 = mA vA2 1 − ( mA + mB )v22 2 2 2
mA vA2 1 − ( mA + mB )v22 = 0.046 m = 4.6 cm k Notice how both conservation laws were needed to solve this problem. (b) Again, both energy and momentum are conserved between “before” and “after.” Energy is 1 1 1 1 mA vA2 1 = mA vA2 3 + mB vB2 3 ⇒ 16 = vA2 3 + vB2 3 2 2 2 2 The spring is no longer compressed, so the energies are purely kinetic. Momentum is ⇒ ∆x max =
m A v A1 = m A v A 3 + m B v B 3 ⇒ 8 = 2 v A 3 + v B 3 We have two equations in two unknowns. From the momentum equation, we can write vB3 = 2( 4 − vA 3 ) and use this in the energy equation to obtain: 2 1 16 = vA2 3 + ⋅ 4 4 − vA2 3 = 3vA2 3 − 16vA 3 + 32 ⇒ 3vA2 3 − 16vA 3 + 16 = 0 2 This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.333 m/s). Using vB3 = 2(4 − vA3), these two roots give vB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each keeps its initial speed. That’s not the situation here. We want the second pair of roots, from which we learn that the blocks’ speeds after the collision are vA 3 = 1.33 m /s and vB3 = 5.33 m /s.
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10.73. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over, the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not change as the balls swing out. Visualize:
In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulum swing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barely touch each other. Solve: As the ball with mass m1, whose string makes an angle θi1 with the vertical, swings through its equilibrium position, it lowers its gravitational energy from m1 gy0 = m1 g( L − L cosθ i1 ) to zero. This change in potential energy transforms into a change in kinetic energy. That is, 1 m1 g( L − L cosθ i1 ) = m1 (v1 )12 ⇒ (v1 )1 = 2 gL(1 − cosθ i1 ) 2 Similarly, (v1 )2 = 2 gL(1 − cosθ i 2 ). Using θ i1 = θ i 2 = 45°, we get (v1 )1 = 2.396 m/s = (v1 )2 . Both balls are moving at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in which ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2 is at rest. Ball 2 is at rest in a frame that moves with ball 2, so choose S′ to have V = –2.396 m/s, with the minus sign because this frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity ( v1′ )1 = (v1)1 – V = 2.396 m/s + 2.396 m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to move with velocities m − m2 1 (v2′ )1 = 1 (v1′ )1 = – ( 4.792 m /s ) = −1.597 m /s m1 + m2 3
(v2′ ) 2 =
2 m1 2 (v1′ )1 = (2.396 m /s ) = 3.194 m /s m1 + m2 3
We can now use v = v´ + V to transform these back into the laboratory frame: (v2 )1 = −1.597 m / s − 2.396 m / s = −3.99 m / s
(v2 ) 2 = 3.195 m / s − 2.396 m / s = 0.799 m / s Having determined the velocities of the two balls after collision, we will once again use the conservation equation K f + U gf = K i + U gi for each ball to solve for the θf1 and θf2.
1 1 m1 (v3 )12 + m1 gL(1 − cosθ f 1 ) = m1 (v2 )12 + 0 J 2 2 Using (v3 )1 = 0, this equation simplifies to gL(1 − cosθ f 1 ) =
1 1 ( −3.99 m /s) 2 ⇒ cosθ f 1 = 1 − ( −3.99 m /s) 2 ⇒ θ f 1 = 79.3° 2 2 gL
The 100 g ball rebounds to 79.3°. Similarly, for the other ball: 1 1 m2 (v3 ) 22 + m2 gL(1 − cosθ f 2 ) = m2 (v2 ) 22 + 0 J 2 2 Using (v3 ) 2 = 0, this equation becomes
1 2 cosθ f 2 = 1 − (0.799) ⇒ θ f 2 = 14.7° 2 gL The 200 g ball rebounds to 14.7°.
10.74. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and gravitational potential energy is conserved during motion. Visualize:
Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry, y1 = Rcos φ . Solve: The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2 1 1 mv1 + mgy1 = mv02 + mgy0 ⇒ mv12 + mgR cos φ = mgR ⇒ v1 = 2 gR(1 − cos φ ) 2 2 2 r (b) If the sled is on the hill, it is moving in a circle and the r-component of Fnet has to point to the center with magnitude Fnet = mv 2 / R. Eventually the speed gets so large that there is not enough force to keep it in a circular trajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis pointing toward the center of the circle, as we usually do for circular motion problems. Newton’s second law along this axis requires: ( Fnet ) r = w cos φ − n = mg cos φ − n = mar = ⇒ n = mg cos φ −
mv 2 R
mv 2 v2 = m g cos φ − R R
The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays on the hill is the speed vmax that makes n → 0. We can see that vmax = gR cos φ . (c) We now know the sled’s speed at angle φ , and we know the maximum speed it can have while remaining on the hill. The angle at which v reaches vmax is the angle φ max at which the sled will fly off the hill. Combining the two expressions for v1 and vmax gives:
2 gR(1 − cos φ ) = gR cos φ ⇒ 2 R(1 − cos φ max ) = R cos φ max ⇒ cos φ max =
2 2 ⇒ φ max = cos −1 = 48.2° 3 3