6.1. Model: We will assume motion under constant-acceleration kinematics in a plane. Visualize:
Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector form. Solve: (a) From the kinematic equation for position, after some manipulation: r r r r r r 2 r1 = r0 + v0 (t1 − t 0 ) + 12 a (t1 − t 0 ) ⇒ (2 s )v0 + 3 s 2 a = 4iˆ − 4 jˆ m
(
( )
)
From the kinematic equation for velocity: r r r r r r r v1 = v0 + a(t1 − t 0 ) ⇒ 5iˆ − 5 jˆ m/s = v0 + a(3 s − 0 s) ⇒ v0 = −(3 s)a + 5iˆ − 5 jˆ m/s r Substituting this form of v0 into the position equation, we find
(
[
)
(
]
r
r
)
(2 s) −(3 s)a + (5iˆ − 5 jˆ) m/s + (3 s 2 )a = (4iˆ − 4 jˆ) m ⇒ a = (2iˆ − 2 jˆ) m/s 2 r
r r Substituting a back into either the velocity equation or the position equation gives v0 = ( −iˆ + jˆ ) m/s . r r r r 2 (b) Using the kinematic equation r2 = r1 + v1 (t 2 − t1 ) + 12 a (t 2 − t1 ) ,
(8iˆ − 2 jˆ) m + [(5iˆ − 5 jˆ) m/s](5 s − 3 s) + [(2iˆ − 2 jˆ) m/s ](5 s − 3 s) 2
1 2
(
[(
)
)
]
2
= (22iˆ − 16 jˆ ) m
r r r v2 = v1 + a(t2 − t1 ) = 5iˆ − 5 jˆ m/s + 2iˆ − 2 jˆ m/s 2 (5 s − 3 s) = (9iˆ − 9 jˆ ) m/s Thus, the speed at t = 5 s is v2 =
( v2 x ) 2 + ( v2 y )
2
=
(9)2 + (−9)2 m/s = 12.7 m/s .
6.2. Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic equations in a plane. Visualize:
Solve:
Resolving the acceleration into its x and y components, we obtain r a = 0.80 m/s 2 cos40°iˆ + 0.80 m/s 2 sin40° jˆ = 0.613 m/s 2 iˆ + 0.514 m/s 2 jˆ r r r From the velocity equation v1 = v0 + a (t1 − t 0 ) , r v1 = (5.0 m/s)iˆ + 0.613 m/s 2 iˆ + 0.514 m/s 2 jˆ (6 s − 0 s) = (8.68 m/s)iˆ + (3.09 m/s) jˆ r The magnitude and direction of v are
(
)
(
[(
) (
v=
)
(
)
) (
]
(8.68 m/s)2 + (3.09 m/s)2
= 9.21 m/s
v 3.09 m/s θ = tan −1 1y = tan −1 = 20° north of east 8.68 m/s v1 x Assess: An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.
)
6.3. Solve: (a)
r r (b) At t = 0 s, x = 0 m and y = 0 m, or r = 0iˆ + 0 jˆ m . At t = 4 s, x = 0 m and y = 0 m, or r = 0iˆ + 0 jˆ m . In
(
)
(
other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for x and y,
r dx ˆ dy ˆ 3 v= i+ j = t 2 − 4t iˆ + (t − 2) jˆ m/s dt dt 2 r r ˆ ˆ ˆ At t = 0 s, v = −2 j m/s , v = 2 m/s . At t = 4 s, v = 8i + 2 j m/s , v = 8.3 m/s . r (c) At t = 0 s, v is along − jˆ , or 90° south of +x. At t = 4 s,
(
)
2 m /s θ = tan −1 = 14° north of +x 8 m /s
)
6.4. Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion. Visualize: Please refer to Figure Ex6.4. r Solve: (a) At t = 2 s, the graphs give vx = 16 cm/s and vy = 30 cm/s. The angle made by the vector v with the x-axis can thus be found as
v 30 cm/s θ = tan −1 y = tan −1 = 62° above the x-axis 16 cm/s vx (b) After t = 5 s, the puck has traveled a distance given by: 5s
x1 = x 0 + ∫ v x dt = 0 m + area under vx-t curve = 0
1 2
(40 cm/s)(5 s) = 100 cm
y1 = y0 + ∫ v y dt = 0 m + area under vy-t curve = (30 cm/s)(5 s) = 150 cm 5s
0
⇒ r1 =
x12 + y12 =
(100 cm)2 + (150 cm)2 = 180 cm
6.5. Model: The model rocket and the target will be treated as particles. Kinematic equations in two dimensions apply. Visualize:
Solve:
For the rocket, Newton’s second law along the y-direction is r r r r Fnet = FR − mg = maR r 1 r 1 FR − mg = ⇒ aR = 15 N − (0.8 kg) 9.8 m /s 2 m 0.8 kg
(
[
)
(
)] = 8.95 m /s
2
Using the kinematic equation y1R = y0R + (v0R ) y (t1R − t 0R ) + 12 aR (t1R − t 0R ) , 2
30 m = 0 m + 0 m +
1 2
(8.95 m/s )(t 2
1R
− 0 s) ⇒ t1R = 2.589 s 2
For the target (noting t1T = t1R),
x1T = x 0T + (v0T ) x (t1T − t 0T ) + 12 aT (t1T − t 0T ) = 0 m + (15 m /s)(2.589 s − 0 s) + 0 m = 38.8 m 2
You should launch when the target is 38.8 m away. Assess: The rocket is to be fired when the target is at x0T. For a net acceleration of approximately 9 m/s2 in the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 38.8 m is reasonable.
6.6. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected. Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s. Solve: For the rocket, Newton’s second law along the y-direction is: r r r r 1 Fnet = ( FR ) − mg = m( aR ) ⇒ a y = (8.0 N) − (0.5 kg) 9.8 m /s 2 0.5 kg
[
(
)]
= 6.2 m /s 2
Thus using y1 = y0 + (v0 ) y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
(20 m) = 0 m + 0 m + 12 (6.2 m/s 2 )(t1R − 0 s) ⇒ (20 m) = (3.1 m/s 2 )t12 ⇒ t1 = 2.54 s 2
Since t1 is also the time for the rocket to move horizontally up to the hoop,
x1 = x 0 + (v0 ) x (t1 − t 0 ) + 12 a x (t1 − t 0 ) = 0 m + (3.0 m /s)(2.54 s − 0 s) + 0 m = 7.62 m 2
Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable.
6.7. Solve: vx, ax, ay, Fx and Fy are constant. Furthermore, ax and Fx are not only constant, they are zero.
6.8. Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic equations of motion in a plane. Visualize:
r (a) We know the velocity v1 = ( 2.0iˆ + 2.0 jˆ ) m/s at t = 1 s. The ball is at its highest point at t = 2 s, so r vy = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus v2 = 2.0iˆ m/s at t = 2 s. We can see that the y-component of velocity changed by ∆vy = –2.0 m/s between t = 1 s and t = 2 s. Because ay is constant, vy changes by –2.0 m/s in any 1-s interval. At t = 3 s, vy is 2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, vy must have been 2.0 m/s more than its value of 2.0 m/s at t = 1 s. Consequently, at t = 0 s, r v0 = ( 2.0iˆ + 4.0 jˆ ) m/s Solve:
At t = 1 s, At t = 2 s, At t = 3 s,
r v0 = ( 2.0iˆ + 2.0 jˆ ) m/s r v0 = ( 2.0iˆ + 0.0 jˆ ) m/s r v0 = ( 2.0iˆ – 2.0 jˆ ) m/s
(b) Because vy is changing at the rate –2.0 m/s per s, the y-component of acceleration is ay = –2.0 m/s2. But ay = –g for projectile motion, so the value of rg on Exidor is g = 2.0 m/s2. (c) From part (a) the components of v0 are v0 x = 2.0 m /s and v0 y = 4.0 m /s . This means
v 4.0 m /s θ = tan −1 0 y = tan −1 = 63.4° above +x 2.0 m /s v0 x Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives an acceleration of −2 m/s2. All the other values obtained above are also reasonable.
6.9. Model: Assume the particle model for the object. Visualize: The object is undergoing projectile motion as illustrated in Figure 6.15. Solve: (a) The components of v0 are
v0 x = v0 cosθ = (50 m /s) cos 36.9° = 40.0 m /s v0 y = v0 sin θ = (50 m /s) sin 36.9° = 30.0 m /s
Since v0x is constant, the position x increases by 40.0 m every second. The y-velocity and the y-position are obtained from the following equations:
vfy = viy + a y (t f − t i )
yf = yi + viy (t f − t i ) + 12 a y (t f − t i )
2
Thus,
v1y = v0 y + a y (t1 − t 0 ) = (30.0 m /s) + ( −9.8 m /s 2 )(1 s − 0 s) = 20.2 m /s y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) = 0 m + (30.0 m /s)(1 s − 0 s) + 2
1 2
(−9.8 m /s )(1 s − 0 s) 2
2
= 25.1 m
v2 y = v0 y + a y (t2 − t 0 ) = (30.0 m /s) + ( −9.8 m /s 2 )(2 s − 0 s) = 10.4 m /s y2 = y0 + v0 y (t2 − t 0 ) + 12 a y (t2 − t 0 ) = 0 + (30.0 m /s)(2 s − 0 s) + 2
1 2
(−9.8 m /s )(2 s − 0 s) 2
2
= 40.4 m
and so on. The table showing x, y, vx, vy and v (from t = 0 s to t = 6 s) is given below. t (s)
x (m)
y (m)
vx (m/s)
vy (m/s)
v (m/s)
0
0
0
40
30.0
50.0
1
40
25.1
40
20.2
44.8
2
80
40.4
40
10.4
41.3
3
120
45.9
40
0.6
40.0
4
160
41.6
40
−9.2
41.0
5
200
27.5
40
−19.0
44.3
6
240
3.6
40
−28.8
49.3
(b)
Assess: As expected the trajectory is a parabola. Given an initial velocity of 50 m/s, a value of x = 240 m for x = 6 s is reasonable.
6.10. Model: The spheres will be treated as particles that move according to the constant-acceleration kinematic equations. Visualize:
Solve:
Using y1A = y0A + (v0A ) y (t1A − t 0A ) + 12 ( aA ) y (t1A − t 0A ) , we get 2
−1.0 m = 0 m + 0 m +
1 2
(−9.8 m/s )(t 2
1A
− 0 s) ⇒ t1A = 0.452 s = t1B 2
Both take the same time to reach the floor. We are now able to calculate x1A and x1B as follows:
x1A = x0A + (v0A ) x (t1A − t0A ) + 12 ( aA ) x (t1A − t0A ) = 0 m + (5.0 m /s)(0.452 s − 0 s) + 0 m = 2.26 m 2
x1B = x0B + (v0B ) x (t1B − t0B ) + 12 ( aB ) x (t1B − t0B ) = 0 m + (2.5 m /s)(0.452 s − 0 s) + 0 m = 1.13 m 2
Assess: Note that t1B = t1A since both the spheres move with the same vertical acceleration and both of them start with zero vertical velocity. The horizontal distance for sphere B is one-half the distance for sphere A because the horizontal velocity of sphere B is one-half that of A.
6.11. Model: The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is neglected. Visualize:
Solve:
(a) Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , we obtain 2
(−2.0 × 10
−2
m) = 0 m + 0 m +
1 2
(−9.8 m/s )(t 2
1
− 0 s) ⇒ t1 = 0.0639 s 2
(b) Using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
(50 m) = 0 m + v0 x (0.0639 s − 0 s) + 0 m ⇒ v0 x = 782 m /s Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s is understandable.
6.12. Model: The ball is treated as a particle and the effect of air resistance is ignored. Visualize:
Solve:
Using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
50 m = 0 m + (25 m/s)(t1 − 0 s) + 0 m ⇒ t1 = 2.0 s
Now, using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
y1 = 0 m + 0 m +
1 2
(−9.8 m/s )(2.0 s − 0 s) 2
2
= −19.6 m
Assess: The minus sign with y1 indicates that the ball’s displacement is in the negative y direction or downward. A magnitude of 19.6 m for the height is reasonable.
6.13. Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with a velocity Vx. Visualize:
Solve: Let vx be the velocity of the boat in frame S, and v′x be the velocity of the boat in S′. Then for travel down the river, 30 km v x = v x′ + Vx = = 10.0 km / hr 3.0 hr For travel up the river,
30 km − v x′ + Vx = − = −6.0 km / hr 5.0 hr Adding these two equations yields Vx = 2.0 km / hr . That is, the velocity of the flowing river relative to the earth is 2.0 km / hr .
6.14. Motion: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the moving sidewalk be S′. Frame S′ moves relative to S with velocity Vx. Solve: Let vx be your velocity in frame S and v′x be your velocity in S′. In the first case, when the moving sidewalk is broken, Vx = 0 m/s and
vx W =
( x1 − x0 )
50 s In the second case, when you stand on the moving sidewalk, v′x = 0 m/s. Therefore, using v x = v x′ + Vx , we get
x1 − x 0 75 s In the third case, when you walk while riding, v x′ = v x W . Using vx = v′x + Vx, we get v x S = Vx =
x1 − x 0 x1 − x 0 x1 − x 0 ⇒ t = 30 s = + t 50 s 75 s Assess: A time smaller than 50 s was expected.
6.15. Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the staple gun be S′. Frame S′ moves relative to S with velocity Vx. Solve: The velocity vx of the parts to be stapled in frame S is +3.0 m/s. Also Vx = −1.0 m/s. Using vx = v′x + Vx, we get v x′ = v x − Vx = +3.0 m /s − ( −1.0 m /s) = 4.0 m /s That is, the velocity of the parts in the frame of the staple gun is 4.0 m/s. In this frame 10 staples are fired per second. That is, distance between two staples distance ⇒ Distance = 0.4 m 4.0 m /s = = 1 time between two staples 10 s Assess: Note that the staple gun is in frame S′ and we had to find the velocity of the moving parts in this frame to solve the problem.
r
r
r
r
r
6.16. Model: The position vectors r and r ′ in frames S and S′ are related by the equation r = r ′ + R , where r
R is the position vector of the origin of frame S′ as measured in frame S. S is Ted’s frame and S′ is Stella’s frame. Visualize:
Solve:
r r The relation between R and the velocity of Stella V is r r R = V = (100 m /s) cos 45°iˆ − (100 m /s) sin 45° jˆ 5s
r r r Because r = r ′ + R ,
r 100 ˆ 100 ⇒ R = (5.0 m ) i− 2 2
(
)
jˆ = 353.6iˆ − 353.6 jˆ m
r r r r ′ = r − R = (200 m )iˆ − (353.6 m )iˆ − (353.6 m ) jˆ = −(154 m )iˆ + (354 m ) jˆ
[
]
r The vector r ′ determines the position of the exploding firecracker as seen by Stella.
r
r
V relative to another frame S and has a displacement R 6.17. Model: If a frame S′ is in motionrwith velocity r r r
relative to S, the and rvelocities ( r and v ) in S are related to the positions and velocities ( r ′ and v ′ ) in r positions r r r S ′ as r = r ′ + R and v = v ′ + V . In the present case, ship A is frame S and ship B is frame S′. Both ships have a common origin at t = 0 s. The position and velocity measurements are made in S and S′ relative to their origins. Solve: (a) The velocity vectors of the two ships are: r v = (20 mph ) cos 30°iˆ − sin 30° jˆ = (17.32 mph )iˆ − (10.0 mph ) jˆ A
r r Since r = v ∆t ,
v r r As rA = rB + R ,
r r r (b) Because vA = vB + V ,
[ [
] ]
r vB = (25 mph ) cos 20°iˆ + sin 20° jˆ = (23.49 mph )iˆ + (8.55 mph ) jˆ
r r rA = vA (2 hr ) = (34.64 miles)iˆ − (20.0 miles) jˆ r r rB = vB (2 hr ) = ( 46.98 miles)iˆ + (17.10 miles) jˆ r r r R = rA − rB = ( −12.34 miles)iˆ − (37.10 miles) jˆ ⇒ R = 39.1 miles r r r V = vA − vB = −(6.17 mph )iˆ − (18.55 mph ) jˆ ⇒ V = 19.5 mph
Assess: The value of the speed is reasonable.
6.18. Model: We will assume that constant-acceleration kinematic equations in a plane apply. Visualize:
r (a) The particle’s position r0 = (9.0 jˆ ) m implies that at t0 the particle’s coordinates are x0 = 0 m and y0 = r 9.0 m. The particle’s position r1 = (20iˆ ) m at time t1 implies that x1 = 20 m and y1 = 0 m. This is the position where the wire hoop is located. Let us find the time t1 when the particle crosses the hoop at x1 = 20 m. From the vx-versus-t curve and using the relation x1 = x0 + area of the vx-t graph, we get Solve:
20 m = 0 m + area of the vx-t graph = area of the vx-t graph From Figure P6.18 we see that the area of the vx-t graph equals 20 m when t = t1 = 3 s. (b) We can now look at the y-motion to find ay. Note that the slope of the vy-t graph (that is, ay) is negative and 2 constant, and we can determine ay by substituting into y3 = y0 + v0 y (t3 − t 0 ) + 12 a y (t3 − t 0 ) :
0 m = 9 m + 0 m + 12 a y (3 s − 0 s) ⇒ a y = −2 m /s 2 2
Therefore,
v4 y = v0 y + a y (t 4 − t 0 ) = 0 m/s + ( −2 m/s 2 )( 4 s − 0 s) = −8 m/s (c)
6.19. Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant-acceleration equations of kinematics. Visualize:
Solve:
(a) The time it will take the asteroid to reach the earth is
displacement 4 × 10 6 km = = 2 × 10 5 s = 55.6 hrs velocity 20 km/s (b) The angle of a line that just misses the earth is
tan θ =
R 6400 km R = 0.0917° ⇒ θ = tan −1 = tan −1 4.0 × 10 6 km y0 y0
(c) When the rocket is fired, the horizontal acceleration of the asteroid is
ax =
5 × 10 9 N = 0.125 m /s 2 4 × 1010 kg
(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely.) The velocity of the asteroid after the rocket has been fired for 300 s is
v x = v0 x + a x (t − t 0 ) = 0 m/s + (0.125 m/s 2 )(300 s − 0 s) = 37.5 m/s After 300 s, the vertical velocity is vy = 2 × 104 m/s and the horizontal velocity is vx = 37.5 m/s. The deflection due to this horizontal velocity is
tan θ = That is, the earth is saved.
37.5 m /s vx ⇒ θ = tan −1 = 0.107° vy 2 × 10 4 m /s
6.20. Model: Assume the particle model for the rocket-powered hockey puck and use the constant-acceleration kinematic equations. Solve: (a) The force and acceleration of the hockey puck are r r F cosθ ˆ F sin θ ˆ r a= i+ j F = F cosθiˆ + F sin θjˆ = ma m m From kinematics:
x = x 0 + v0 x (t − t 0 ) + 12 a x (t − t 0 ) = 0 m + 0 m + 12 a x t 2 = 12 a x t 2 2
y = y0 + v0 y (t − t 0 ) + 12 a y (t − t 0 ) = 0 m + v0 t + 12 a y t 2 = v0 t + 12 a y t 2 2
From the x-equation t = 2 x a x . Substituting this into the y-equation, we get
y = v0
2x 1 2x 2 mx + 2 ay = v0 + x tan θ ax ax F cosθ
(b) Substituting the known values into the expression from part (a),
y = (2.0 m /s)
2(1.0 kg) x + x tan 45° = 2.38 m /s (2.0 N) cos 45°
when θ = 45° and when x and y are in meters. The equation for θ = –45° is y = 2.38 m /s
x −x
x+x
6.21. Model: The ions are particles that move in a plane. They have vertical acceleration while between the acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so small, because of the large speeds, that we’ll ignore any deflection due to gravity. Visualize:
Solve: There’s never a horizontal acceleration, so the horizontal motion is constant velocity motion at vx = 5.0 × 106 m/s. The times to pass between the 5.0-cm-long acceleration plates and from the plates to the tumor are
0.050 m = 1.00 × 10 −8 s 5.0 × 10 6 m /s 1.50 m t2 − t1 = = 3.00 × 10 −7 s 5.0 × 10 6 m /s t1 − t 0 = t1 =
Upon leaving the acceleration plates, the ion has been deflected sideways to position y1 and has velocity v1y . These are
y1 = y0 + v0 y t1 + 12 a y t12 = 12 a y t12 v1y = v0 y + a y t1 = a y t1 In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position
y2 = y1 + v1y (t2 − t1 ) = 12 a y t12 + ( a y t1 )(t2 − t1 ) = ( 12 t12 + t1 (t2 − t1 )) a y We know y2 = 2.0 cm = 0.020 m, so we can solve for the acceleration ay that the ion had while between the plates:
ay =
y2 0.020 m = 1 = 6.56 × 1012 m / s 2 −8 2 t + t1 (t2 − t1 ) 2 (1.00 × 10 s) + (1.00 × 10 −8 s)(3.00 × 10 −7 s)
1 2 2 1
Assess: This acceleration is roughly 10 12 times larger than the acceleration due to gravity. This justifies our assumption that the acceleration due to gravity can be neglected.
6.22. Solve: From the expression for R, Rmax = v02 g . Therefore, R=
Rmax v02 sin 2θ 1 = ⇒ sin 2θ = ⇒ θ = 15° and 75° 2 2 g
Assess: The discussion of Figure 6.21 explains why launch angles θ and (90° − θ ) give the same range.
6.23. Model: Assume particle motion in a plane and constant-acceleration kinematics for the projectile. Visualize:
Solve:
(a) We know that v0 y = v0 sin θ , a y = − g , and v1 y = 0 m/s. Using v12y = v02y + 2 a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02 sin 2 θ + 2( − g)h ⇒ h =
v02 sin 2 θ 2g
(b) Using Equation 6.19 and the above expression for θ = 30°:
h=
( x2 − x0 ) =
(33.6 m /s)2 sin 2 30° = 14.4 m 2(9.8 m /s 2 )
v02 sin 2θ (33.6 m /s) sin(2 × 30°) = = 99.8 m g (9.8 m /s2 ) 2
For θ = 45°:
h=
(33.6 m /s)2 sin 2 45° = 28.8 m 2(9.8 m /s 2 )
( x2 − x0 ) =
(33.6 m /s)2 sin(2 × 45°) = 115.2 m
(9.8 m /s ) 2
For θ = 60°:
h=
( x2 − x0 ) =
(33.6 m /s)2 sin 2 60° = 43.2 m 2(9.8 m /s 2 )
(33.6 m /s)2 sin(2 × 60°) = 99.8 m 2(9.8 m /s 2 )
Assess: The projectile’s range, being proportional to sin(2θ), is maximum at a launch angle of 45°, but the maximum height reached is proportional to sin2(θ). These dependencies are seen in this problem.
6.24. Model: Assume the particle model for the spyglass, and the Galilean transformation of position equations. Visualize:
Solve: (a) In the ship’s frame, the spyglass lands at the base of the mast directly below the sailor. This is because everything in this frame, including the falling spyglass, has a forward velocity of 4.0 m/s. (b) Let S be the dinghy’s frame and S′ be the ship’s frame. S′ moves along the x-direction with velocity Vx relative to S. According to Equation 6.22, x = x ′ + Vx (t1 − t 0 ) where t1 is the time during which the origin of S′ has moved by a certain distance. t1 can be obtained from the vertical falling motion as follows:
y1 = y0 + v0 y t + 12 a y (t1 − t 0 ) ⇒ 0 m = 15m + 0 m + 2
1 2
(−9.8 m/s )(t 2
1
− 0 s) ⇒ t1 = 1.749 s 2
Therefore, x1 = 0 m /s + ( 4.0 m /s)(1.749 s) = 7.0 m . The spyglass lands 7.0 m to the right of the fisherman.
6.25. Model: Assume the particle model for the projectile and motion in a plane. Visualize:
Solve:
(a) Using y2 = y0 + v0 y (t2 − t 0 ) + 12 a y (t2 − t 0 ) , 2
y2 = 0 m + (30 m/s) sin 60°(7.5 s − 0 s) +
1 2
(−9.8 m/s )(7.5 s − 0 s) 2
2
= −80.8 m
Thus the launch point is 80.8 m higher than where the projectile hits the ground. (b) Using v12y = v02y + 2 a y ( y1 − y0 ) ,
0 m 2 /s 2 = (30 sin 60° m/s) + 2( −9.8 m/s 2 )( y1 − 0 m ) ⇒ y1 = 34.4 m 2
(c) The x-component is v2 x = v0 x = v0 cos 60° = (30 m /s) cos 60° = 15 m /s . The y-component is
v2 y = v0 y + a y (t2 − t 0 ) = v0 sin 60° − g(t2 − t1 ) = (30 m/s) sin 60° − (9.8 m/s 2 )(7.5 s − 0 s) = −47.52 m /s ⇒v=
(15 m /s)2 + (−47.52 m /s)2
θ = tan −1
v2 y v2 x
= tan −1
= 49.8 m /s
47.52 = 72.5° below +x 15
Assess: An angle of 72.5° made with the ground, as the projectile hits the ground 80.8 m below its launch point, is reasonable in view of the fact that the projectile was launched at an angle of 60°.
6.26. Model: Assume the particle model and motion under constant-acceleration kinematic equations in a plane. Visualize:
Solve:
(a) Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
(−9.8 m /s )(t − 0 s) = 1.8 m + (7.713 m /s)t − ( 4.9 m /s )t ⇒ t = − 0.206 s and 1.780 s
0 m = 1.8 m + v0 sin 40°(t1 − 0 s) +
2
2
1 2
1
2
1
2 1
1
The negative value of t1 is unphysical for the current situation. Using t1 = 1.780 s and x1 = x 0 + v0 x (t1 − t 0 ) , we get
x1 = 0 + (v0 cos 40° m/s)(1.780 s − 0 s) = (12 m /s) cos 40°(1.78 s) = 16.36 m (b) We can repeat the calculation for each angle. A general result for the flight time at angle θ is
(
)
t1 = 12 sin θ + 144 sin 2 θ + 35.28 / 9.8 s and the distance traveled is x1 = 12cosθ × t1. We can put the results in a table.
θ
t1
x1
40.0°
1.780 s
16.36 m
42.5°
1.853 s
16.39 m
45.0°
1.923 s
16.31 m
47.5°
1.990 s
16.13 m
Maximum distance is achieved at θ ≈ 42.5°. Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is launched and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the angle from 40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.
6.27. Model:
The golf ball is a particle following projectile motion.
Visualize:
(a) The distance traveled is x1 = v0xt1 = v0 cosθ × t1. The flight time is found from the y-equation, using the fact that the ball starts and ends at y = 0: 2v sin θ y1 − y0 = 0 = v0 sin θ t1 − 12 gt12 = (v0 sin θ − 12 gt1 ) t1 ⇒ t1 = 0 g
Thus the distance traveled is x1 = v0 cosθ ×
2v0 sin θ 2v02 sin θ cosθ = g g
For θ = 30°, the distances are
( x1 ) earth =
2v02 sin θ cosθ 2(25 m / s) 2 sin30°cos30° = = 55.2 m 9.80 m / s 2 gearth
( x1 ) moon =
2v02 sin θ cosθ 2v02 sin θ cosθ 2v 2 sin θ cosθ = =6× 0 = 6( x1 ) earth = 331.2 m 1 gmoon gearth 6 gearth
The golf ball travels 331.2 m – 55.2 m = 276 m farther on the moon than on earth. (b) The flight times are 2v sin θ (t1 ) earth = 0 = 2.55 s gearth
2v0 sin θ 2v0 sin θ = 1 = 6 (t1 ) earth = 15.30 s gmoon 6 gearth The ball spends 15.30 s – 2.55 s = 12.75 s longer in flight on the moon. (t1 ) moon =
6.28. Model: The particle model for the ball and the constant-acceleration equations of motion are assumed. Visualize:
Solve:
(a) Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
( )
(b) Using v
h = 0 m + (30 m/s) sin 60°( 4 s − 0 s) +
2 y top
(
)
1 2
(−9.8 m/s )(4 s − 0 s) 2
2
= 25.5 m
= v + 2 a y ytop − y0 , 2 y
( )
0 m /s = (v0 sin θ ) + 2( − g) ytop ⇒ ytop 2
2
2
2 v0 sin θ ) ( [(30 m /s) sin 60°] = =
2
2g
2(9.8 m /s 2 )
= 34.4 m
(c) The x and y components are
v1y = v0 y + a y (t1 − t 0 ) = v0 sin θ − gt1 = (30 m/s) sin 60° − (9.8 m/s 2 ) × ( 4.0 s) = −13.22 m / s v1 x = v0 y = v0 cos 60° = (30 m /s) cos 60° = 15.0 m /s
⇒ v1 = v12x + v12y = 20.0 m /s Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable.
6.29. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize:
Solve:
The initial velocity is
v0 x = v0 cos 5° = (20 m /s) cos 5° = 19.92 m /s v0 y = v0 sin 5° = (20 m /s) sin 5° = 1.743 m /s The time it takes for the ball to reach the net is
x1 = x 0 + v0 x (t1 − t 0 ) ⇒ 7.0 m = 0 m + (19.92 m/s)(t1 − 0 s) ⇒ t = 0.351 s The vertical position at t1 = 0.351 s is
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 )
2
= (2.0 m ) + (1.743 m /s)(0.351 s − 0 s) +
1 2
(−9.8 m /s )(0.351 s − 0 s) 2
2
= 2.01 m
Thus the ball clears the net by 1.01 m. Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial vertical velocity and the ball clears by a larger distance. The above result is reasonable.
6.30. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize:
Solve:
(a) The time for the ball to fall is calculated as follows:
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒0m=4m+0m+
1 2
(−9.8 m/s )(t 2
1
2
− 0 s) ⇒ t1 = 0.9035 s 2
Using this result for the horizontal velocity:
x1 = x 0 + v0 x (t1 − t 0 ) ⇒ 25 m = 0 m + v 0 x (0.9035 s − 0 s) ⇒ v0x = 27.7 m/s (b) We have v0 y = ± v0 sin θ , where we will use the plus sign for up 5° and the minus sign for down 5°. We can write
y1 = y0 ± v0 sin θ (t1 − t 0 ) −
g g 2 t1 − t 0 ) ⇒ 0 m = 4 m ± v0 sin θ t1 − t12 ( 2 2
Let us first find t1 from x1 = x 0 + v0 x (t1 − t 0 ) :
25 m = 0 m + v0 cosθ t1 ⇒ t1 =
25 m v0 cosθ
Now substituting t1 into the y-equation above yields
25 m g 25 m 0 m = 4 m ± v0 sin θ − v0 cosθ 2 v0 cosθ
⇒ v02 =
2
2 g(25 m ) 1 = 22.3 m /s and 44.2 m/s 2 2 cos θ 4 m ± (25 m ) tan θ
The range of speeds is 22.3 m /s to 41.2 m /s , which is the same as 50 mph to 92 mph. Assess: These are reasonable speeds for baseball pitchers.
6.31. Model: We will use the particle model and the constant-acceleration kinematic equations in a plane. Visualize:
Solve:
The x- and y-equations of the ball are
x1B = x 0B + (v0B ) x (t1B − t 0B ) + y1B = y0B + (v0B ) y (t1B − t 0B ) +
1 2
1 2
(aB ) x (t1B − t0B )
(aB ) y (t1B − t0B )
2
2
⇒ 65 m = 0 m + (v0B cos30°)t1B + 0 m
⇒ 0 m = 0 m + (v0B sin30°)t1B +
From the y-equation,
v0B =
gt1B 2 sin 30°) (
Substituting this into the x-equation yields 2 g cos 30° t1B 2 sin 30° = 2.77 s
65 m = ⇒ t1B For the runner:
t1R =
20 m = 2.50 s 8.0 m /s
Thus, the throw is too late by 0.27 s. Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.
1 2
(− g)t1B2
6.32. Model: We will use the particle model and the constant-acceleration kinematic equations for the car. Visualize:
Solve:
(a) The initial velocity is
v0 x = v0 cosθ = (20 m /s) cos 20° = 18.79 m /s v0 y = v0 sin θ = (20 m /s) sin 20° = 6.840 m /s Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
0 m = 30 m + (6.840 m/s)(t1 − 0 s) +
1 2
(−9.8 m/s )(t 2
1
− 0 s) ⇒ 4.9t12 − 6.840t1 − 30 = 0 2
The positive root to this equation is t1 = 3.269 s . The negative root is physically unreasonable in the present case.
Using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , we get 2
x1 = 0 m + (18.79 m /s)(3.269 s − 0 s) + 0 = 61.4 m (b) The components of the final velocity are v1 x = v0 x = 18.79 m /s and
v1y = v0 y + a y (t1 − t 0 ) = 6.840 m/s − (9.8 m/s 2 )(3.269 s − 0 s) = −25.2 m /s ⇒v=
(18.79 m /s) + (−25.2 m /s)2
= 31.4 m /s
Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of approximately 200 feet (61.4 m). This distance is reasonable.
6.33. Model: We will use the particle model for the ball’s motion under constant-acceleration kinematic equations. Note that the ball’s motion on the smooth, flat board is ay = − g sin 20° = −3.352 m /s 2 . Visualize:
Solve:
The ball’s initial velocity is v0 x = v0 cosθ = (3.0 m / s) cosθ
v0 y = v0 sin θ = (3.0 m /s) sin θ
Using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
2.5 m = 0 m + (3.0 m/s) cosθ (t1 − 0 s) + 0 m ⇒ t1 =
(2.5 m)
(3.0 m/s) cosθ
=
0.833 s cosθ
Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) and the above equation for t1, 2
0.833 s 1 (0.833 s) − 3.352 m/s 2 ) 0 m = 0 m + (3.0 m/s) sin θ cosθ 2 ( cos 2 θ sin θ 1.164 ⇒ (2.5 m ) = ⇒ 2.5 sin θ cosθ = 1.164 ⇒ 2θ = 68.6° ⇒ θ = 34.3° cosθ cos2 θ 2
6.34. Model: Use the particle model for the ball and the constant-acceleration kinematic equations. Visualize:
Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this distance is 5.0 m. Using the kinematic equation v12y = v02y + 2 a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02y + 2( −9.8 m/s 2 )(5.0 m − 0 m ) ⇒ v0 y = 9.899 m /s
The time for up and down motion is calculated as follows:
y2 = y0 + v0 y (t2 − t 0 ) + 12 a y (t2 − t 0 ) ⇒ 0 m = 0 m + (9.899 m/s)t2 − 2
1 2
(9.8 m/s )t 2
2 2
⇒ t2 = 0 s and 2.02 s
The zero solution is not of interest. Having found the time t2 = 2.02 s, we can now find the horizontal velocity needed to cover a displacement of 18.0 m:
x 2 = x 0 + v0 x (t2 − t 0 ) ⇒ 18.0 m = 0 m + v0 x (2.02 s − 0 s) ⇒ v0 x = 8.911 m /s ⇒ v0 =
r (b) The direction of v0 is given by
(8.911 m /s)2 + (9.899 m /s)2
= 13.3 m /s
v0 y 9.899 = tan −1 = 48.0° v0 x 8.911 Assess: Since the maximum range corresponds to an angle of 45°, the value of 48° corresponding to a range of 18 m and at a modest speed of 13.3 m/s is reasonable.
θ = tan −1
6.35. Model: We will use the particle model for the food package and the constant-acceleration kinematic equations of motion. Visualize:
Solve:
For the horizontal motion,
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) = 0 m + (150 m/s)(t1 − 0 s) + 0 m = (150 m /s)t1 2
We will determine t1 from the vertical y-motion as follows:
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 )
⇒ 0 m = 100 m + 0 m +
1 2
(−9.8 m/s )t 2
2 1
⇒ t1 =
2
200 m = 4.518 s 9.8 m / s 2
From the above x-equation, the displacement is x1 = (150 m /s)( 4.518 s) = 678 m . Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.
6.36. Model: We will use the particle model for the gun shell and the plane. The motion is assumed to be governed by constant-acceleration kinematic equations without air resistance. Visualize:
The gun shell’s trajectory is parabolic. The airplane will cross such a trajectory twice – when the trajectory’s slope is upward or positive (the first intersection point) and when the slope is downward or negative (the second intersection point). The gun can therefore be fired at two times depending upon the horizontal position of the plane. Solve: The x position of the plane flying at a height of 500 m can be found from the following equations
x1P = x 0P + (v0 x ) P (t1P − t 0P ) = x 0P − (300 m/s)t1P
x 2P = x 0P + (v0 x ) P (t2P − t 0P ) = x 0P − (300 m/s)t2P The velocity of the shell is
(v0S ) x = v0S cos 60° = (200 m/s) cos 60° = 100.0 m/s (v0S ) y = v0S sin 60° = (200 m/s) sin 60° = 173.2 m/s Times t1S and t2S can be found from the y-motion as
y1S = y0S + (v0S ) y (tS − t 0S ) − 12 g(tS − t 0S )
⇒ 500 m = 0 m + (173.2 m/s)tS −
1 2
2
(9.8 m/s )t 2
2 S
⇒ ( 4.9 m/s 2 )tS2 − (173.2 m/s)tS + 500 m = 0
The two roots of this quadratic equation are t1S = 3.15 s and t2S = 32.18 s. The shell’s positions are therefore at:
x1S = (v0S ) x t1S = (100.0 m/s)(3.151 s) = 315 m x 2S = (v0S ) y t2S = (100.0 m/s)(32.18 s) = 3218 m Let us now go back to the x-positions of the plane at time t1P = t1S = 3.151 s and at time t2P = t2S = 32.18 s:
x1P = x 0P − (300 m /s)t1P = 315 m = x1S
x 0P = 315 m + (300 m/s)(3.151 s) = 1260 m x 2P = x 0P − (300 m /s)t2P = 3218 m = x 2S
x 0P = 3218 m + (300 m/s)(32.18 s) = 12, 900 m The gun shell will hit the plane if it is fired when either the plane is 1,260 m away or when it is 12,900 m away. Assess: The distances seem a little too large, but note that the plane’s speed is nearly 670 mph and the times involved are of the order of a few seconds.
6.37. Model: We will assume a particle model for the cannonball, and apply the constant-acceleration kinematic equations. Visualize:
Solve:
(a) The cannonball that was accidentally dropped can be used to find the height of the wall:
y1A = y0A + (v0A ) y (t1A − t 0A ) +
⇒ y0A =
1 2
1 2
(aA ) y (t1A − t0A )
(9.8 m/s )(1.5 s) 2
2
2
2 ⇒ 0 m = y0A + 0 m − 12 gt1A
= 11.03 m
For the cannonball that was shot:
(v0S ) x = v0S cos 30° = (50 m/s) cos 30° = 43.30 m/s (v0S ) y = v0s sin 30° = (50 m/s) sin 30° = 25.0 m/s
We can now find the time it takes the cannonball to hit the ground:
y2S = y0S + (v0S ) y (t2S − t 0S ) +
1 2
(aS ) y (t2S − t0S )2
⇒ 0 m = (11.03 m ) + (25.0 m/s)t2 S −
(9.8 m/s ) t 2
2 2S
2 ⇒ ( 4.9 m/s 2 )t22S − (25.0 m/s)t2S − (11.03 m ) = 0 ⇒ t2s = 5.51 s There is also an unphysical root t2S = − 0.41 s. Using this time t2S, we can now find the horizontal distance from the wall as follows: x 2s = x 0s + (v0s ) x (t2s − t 0s ) = 0 m + ( 43.30 m /s)(5.51 s) = 239 m (b) At the top of the trajectory (v1S)y = 0 m/s. Using (v1S ) y = (v0S ) y − 2 g( y1S − y0S ) , 2
2
0 m 2 /s 2 = (25.0 m/s) − 2(9.8 m/s 2 )( y1s − 11.03 m ) ⇒ y1s = 42.9 m 2
Assess: In view of the fact that the cannon ball has a speed of approximately 110 mph, a distance of 239 m for the cannonball to hit the ground is reasonable.
6.38. Model: Apply the particle model and the constant-acceleration equations of kinematics. Visualize:
Solve:
For the football:
y1F = y0F + (v0F ) y (t1F − t 0F ) +
⇒ 0 m = 0 m + (v0F sin 40°)t1F −
1 2
(g)t1F2
1 2
(aF ) y (t1F − t0F )
2
⇒ t1F = 0 m /s and t 1F =
2v0F sin 40° g
Thus
x1F = x 0F + (v0F ) x (t1F − t 0F ) + = (v0F cos 40°)
1 2
(aF ) x (t1F − t0F )
2
= 0 m + (v0F cos 40°)t1F + 0 m
2 2v0F sin 40° 2v0F = sin 40° cos 40° g g
For Doug:
x1D = x 0D + (v0D ) x (t1D − t 0D ) +
1 2
(aD ) x (t1D − t0D )
2
2v = 20 m + (6.0 m /s)t1D + 0 m = 20 m + (6.0 m /s) 0F sin 40° g
Since Doug and the football meet together, x1F = x1D. Equating the expressions for Doug and the football yields: 2 2v0F 2v sin 40° cos 40° = 20 m + (6.0 m/s) 0F sin 40° g g 2 ⇒ 0.1 v0F − (0.787 m /s) v0F − (20 m /s 2 ) = 0 ⇒ v0F = 18.6 m/s and −10.7 m /s
The negative solution is unphysical. Assess: A speed of 18.6 m/s or 42 mph for the ball is reasonable. The mass of the ball, which was given in the problem, was not needed.
6.39. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. Visualize:
Solve:
Using v1 y = v0 y + a y (t1 − t 0 ) , we get
v1y = 0 m /s − gt1 ⇒ v1y = –gt1 Also using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
60 m = 0 m + v0xt1 + 0 m ⇒ v0 x =
60 m = v1 x t1
Since v1 y / v1 x = − tan 3° = − 0.0524 , using the components of v0 gives
− gt1 = − 0.0524 ⇒ t1 = (60 m / t1 )
(0.0524)(60 m)
(9.8 m /s ) 2
= 0.566 s
Having found t1, we can go back to the x-equation to obtain v0 x = 60 m / 0.566 s = 106 m /s . Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of 106 m/s or 237 mph for the arrow is understandable.
6.40. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We will assume that the archer shoots from 1.75 m above the slope (about 5′ 9′′). Visualize:
Solve:
For the y-motion:
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ y1 = 1.75 m + (v0 sin 20°)t1 − 12 gt12 2
⇒ y1 = 1.75 m + (50 m /s) sin 20°t1 − 12 gt12 For the x-motion: 2 x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) = 0 m + (v0 cos 20°)t1 + 0 m = (50 m /s)(cos 20°)t1 Because y1 x1 = − tan 15° = − 0.268 ,
1.75 m + (50 m/s)(sin 20°)t1 − 12 gt12 = − 0.268 ⇒ t1 = 6.12 s and −0.058 s (unphysical) (50 m/s)(cos 20°)t1 Using t1 = 6.12 s in the x- and y-equations above, we get y1 = –77.0 m and x1 = 287 m. This means the distance down the slope is x12 + y12 = (287 m ) + ( −77.0 m ) = 297 m. Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15° slope at an angle of 20° above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers. 2
2
6.41. Model: The object is treated as a particle in the model of kinetic friction with its motion governed by constant-acceleration kinematics. Visualize:
Solve: The velocity v1x as the object sails off the edge is related to the initial velocity v0x by v12x = v02x + 2 a x ( x1 − x 0 ) . Using Newton’s second law to determine ax gives ∑ Fx = − fk = ma x ⇒ ∑ Fy = n − mg = 0 N ⇒ n = mg Using this result and the model of kinetic friction − µ k mg = ma x . This implies
( f k = µ k n) ,
the x-component equation can be written as
a x = − µ k g = −(0.5)(9.8 m/s 2 ) = −4.9 m/s 2
Kinematic equations for the object’s free fall can be used to determine v1x: 2 g 2 y2 = y1 + v1y (t2 − t1 ) + 12 ( − g)(t2 − t1 ) ⇒ 0 m = 1.0 m + 0 m − (t2 − t1 ) ⇒ (t2 − t1 ) = 0.4518 s 2
x 2 = x1 + v1 x (t2 − t1 ) = 2.3 m = 2.0 m + v1 x (0.4518 s) ⇒ v1 x = 0.664 m /s
Having determined v1x and ax, we can go back to the velocity equation v12x = v02x + 2 a x ( x1 − x 0 ) :
(0.664 m/s)2 = v02x + 2(−4.9 m/s 2 )(2.0 m) ⇒ v0x = 4.48 m/s
Assess: v0x = 4.48 m/s is about 10 mph and is a reasonable speed.
6.42. Model: We will assume a particle model for the sand, and use the constant-acceleration kinematic equations. Visualize:
Solve:
Using the equation x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
x1 = 0 m + (v0 cos15°)(t1 − 0 s) + 0 m = (60 m /s)(cos15°)t1
We can find t1 from the y-equation, but note that v0y = –v0 sin15° because the sand is launched at an angle below horizontal.
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ 0 m = 3.0 m − (v0 sin 15°)t1 − 12 gt12 2
= 3.0 m − (6.0 m/s)(sin 15°)t1 −
1 2
(9.8 m/s )t 2
2 1
⇒ 4.9t12 + 1.55t1 − 3.0 = 0 ⇒ t1 = 0.6399 s and − 0.956 s (unphysical) Substituting this value of t1 in the x-equation gives the distance d = x1 = (6.0 m /s) cos15°(0.6399 s) = 3.71 m
6.43. Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.
Solve:
Sam’s acceleration up the slope is given by Newton’s second law: ( Fnet ) s = F − mg sin 10° = ma0
F 200 N − g sin 10° = − (9.8 m/s 2 ) sin 10° = 0.965 m/s 2 m 75 kg The length of the slope is s1 = (50 m)/sin10° = 288 m. His velocity at the top of the slope is a0 =
v12 = v02 + 2 a0 ( s1 − s0 ) = 2 a0 s1 ⇒ v1 = 2(0.965 m/s 2 ) (288 m) = 23.6 m/s This is Sam’s initial speed into the air, giving him velocity components v1x = v1 cos10° = 23.2 m/s and v1y = v1sin10° = 4.10 m/s. This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for Sam’s acceleration is (F ) (200 N) cos10° a1 x = net x = = 2.63 m /s 2 75 kg m
a1y =
( Fnet ) y m
=
(200 N) sin 10° − (75 kg)(9.80 m /s 2 ) = −9.34 m /s 2 75 kg
The y-equation of motion allows us to find out how long it takes Sam to reach the ground: y2 = 0 m = y1 + v1y t2 + 12 a1y t22 = 50 m + ( 4.10 m /s) t2 − ( 4.67 m /s 2 ) t22 This quadratic equation has roots t2 = –2.86 s (unphysical) and t2 = 3.74 s. The x-equation of motion—this time with an acceleration—is x 2 = x1 + v1 x t2 + 12 a1 x t22 = 0 m + (23.2 m /s) t2 − 12 (2.63 m /s 2 ) t22 = 105 m Sam lands 105 m from the base of the cliff.
6.44. Model: Treat the skateboarder as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she becomes airborne.
Solve: Without friction, the skateboarder’s acceleration on the ramp is a0 = –gsin30° = –4.90 m/s2 . The length of the ramp is s1 = (1.0 m)/sin30° = 2.0 m. We can use kinematics to find her speed at the top of the ramp:
v12 = v02 + 2 a0 ( s1 − s0 ) = v02 + 2 a0 s1 ⇒ v1 = (7.0 m/s) 2 + 2( −4.90 m/s 2 ) (2.0 m) = 5.42 m/s This is the skateboarder’s initial speed into the air, giving her velocity components v1x = v1 cos30° = 4.69 m/s and v1y = v1sin30° = 2.71 m/s. We can use the y-equation of projectile motion to find her time in the air:
y2 = 0 m = y1 + v1y t2 + 12 a1y t22 = 1.0 m + (2.71 m /s) t2 − ( 4.90 m /s 2 ) t22 This quadratic equation has roots t2 = –0.253 s (unphysical) and t2 = 0.806 s. The x-equation of motion is thus
x 2 = x1 + v1 x t2 = 0 m + ( 4.69 m /s) t2 = 3.78 m She touches down 3.78 m from the end of the ramp.
6.45. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne.
Solve:
The motorcycle’s acceleration on the ramp is given by Newton’s second law: ( Fnet ) s = − fr − mg sin 20° = − µ r n − mg sin 20° = − µ r mg cos 20° − mg sin 20° = ma0
a0 = − g( µ r cos 20° + sin 20°) = −(9.8 m /s 2 ) ((0.02)cos 20° + sin 20°) = −3.536 m /s 2 The length of the ramp is s1 = (2.0 m)/sin20° = 5.85 m. We can use kinematics to find its speed at the top of the ramp:
v12 = v02 + 2 a0 ( s1 − s0 ) = v02 + 2 a0 s1 ⇒ v1 = (11.0 m/s) 2 + 2( −3.536 m/s 2 ) (5.85 m) = 8.92 m/s This is the motorcycle’s initial speed into the air, with velocity components v1x = v1cos20° = 8.38 m/s and v1y = v1sin20° = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air:
y2 = 0 m = y1 + v1y t2 + 12 a1y t22 = 2.0 m + (3.05 m /s) t2 − ( 4.90 m /s 2 ) t22 This quadratic equation has roots t2 = –0.399 s (unphysical) and t2 = 1.021 s. The x-equation of motion is thus
x 2 = x1 + v1 x t2 = 0 m + (8.38 m /s) t2 = 8.56 m 8.56 m < 10.0 m, so it looks like crocodile food.
6.46. Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket. r Solve:
r (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law F = ma : 140, 700 N = (5000 kg)a ⇒ a = 28.14 m /s 2
Therefore, a x = a cos 44.7° = 20.0 m /s 2 and a y = a sin 44.7° = 19.8 m /s 2 . The net acceleration in the y-direction is thus
(anet ) y = ay − g = (19.8 − 9.8) m /s 2 = 10.0 m /s 2 With this acceleration, we can write the equations for the x- and y-motion of the rocket.
y = y0 + v 0 y (t − t 0 ) +
x = x 0 + v 0 x (t − t 0 ) +
1 2
1 2
(anet ) y (t − t0 )
(anet ) x (t − t0 )
2
2
=0m+0m+ =0m+0m+
1 2
1 2
(10.0 m/s )t = (5.0 m/s )t 2
2
2
2
2 2 x (10.0 m /s )t =2 = y (5.0 m /s2 )t 2
The equation that describes the rocket’s trajectory is y = 12 x . (b) It is a straight line with a slope of 12 . (c) In general,
v y = v0 y + ( anet ) y (t1 − t 0 ) = 0 + (10.0 m/s 2 )t1
v x = v0 x + ( anet ) x (t1 − t 0 ) = 0 + (20.0 m/s 2 )t1
(10.0 m /s ) t + (20.0 m /s ) t 2 2 2 1
2 2 2 1
2
(20.0 m/s )t = (10.0 m/s )t
From these two equations,
v=
2
= (22.36 m /s 2 )t1
The time required to reach the speed of sound is calculated as follows:
330 m/s = (22.36 m/s 2 )t1 ⇒ t1 = 14.76 s We can now obtain the elevation of the rocket. From the y-equation,
y = (5.0 m/s 2 )t12 = (5.0 m/s 2 )(14.76 s) = 1090 m 2
2
2
6.47. Model: The hockey puck will be treated as a particle whose motion is determined by constantacceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to free fall. Visualize:
Solve:
Newton’s second law is:
Fx = ma x ⇒ a x =
Fx 2.0 N = = 2.0 m /s 2 m 1.0 kg
The kinematic equation v12x = v02x + 2 a x ( x1 − x 0 ) yields:
v12x = 0 m 2 /s 2 + 2(2.0 m/s 2 )( 4.0 m ) ⇒ v1 x = 4.0 m /s Let us now find the time of free fall (t2 − t1 ) :
y2 = y1 + v1y (t2 − t1 ) + 12 a1y (t2 − t1 ) ⇒0 m=2m+0 m+
1 2
(−9.8 m/s )(t
2
− t1 ) ⇒ (t2 − t1 ) = 0.639 s 2
2
2
Having obtained v1x and (t2 − t1 ) , we can now find (x2 – x1) as follows:
x 2 = x1 + v1 x (t2 − t1 ) + 12 a x (t2 − t1 )
⇒ x 2 − x1 = ( 4.0 m/s)(0.639 s) +
1 2
2
(2.0 m/s )(0.639 s) 2
2
= 2.96 m
Assess: For a modest horizontal thrust of 2.0 N, a landing distance of 2.96 m is reasonable.
6.48. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize:
Solve: As the rocket is accidentally bumped v0 x = 0.5 m /s and v0 y = 0 m /s . On the other hand, when the engine is fired
Fx = ma x ⇒ a x =
Fx 20 N = = 40 m /s 2 m 0.5 kg
(a) Using y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) , 2
0 m = 40 m + 0 m +
1 2
(−9.8 m/s )t 2
2 1
⇒ t1 = 2.857 s
The distance from the base of the wall is
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) = 0 m + (0.5 m/s)(2.857s) + 2
1 2
(40 m/s )(2.857 s) 2
2
= 165 m
(b) The x- and y-equations are
y = y0 + v0 y (t − t 0 ) + 12 a y (t − t 0 ) = 40 − 4.9t 2 2
x = x 0 + v0 x (t − t 0 ) + 12 a x (t − t 0 ) = 0.5t + 20t 2 2
Except for a brief interval at t = 0, 20t2 >> 0.5t. Thus x ≈ 20t 2 , or t2 = x/20. Substituting this into the y-equation gives y = 40 – 0.245x This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
6.49. Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure, based on past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf ball took flight. She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the ball? (b) From the second equation, ( 4.9 m /s 2 )t12 − (50 sin θ m /s) t1 = 0 ⇒ t1 = 0 s and t1 =
(50 m /s)sin θ 4.9 m /s 2
Using the above value for t1 in the first equation yields:
100 m =
(50 cosθ )(50 sin θ ) m 2 /s 2 4.9 m /s 2
⇒ 2 cosθ sin θ = sin 2θ =
9.8 = 0.392 ⇒ 2θ = 23.1 ⇒ θ = 11.5° 25
Assess: Although the original speed is reasonably high (50 m/s = 112 mph), the ball travels a distance of only 100 m, implying either a small launch angle around 10° or an angle closer to 80°. The calculated angle of 11.5° is thus pretty reasonable.
6.50. Solve: While visiting the Black Canyon of the Gunnison you come to a place where a sign tells you that the Canyon is 300 m deep at that point. To check the depth and determine how far it is to the river below you throw a rock horizontally with a speed you judge to be 30 m/s. How long should it take for the rock to hit the river and how far away from you is that point on the river? (b) From the second equation 300 m = ( 4.9 m /s 2 )t12 . This means t1 = 7.82 s. This is the time before splashing. Using this value of time t1 in the first equation, we get for the distance to the splash x = 30 m /s(7.82 s) = 235 m .
6.51. Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The submarine’s torpedo tube happens to be stuck in a position pointing 45° west of north. The tube fires a torpedo with a speed of 6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it fires? (b) Relative to the water, the torpedo will have velocity components v x = −6.0 cos 45° m /s + 3.0 m /s = −4.24 m /s + 3 m /s = −1.24 m /s v y = +6.0 cos 45° m /s = +4.24 m /s The time to travel north to the ship is
100 m = ( 4.24 m / s) t1 ⇒ t1 = 25.6 s
Thus, x = (1.24 m /s)(23.6 s) = −29.2 m . That is, the ship should be 29.2 m west of the submarine.
6.52. Solve: You decide to test fly your model airplane off of a 125 m tall building. The model’s engine starts fine and gets the airplane moving at 4.0 m/s but quits just as it gets to the edge of the building. The model proceeds to fall “like a rock.” How far from the edge of the building will it crash into the ground? (Assume g = 10 m/s2 for easier calculation.) Visualize:
Using the equation y1 = y0 + v0 y t + 12 a y (t 0 − t1 ) , we get 2
y1 = −(5 m /s 2 ) t12 = −125 m ⇒ t1 = The distance x1 = ( 4 m /s)(5 s) = 20 m .
125 m = 25 s 2 = 5 s 5 m / s2
6.53. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. The frames S and S′ have their origins coincident at t = 0 s. r r r r r Solve: (a) According to the Galilean transformation of position: r = r ′ + R = r ′ + tV . We need to find Mary’s r position vector r from the earth frame S. The observer in frame S′ will observe the boat move straight north and will r r find its position as r ′ = (2.0t ) jˆ m/s . We also know that V = (3.0)iˆ m/s . Since r ′ = 100 m = (2.0 m /s)t and t = 50 s , we have r r = (2.0 m/s)(50 s) jˆ + (50 s)(3.0 m/s)t iˆ = (150 m )iˆ + (100 m ) jˆ Thus she lands 150 m east of the point that was straight across the river from her when she started. (b)
r r r Note that r ′ is the displacement due to rowing, R is the displacement due to the river’s motion, and r is the net displacement.
6.54. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. r r r r r r r Solve: (a) According to the Galilean transformation of velocity v = v ′ + V or v ′ = v − V . Because V = (2.0 m/s)iˆ r and v = (3.0 m/s) jˆ , we have r v ′ = (3.0 m/s) jˆ − (2.0 m/s)iˆ = v x′ iˆ + v y′ jˆ
v′ 2.0 m /s θ = tan −1 x = tan −1 = 33.7° west of north 3.0 m /s v y′ (b) The time to cross is
t=
100 m 100 m = = 33.3 s v 3.0 m /s
6.55. Model: Let Mike’s frame be S and Nancy’s frame be S′. Frame S′ moves relative to S with velocity V. x is the horizontal direction and y is the vertical direction for motion. The frames S and S′ coincide at t = 0 s. r r Solve: (a) According to the Galilean transformation of velocity v = v ′ + V . Mike throws the ball with velocity r r v = (22 m/s) cos 63°iˆ + (22 m/s)sin63° jˆ , and V = (30 m/s)iˆ . Thus in Nancy’s frame r r r v ′ = v − V = (22 cos 63° − 30)iˆ m/s + (22 sin 63°) jˆ m/s = −20.0iˆ + 19.6 jˆ m /s
(
θ = tan −1
v y′
= tan −1
v x′
)
19.6 m / s = 44.4° 20.0 m / s
The direction of the angle is 44.4° above the − x ′ axis (in the second quadrant). 2 (b) In Nancy’s frame S′ the equation x ′ = x 0′ + v x′ (t ′ − t 0′ ) + 12 a x′ (t ′ − t 0′ ) becomes
x ′ = −(20.0 m / s)t ′ and the equation y ′ = y0′ + v y′ (t ′ − t 0′ ) + 12 a y′ (t ′ − t 0′ ) becomes 2
y ′ = 0 m + (19.6 m/s)t ′ −
1 2
(9.8 m/s )t ′ = (19.6 m/s)t ′ − (4.9 m/s )t ′ 2
2
2
2
6.56. Model:r Let the earth frame be S and a frame attached to the sailboat be S′. Frame S′ moves relative to S with velocity V = 8.0iˆ mph.
r r r r Solve: (a) From Equation 6.24, v = v ′ + V , where v ′ = (12.0) cos 45°iˆ mph + (12.0) sin 45° jˆ mph is the velocity of the wind in S′, or the apparent wind velocity. Thus, r v = (12.0) cos 45°iˆ mph + (12.0) sin 45° jˆ mph + (8.0)iˆ mph
(
)
= (8.0 + 12.0cos45°)iˆ mph + (12.0sin45°) jˆ mph = 16.48iˆ + 8.48 jˆ mph
θ = tan −1
8.48 mph = 27.2° 16.48 mph
The wind speed is v = 18.5 mph and the direction is from 27.2° south of west. r r (b) In this case, v ′ = −(12.0 mph ) cos 45°iˆ − (12.0 mph ) sin 45° jˆ and V = (8.0 mph )iˆ . So, r v = −(12.0 mph ) cos 45° + 8.0 mph iˆ − (12.0 mph ) sin 45° jˆ = −0.485iˆ − 8.485 jˆ
[
]
θ = tan −1
8.485 mph = 86.7° 0.485 mph
The speed is 8.5 mph and the direction is from 86.7° north of east or from 3.3° east of north.
6.57. Model:
Let the earth be frame S and the river be frame S′. Assume the river flows toward the east, which is the x and x′-axis. Visualize:
Solve: In frame S′, the river’s frame, the child is at rest. The boat can go directly to the child at angle θ′ = tan–1 (200/1500) = 7.595°. The boat’s speed is 8.0 m/s, so the components of the boat’s velocity in S′ are vx′ = –(8.0 m/s)cos7.595° = –7.93 m/s
vy′ = (8.0 m/s)sin7.595° = 1.06 m/s →
The river flows with velocity V = 2.0iˆ m/s relative to the earth. In the earth’s frame, which is also the frame of the riverbank and the boat dock, the boat’s velocity is vx = vx′ + Vx = –5.93 m/s and vy = vy′ + Vy = 1.06 m/s Thus the boat’s angle with respect to the riverbank is θ = tan–1(5.93/1.06) = 10.1°. Assess: The boat, like the child, is being swept downstream. This moves the boat’s angle away from the shore.
6.58. Model: Let Alberto’s frame be S and Fred’s frame be S′. Frame S′ is moving relative to S with velocity V. Assume that “upfield” is in the x-direction. Visualize:
r r r r r The Galilean transformation of velocity is v = v ′ + V , where V = −( 4.0 m/s) jˆ and v ′ = (20 m/s)cosθiˆ + (20 m/s)sinθjˆ . Thus, r v = (20 m/s) cosθiˆ + (20 m/s) sin θ − ( 4.0 m/s) jˆ
Solve:
[
]
According to Alberto, Fred released the ball as he was directly upfield from Alberto. He observes the ball thrown toward him along the x-direction. That is vy = 0 m/s. This means 1 (20 m /s) sin θ − (4.0 m /s) = 0 m/s ⇒ θ = sin −1 = 11.5° 5 Alberto must throw the ball 11.5° toward the side opposite the one he is running toward.
6.59. Model: Let the ground be frame S and the pickup be frame S′. S′ moves relative to S with velocity V along the y-direction. x is the direction perpendicular to the pickup’s motion. Visualize:
r r r r r (a) The Galilean transformation of velocity is v = v ′ + V , where V = (10 mph ) jˆ and v ′ = (20 mph ) cosθiˆ − (20 mph) sin θjˆ . Thus,
Solve:
r v = (20 mph ) cosθiˆ + (10 mph ) − (20 mph ) sin θ jˆ
[
]
Since the paper is thrown just as the pickup passes the driveway, the observer in the driveway will find the paper to have no y-component of the velocity. That is, 1 vy = 0 mph = (10 mph ) − (20 mph ) sin θ ⇒ θ = sin −1 = 30° 2 He must throw the paper 30° to the side of the driveway opposite his motion. r (b) Since v = (20 mph ) cos 30°iˆ + (0 mph ) jˆ = (17.3 mph )iˆ, the speed is 17.3 mph.
6.60. Model: Let the ground frame be S and the car frame be S′. S′ moves relative to S with a velocity V along the x-direction. r r r r r Solve: The Galilean transformation of r velocity is v = v ′ + V where v and v ′ are the velocities of the raindrops in frames S and S′. While driving north, V = (25 m/s)iˆ and v = − vR cosθ jˆ − vR sin θ iˆ . Thus, r r r v ′ = v − V = ( − vR sin θ − 25 m/s)iˆ − vR cosθjˆ Since the observer in the car finds the raindrops making an angle of 38° with the vertical, we have vR sin θ + 25 m /s = tan 38° vR cosθ r r While driving south, V = −(25 m/s)iˆ , and v = − vR cosθ jˆ − vR sin θ iˆ . Thus, r v ′ = ( − vR sin θ + 25 m/s)iˆ − vR cosθ jˆ Since the observer in the car finds the raindrops falling vertically straight, we have
− vR sin θ + 25 m /s = tan 0° = 0 ⇒ vR sin θ = 25 m /s vR cosθ Substituting this value of vR sinθ into the expression obtained for driving north yields:
25 m /s + 25 m /s 50 m /s = tan 38° ⇒ vR cosθ = = 64.0 m /s vR cosθ tan 38° Therefore, we have for the velocity of the raindrops:
(vR sin θ )2 + (vR cosθ )2 = (25 m/s)2 + (64.0 m/s)2 ⇒ vR2 = 4721(m/s)2 ⇒ vR = 68.7 m /s tan θ =
vR sin θ 25 m /s = ⇒ θ = 21.3° v R cosθ 64 m /s
The raindrops fall at 68.7 m/s while making an angle of 21.3° with the vertical.
6.61. Model: Let the ground be frame S and the wind be frame S′. S′ moves relative to S. The ground frame S has the x-axis along the direction of east and the y-axis along the direction of north. Visualize:
Solve:
r r r (a) The Galilean velocity transformation is v = v ′ + V , where r V = (50 mph ) cos 30°iˆ + (50 mph ) sin 30° jˆ r v ′ = (200 mph ) cosθiˆ − (200 mph ) sin θjˆ
[
]
r r Thus, v = (50 cos 30° + 200 cosθ )iˆ + (50 sin 30° − 200 sin θ ) jˆ mph . Because v should have no j-component, 50 sin 30° − 200 sin θ = 0 ⇒ θ = 7.18°
r r (b) The pilot must head 7.18° south of east. Substituting this value of θ in the v equation gives v = (242)iˆ mph , 600 mi along the direction of east. At a speed of 242 mph, the trip takes t = = 2.48 hours. 242 mph
6.62. Model: Let Susan’s frame be S and Shawn’s frame be S′. S′ moves relative to S with velocity V. Both Susan and Shawn are observing the intersection point from their frames. r r r r Solve: The Galilean transformation of velocity is v = v ′ + V , where v is the velocity of the intersection r point r from Susan’s reference frame, v ′ is the velocity of the intersection point from Shawn’s frame S′, and V is the r r velocity of S′ relative to S or Shawn’s velocity relative to Susan. Because v = −(60 mph ) jˆ and v ′ = −( 45 mph )iˆ , r r we have V = v − v ′ = ( 45 mph )iˆ − (60 mph ) jˆ . This means that Shawn’s speed relative to Susan is
V=
(45 mph)2 + (−60 mph)2
= 75 mph
6.63. Solve: All light travels through a vacuum at 3 Ă&#x2014; 108 m/s in any inertial frame of reference. This seems impossible, but the relativity chapter will discuss why this is true.
6.64. Model: Assume the particle model and apply the constant-acceleration kinematic equations. Visualize:
Solve:
(a) Newton’s second law for the projectile is r −F Fnet = − Fwind = ma x ⇒ a x = x m where Fwind is shortened to F. For the y-motion:
( )
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ 0 m = 0 m + (v0 sin θ )t1 − 12 gt12 ⇒ t1 = 0 s and t1 = 2
2v0 sin θ g
Using the above expression for t1 and defining the range as R we get from the x motion:
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 )
2
2v sin θ F 2v0 sin θ F − ⇒ x1 − x 0 = R = v0 x t1 + 12 − t12 = (v0 cosθ ) 0 m g 2m g =
2
2v02 2v 2 F cosθ sin θ − 0 2 sin 2 θ g mg
We will now maximize R as a function of θ by setting the derivative equal to 0:
2 Fv02 dR 2v02 = cos 2 θ − sin 2 θ ) − 2 sin θ cosθ = 0 ( dθ g mg 2 2 Fv02 g mg ⇒ cos 2 θ − sin 2 θ = cos 2θ = sin 2θ ⇒ tan 2θ = 2 2 F mg 2v0 Thus the angle for maximum range is θ = 12 tan −1 ( mg / F ) (b) We have 2 mg (0.50 kg)(9.8 m/s ) = 8.167 ⇒ θ = 12 tan −1 (8.167) = 41.51° = 0.60 N F
The maximum range without air resistance is
R′ =
2v02 sin 45° cos 45° v02 = g g
Therefore, we can write the equation for the range R as 2F R = 2 R ′ sin 41.51° cos 41.51° − R ′ sin 2 41.51° = R ′(0.9926 − 0.1076) = 0.885 R ′ mg
⇒
R′ − R R = 1 − 0.8850 = 0.115 = 0.8850 ⇒ R′ R′
Thus R is reduced from R′ by 11.5%. Assess: The condition for maximum range ( tan 2θ = mg / F ) means 2θ → 90° as F → 0. That is, θ = 45° when F = 0, as is to be expected.
6.65. Model: The airplane, car, and the package are modeled as particles. The motion of the package is governed by kinematics. Ignore any air resistance effects. Visualize:
Since the plane and the car are in motion with constant velocities, we can solve this problem from the car’s reference frame S′. The earth is frame S. Solve: The velocity V of S′ relative to S is 30 m/s. The velocity v of the airplane relative to S is 50 m/s. The velocity v′ of airplane relative to S′ is given by v′ = v – V = 50 m/s – 30 m/s = 20 m/s From the car’s reference frame, the package falls 60 m. Thus
y1′ = y0′ + v0′ y (t1 − t 0 ) + 12 a y′ (t1 − t 0 ) ⇒ 0 m = 60 m + 0 m + 2
1 2
(−9.8 m/s )t 2
2 1
⇒ t1 = 3.5 s
We will find the horizontal distance the package will travel in 3.5 s as follows:
x1′ = x 0′ + v0′ x (t1 − t 0 ) + 12 a x′ (t1 − t 0 ) = 0 m + (20 m /s)(3.5 s − 0 s) + 0 m = 70.0 m 2
The angle below the horizontal at which the package is dropped is thus
tan θ =
60 m 60 ⇒ θ = tan −1 = 40.6° 70 70 m
Assess: In view of the other numbers for the velocities and the altitude, the obtained values of 70.0 m and 40.6° are reasonable. Also note that the Galilean transformation of acceleration is a y′ = a y .
6.66. Model: Use the particle model for the cat and apply the constant-acceleration kinematic equations. Visualize:
Solve:
The relative velocity of the cat from the mouse’s reference frame is (4.0 cos 30° − 1.5) m /s = 1.964 m /s = v0
Thus, v0 x = v0 = 1.964 m /s and v0 y = v0 sin 30° = ( 4.0 m /s) sin 30° = 2.0 m /s . The time for the cat to land on the floor is found as follows:
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ 0 m = 0 m + (2.0 m/s)t1 + 2
1 2
(−9.8 m/s )t 2
⇒ t1 = 0 s (trivial solution) and 0.408 s The horizontal distance covered in time t1 is
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) = 0 m + (1.964 m/s)(0.408 s) = 0.802 m 2
That is, the cat should leap when he is 0.802 m behind the mouse.
2 1
6.67. Model: Treat the ball as a particle and apply the constant-acceleration equations of kinematics. Visualize:
Solve: After the first bounce, the ball leaves the surface at 40° relative to the vertical or 50° relative to the horizontal. We first calculate the time t1 between the second bounce and the first bounce as follows:
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) ⇒ 3.0 m = 0 m + (v0 cos 50°)t1 + 0 m ⇒ t1 = 2
3.0 m v0 cos 50°
In this time, the ball undergoes a vertical displacement of y1 – y0 = –(3.0 m)tan 20° = –1.092 m. Substituting these values in the equation for the vertical displacement yields:
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 )
2
3.0 m −1.092 m = 0 m + (v0 sin 50°)t1 − 12 gt12 = (v0 sin 50°) − v0 cos 50°
1 2
−106.73 m 3 / s 2 ⇒ v0 = 4.78 m/s ⇒ −1.092 m − 3.575 m = v02 Assess: A speed of 4.78 m/s or 10.7 mph on the first bounce is reasonable.
m (9.8 m /s ) v 3.0 cos 50° 2
0
2
6.68. Model: Use the particle model for the motorcycle daredevil and apply the kinematic equations of motion. Visualize:
Solve:
We need to find the coordinates of the landing ramp (x1, y1). We have
x1 = x 0 + v0 x (t1 − t 0 ) = 0 m + ( 40 m/s)t1 y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) = 0 m + 0 m + 12 ( −9.8 m/s 2 )t12 = −( 4.9 m/s 2 )t12 r This means we must find t1. Since v1 makes an angle of 20° below the horizontal we can find v1y as follows: 2
v1y v1 x
= tan 20° ⇒ v1y = − v1 x tan 20° = −( 40 m /s) tan 20° = −14.56 m /s
We now use this value of v1y, a y = −9.8 m /s 2 , and v0y = 0 m/s in the following equation to obtain t1:
v1y = v0 y + a y (t1 − t 0 ) ⇒ −14.56 m/s = 0 m/s − (9.8 m/s 2 )t1 ⇒ t1 = 1.486 s Now, we are able to obtain x1 and y1 using the above x- and y-equations:
x1 = ( 40 m /s)(1.486 s) = 59.4 m
y1 = −( 4.9 m/s 2 )(1.486 s) = −10.82 m 2
That is, the landing ramp should be placed 10.82 m lower and 59.4 m away from the edge of the horizontal platform.
6.69. Model: Use the particle model and apply the constant-acceleration kinematic equations of motion. Visualize:
Solve: (a) The thrust of the rocket burn must be such that vx decreases from 2.0 km/s to 0 during the same interval that vy increases from 0 to 1.0 km/s. The forward distance in which to accomplish this is x1 = (500 km)cos30° = 433 km. We can find the burn time by combining two kinematic equations:
v1 x = 0 = v0 x + a x t1 ⇒ a x t1 = − v0 x x1 = 0 + v0 x t1 + 12 a x t12 = v0 x t1 + 12 ( − v0 x )t1 = 12 v0 x t1 ⇒ t1 =
2 x1 2( 433 km) = = 433 s v0 x 2.0 km /s
The required acceleration is ax = –v0y/t1 = –(2000 m/s)/(433 s) = –4.62 m/s. During this interval, vy increases from 0 to 1.0 km/s, thus
v1y = 0 + a y t1 ⇒ a y =
1000 m /s = 2.31 m /s 2 433 s
The thrust force is given by Newton’s second law: v r Fthrust = ma = (20, 000 kg)(–4.62iˆ + 2.31 jˆ m/s 2 ) = ( −92, 400iˆ + 46, 200 jˆ ) N The magnitude of the thrust force is 103,300 N and it is at angle θ = tan–1(46,200/92,400) = 26.6° below the +x-axis (i.e., to the right and down). In order to point the thrusters in this direction, you must rotate the rocket counterclockwise 180° – θ = 153.4°. To make it home with one rocket burn, you need to rotate your rocket to 153.4° and fire with a thrust of 103,300 N for 433 s. (b) The x- and y-positions during the burn are
x = (2000 m /s) t − 12 ( 4.62 m /s 2 ) t 2 y = 12 (2.31 m /s 2 ) t 2 The y-position at t1 = 433 s is 216,500 m = 216.5 km, 33.5 km away from the entrance. With the rocket off, you’re now coasting straight toward the entrance at 1.0 km/s and will cover this distance in 33.5 s. Thus you pass through the entrance at 433 s + 33.5 s = 466.5 s. The following table calculates your position at intervals of 50 s and, for accuracy, at t1 = 433 s when you end the burn. The trajectory is a parabola that intersects the x = 433 km line at y = 216.5 km, then a vertical line into the entrance.
6.70. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S. r Frame S′ moves relative to frame S with velocity V . Visualize:
r r r r Solve: According to the Galilean transformation of velocity v = v ′ + V , where v r is the velocity of Val’s plane r relative to the Earth, v ′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth. We have r v = −(500 mph ) cos 30°iˆ + (500 mph ) sin 30° jˆ r V = −(500 mph ) cos 20°iˆ − (500 mph ) sin 20° jˆ r r r v ′ = v − V = (36.8 mph )iˆ + ( 421 mph ) jˆ
421 θ = tan −1 = 85° 36.8 The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east. Thus the angle between the fuselage and the direction of motion is
φ = 180° − 30° − 85° = 65°