Physics 107 HOMEWORK ASSIGNMENT #7

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Physics 107 HOMEWORK ASSIGNMENT #7. Cutnell & Johnson, 7 th edition ... * 19 To view an interactive solution to a problem that is very similar to this one, go ...
Physics 107 HOMEWORK ASSIGNMENT #7 Cutnell & Johnson, 7th edition Chapter 7: Problems 10, 16, 19, 34, 39 *10 A student (m=63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.010 s. The average force exerted on him by the ground is + 18 000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. 16 A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with a velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water. *19 To view an interactive solution to a problem that is very similar to this one, go to and select Interactive Solution 7.19 A fireworks rocket is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2, as shown in the drawing. What is the magnitude of (a) v1, and (b) v2 ? *34 A mine car (mass= 440 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 150-kg chunk of coal has a speed of 0.80 m/s when it leaves the chute. Determine the velocity of the car/coal system after the coal has come to rest in the car. Problem 19 **39 Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50-kg ball swings downward and strikes a 4.60-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.50-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

Problem 34

Problem 39

10. REASONING This solution can be divided into two parts. First, the student falls freely from rest and attains a certain velocity (which is unknown) just before hitting the ground. This impact velocity depends on the height H from which the student falls, and we will use an equation of kinematics from Chapter 3 to relate the height to the impact velocity. Second, the student then collides with the ground and comes to rest. We will employ the impulsemomentum theorem to determine the impact velocity from a knowledge of the average impact force and the time of impact, both of which are known. SOLUTION Since the student falls freely from rest, we know that v0y = 0 m/s, and ay = 2 −9.8 m/s . The acceleration is negative, because the downward direction is taken to be the negative direction. The displacement y of the student is y = −H, where H is the height; the minus sign indicates that the student falls downward. These variables are related by Equation 3.6b from the equations of kinematics as 2

v 2y = v02y + 2a y y = ( 0 m/s ) + 2a y ( − H ) where vy is the velocity of the student just before hitting the ground. Solving this expression for the height yields −v 2y H= (1) 2a y

At this point we do not know the impact velocity vy. However, it can be determined by examining the collision of the student with the ground. Just before the collision, the velocity of the student is vy. Since the student comes to rest, the final velocity is vf = 0 m/s. The average force exerted on the student by the ground is F = +18 000 N, and the time of collision is ∆t = 0.010 s. The impulse-momentum theorem, Equation 7.4, relates these variables: F ∆t = mvf − mv0 = m ( 0 m/s ) − mv0 = − mv0 (2) Now v0 is the impact velocity of the student, which was labeled as vy in Equation (1); therefore, v0 = vy. Thus, Equation (2) becomes F ∆t = − mv y . Solving this equation for vy and substituting the result into Equation (1) yields 2

 − F ∆t  − 2 2 2  − F 2 ( ∆t ) − ( +18 000 N ) ( 0.010 s )  m  H= = = = 0.42 m 2 2a y 2a y m 2 2 −9.80 m/s 2 ( 63 kg )

(

)

16. REASONING The sum of the external forces acting on the swimmer/raft system is zero, because the weight of the swimmer and raft is balanced by a corresponding normal force and friction is negligible. The swimmer and raft constitute an isolated system, so the principle of conservation of linear momentum applies. We will use this principle to find the recoil velocity of the raft. SOLUTION As the swimmer runs off the raft, the total linear momentum of the swimmer/raft system is conserved:

ms vs + mr vr = 0  Total momentum

Total momentum after swimmer runs off raft

before swimmer starts running

where ms and vs are the mass and final velocity of the swimmer, and mr and vr are the mass and final velocity of the raft. Solving for vr gives vr = −

ms vs mr

=−

( 55 kg )( +4.6 m/s ) 210 kg

= −1.2 m/s

19. REASONING During the breakup¸ the linear momentum of the system is conserved, since the force causing the breakup is an internal force. We will assume that the +x axis is along the original line of motion (before the breakup), and the +y axis is perpendicular to this line and points upward. We will apply the conservation of linear momentum twice, once for the momentum components along the x axis and again for the momentum components along the y axis. SOLUTION The mass of each piece of the rocket after breakup is m, and so the mass of the rocket before breakup is 2m. Applying the conservation of momentum theorem along the original line of motion (the x axis) gives

mv1 cos 30.0° + mv2 cos 60.0° = 2mv0    

or

v1 cos 30.0° + v2 cos 60.0° = 2v0

(1)

P0, x

Pf ,x

Applying the conservation of momentum along the y axis gives

mv1 sin 30.0° − mv2 sin 60.0° =  0    Pf , y

P0, y

or

v2 =

v1 sin 30.0° sin 60.0°

(2)

a. To find the speed v1 of the first piece, we substitute the value for v2 from Equation (2) into Equation (1). The result is

 v sin 30.0°  v1 cos 30.0° +  1  cos 60.0° = 2v0  sin 60.0°  Solving for v1 and setting v0 = 45.0 m/s yields

v1 =

2v0  sin 30.0°  cos 30.0° +   cos 60.0°  sin 60.0° 

=

2 ( 45.0 m/s )  sin 30.0°  cos 30.0° +   cos 60.0°  sin 60.0° 

= 77.9 m/s

b. The speed v2 of the second piece can be found by substituting v1 = 77.9 m/s into Equation (2): v sin 30.0° ( 77.9 m/s ) sin 30.0° v2 = 1 = = 45.0 m/s sin 60.0° sin 60.0°

34. REASONING AND SOLUTION Momentum is conserved in the horizontal direction during the "collision." Let the coal be object 1 and the car be object 2. Then (m1 + m2 )vf = m1v1 cos 25.0° + m2v2 vf =

m1v1 cos 25.0° + m2v2 m1 + m2

=

(150 kg)(0.80 m/s) cos 25.0° + (440 kg)(0.50 m/s) = 0.56 m/s 150 kg + 440 kg

The direction of the final velocity is to the right .

39. SSM REASONING The two balls constitute the system. The tension in the wire is the only nonconservative force that acts on the ball. The tension does no work since it is perpendicular to the displacement of the ball. Since Wnc = 0 J, the principle of conservation of mechanical energy holds and can be used to find the speed of the 1.50-kg ball just before the collision. Momentum is conserved during the collision, so the principle of conservation of momentum can be used to find the velocities of both balls just after the collision. Once the collision has occurred, energy conservation can be used to determine how high each ball rises. SOLUTION a. Applying the principle of energy conservation to the 1.50-kg ball, we have 1 mv f2 + mgh f = 21 mv 02 + mgh0 2        Ef E0 If we measure the heights from the lowest point in the swing, hf = 0 m, and the expression above simplifies to

1 2

mv f2 = 21 mv 02 + mgh0

Solving for vf , we have v f = v 02 + 2 gh0 = ( 5.00 m / s) 2 + 2(9.80 m / s 2 )(0.300 m) = 5.56 m / s

b. If we assume that the collision is elastic, then the velocities of both balls just after the collision can be obtained from Equations 7.8a and 7.8b:

v f1 =

F m −m Iv GH m + m JK 1 1

2

01

and

v f2 =

2

F 2m I v GH m + m JK 1

1

01

2

Since v01 corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the quantity vf calculated in part (a). With the given values of m1 = 1.50 kg and m2 = 4 .60 kg , and the value of v 01 = 5.56 m / s obtained in part (a), Equations 7.8a and 7.8b yield the following values: v f1 = –2.83 m / s

and

v f2 = +2.73 m / s

The minus sign in vf1 indicates that the first ball reverses its direction as a result of the collision. c. If we apply the conservation of mechanical energy to either ball after the collision we have 1 mv f2 + mgh f = 21 mv 02 + mgh0 2        Ef E0 where v0 is the speed of the ball just after the collision, and hf is the final height to which the ball rises. For either ball, h0 = 0 m, and when either ball has reached its maximum height, vf = 0 m/s. Therefore, the expression of energy conservation reduces to gh f = 21 v 02

or

hf =

v 02 2g

Thus, the heights to which each ball rises after the collision are 1.50 - kg ball 4.60 - kg ball

v 02

(2.83 m / s) 2 hf = = = 0.409 m 2 g 2 (9.80 m / s 2 ) v 02 (2.73 m / s) 2 hf = = = 0.380 m 2 g 2 (9.80 m / s 2 )