A racquet ball with mass m = 0.252 kg is moving toward the wall at v = 18.3 m/s

Question

A racquet ball with mass m = 0.252 kg is moving toward the wall at v = 18.3 m/s and at an angle of = 26° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.06 s.

1) What is the magnitude of the initial momentum of the racquet ball?

kg-m/s

2) What is the magnitude of the change in momentum of the racquet ball?

kg-m/s

3) What is the magnitude of the average force the wall exerts on the racquet ball?

N

4)

Now the racquet ball is moving straight toward the wall at a velocity of vi = 18.3 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -11.3 m/s. The ball exerts the same average force on the ball as before.

What is the magnitude of the change in momentum of the racquet ball?

kg-m/s

5) What is the time the ball is in contact with the wall?

s

6) What is the change in kinetic energy of the racquet ball?

J

Explanation / Answer

1) Pi = m*vi

= 0.252*18.3

= 4.61 kg.m/s

2) change in momentum of the rocket ball = m*(vix - vfx)

= 0.252*(18.3 - (-18.3))*cos(26)

= 8.29 kg.m/s

3) use Impulse momentum theorem,

Impulse = change in momentum

F*dt = change in momentum

F = change in momentum/dt

= 8.29/0.06

= 138 N

4) magnitude of change in momentum = m*(vi - vf)

= 0.252*(18.3 - (-11.3))

= 7.46 kg.m/s

5) use, impulse = change in momentum

F*dt = change in momentum

dt = change in momentum/F

= 7.46/138

= 0.054 s

6) change in kinetic energy of the ball = (1/2)*m*(vf^2 - vi^2)

= (1/2)*0.252*((-11.3)^2 - 18.3^2)

= -26.1 J

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