A racquet ball with mass m = 0.245 kg is moving toward the wall at v = 15.4 m/s

Question

A racquet ball with mass m = 0.245 kg is moving toward the wall at v = 15.4 m/s and at an angle of = 28° 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 = 0.069 s.

Now the racquet ball is moving straight toward the wall at a velocity of vi = 15.4 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -9.7 m/s. The wall exerts the same average force on the ball as before.

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

2.What is the time the ball is in contact with the wall?

3.What is the change in kinetic energy of the racquet ball?

Explanation / Answer

here,

Velocity of ball , vi = 15.4 m/s
Velocity of ball after coliision, vf = -9.7 m/s

Part A:
Momentum before collision, Pi
Pi = m*Vi
Pi = 0.245 * 15.4
Pi = 3.773 Kg.m/s

Momentum After collision, Pf
Pi = m*Vf
Pi = -0.245 * 9.7
Pi = -2.377 Kg.m/s

Change in Momentum,
P = Pf - Pi
P = -2.377 - 3.773
P = - 6.15 Kg.m/s

Part B:

Average Force = Impulse/time of contact
Favg = 0.245(15.4) / 0.069
Favg = 54.681 N

Therefore, Time of Constact when ball travel straight to wall

t = P/Favg
t = 6.15 / 54.681
t = 0.112 s

Part C:
KE = 0.5 * m * ( Vf - Vi )^2
KE = 0.5 * 0.245*( 9.7 - 15 )^2
KE = 3.441 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects