A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s

Question

A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s and at an angle of ? = 29 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. While the ball is in contact with the wall, the average force the wall exerts on the ball has a magnitude of F = 76 N.

3)

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

s

4)

Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.5 m/s. The ball is in contact with the wall for the same amount of time as in the elastic collision above.

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

kg-m/s

5)

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

N

6)

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

Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.5 m/s. The ball is in contact with the wall for the same amount of time as in the elastic collision above. What is the magnitude of the change in momentum of the racquet ball? kg-m/s 5) What is the magnitude of the average force the wall exerts on the ball? N 6) What is the change in kinetic energy of the racquet ball? A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s and at an angle of ? = 29 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. While the ball is in contact with the wall, the average force the wall exerts on the ball has a magnitude of F = 76 N. 3) What is the time the ball is in contact with the wall? s 4)

Explanation / Answer

Change in momentum = m * (chhange in velocity) = 0.255*2*11.5*cos29 = 5.129kgm/s

The force exerted by wall on ball = change in momentum/time of contact

time of contact = 5.129/76 = 0.0674s

so in first case change in momo = 5.129kg-m/s over 0.0674s

In second case change in mom = 4.845 for same duration

Thus force = 71.8N

change in KE = 0.5m*change in square of velocity. = 9.69J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.