1-What is the magnitude of the initial momentum of the racquet ball? 2- What is

Question

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

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

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

4- Now the racquet ball is moving straight toward the wall at a velocity of vi = 18.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -12.4 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?

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

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

Ball Hits Wall 1 234567 A racquet ball with mass m 0.233 kg is moving toward the wall at v 18.5 m/s and at an angle of e 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 0.071 s.

Explanation / Answer

Let the mass of the raquet ball m = 0.233 kg

Let the initial velocity of the ball be v = 18.5 m/s

Time for which ball is in contact with wall be t = 0.071 sec

Angle made by the ball with horizontal be = 260

1)Initial momentum of the ball

Let the initial momentum of the ball be P

Then P = m * v = 0.233 * 18.5 = 4.31 kg m/s

2)Change in momentum of the ball

Change in the momentum is given by
P = m(vf – vi)
P = m * [ vcos - ( - vcos ) ]
P = 2 * m* vcos
P = 2 * (0.233 ) * ( 18.5 * cos26 )
P = 2* 0.233 * 18.5 * 0.8987

P = 7.74 kg•m/s

3)Magnitude of the average force the wall exerts on the racquet ball
An impulse equals the change in momentum             

   m * v = F * t

F = m v / t
F = P / t
F = 7.74 / 0.071
F = 109 N

4)Magnitude of the change in momentum of the racquet ball

Final velocity is given by Vf = -12.4 m/s
P = m * ( vf – vi )

     = m * ( Vf – v )

     = 0.233 * ( - 12.4 – 18.5 )

     = 0.233 * ( - 30.9 )

     = - 7.1997 kg m/s
5)Time the ball is in contact with the wall
Since force is the same we can write ,

          F = P / t

         t = P / F

             = 7.1997 / 109

             = 0.06605 sec
so the ball is in contact with wall for t = 0.06605 sec
6)Change in kinetic energy of the racquet ball
Let the change in the kinetic energy be K

The change in the kinetic energy of the ball is given by
K = ½m * vf² - ½m * vi²
K = ½m(Vf² - v²)

K = ½ * 0.233 * ( 12.42 – 18.52 )
K = ½ * 0.233 * ( 153.76 - 342.25 )

K = - 21.95 J

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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