Question

A racquet ball with mass m = 0.229 kg is moving toward the wall at v = 16.9 m/s and at an angle of θ = 32° 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 = 0.065 s.What is the magnitude of the initial momentum of the racquet ball?What is the magnitude of the change in momentum of the racquet ball?What is the magnitude of the average force the wall exerts on the racquet ball?Now the racquet ball is moving straight toward the wall at a velocity of vi = 16.9 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.5 m/s. The wall exerts the same average force on the ball as before.What is the magnitude of the change in momentum of the racquet ball?What is the time the ball is in contact with the wall?What is the change in kinetic energy of the racquet ball?

Answer 1

1) p = 3.8701 kg* m/s

2) Δp = 6.564 kg*m/s

3) F = 100.98 N

4) Δp = 6.27 kg*m/s

5) Δt = 0.062 s

6) ΔKE = -20.08 J

Step-by-step explanation:

Step 1: Data given

Mass of the ball = 0.229 kg

The ball is moving at v =16.9 m/s

Angle = θ = 32°

The ball is in contact with the wall for t = 0.065 s.

Question 1. What is the magnitude of the initial momentum of the racquet ball

p=M*V

p = 0.229 Kg * 16.9 m/s

p = 3.8701 kg* m/s

2. What is the magnitude of the change in momentum of the racquetball?

Momentum is a vector, so only the horizontal portion of velocity changes direction. The velocity parallel to the wall does not change.

Let's consider 'away from the wall' to be the positive direction.

Change in momentum Δp will be:

Δp = m(vf - vi)

⇒ with m = the mass of the ball

⇒ with vf = the final speed

⇒ with vi = the initial speed

Δp = m(vcosθ - (-vcosθ))

Δp = m(vcosθ +vcosθ)

Δp = 2mvcosθ

⇒ with θ = 32°

Δp = 2*(0.229)*(16.9*cos32)

Δp = 6.564 kg*m/s

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

Impulse = Δp

An impulse equals a change in momentum

m*Δv = F*Δt

OR

F = mΔv / Δt

⇒ with mΔv = Δp = 6.564 kg*m/s

⇒ with Δt = the time of the contact ball- wall = 0.065s

F = 6.564 / 0.065

F = 100.98 N

4. What is the magnitude of the change in momentum of the racquetball?

Δp = m*(Δv)

Δp = 0.229(10.5 - (-16.9))

Δp = 6.27 kg*m/s

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

F = Δp/ Δt

Δp =F* Δt

6.27 kg*m/s = 100.98*Δt

Δt = 0.062 s

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

ΔKE = ½m*(vf)² - ½m*(vi)²

ΔKE = ½m(vf² - vi²)

ΔKE = ½(0.229)(110.25 - 285.61)

ΔKE = -20.08 kg*m²/s² = -20.08 J