Question

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 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. The ball is in contact with the wall for t = 0.066 s.1)What is the magnitude of the initial momentum of the racquet ball?kg-m/s2) What is the magnitude of the change in momentum of the racquet ball?kg-m/s3) What is the magnitude of the average force the wall exerts on the racquet ball?N4) ballhitswall2 Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.8 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.8 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/s5) What is the time the ball is in contact with the wall?s6). What is the change in kinetic energy of the racquet ball?J

Answer 1

Part a)

`P = 5.72 kg m/s`

Part b)

`\Delta P = 2.93 kg m/s`

Part c)

`F = 44.4 N`

Part d)

`\Delta P = 5.02 kg m/s`

Part e)

`\Delta t = 0.113 s`

Part f)

`\Delta K = 0`

Step-by-step explanation:

As we know that initial velocity of the ball is given as

`v = 11.8 cos29 \hat i + 11.8 sin29 \hat j`

`v_i = 10.3 \hat i + 5.72 \hat j`

Now final velocity of the system is given as

`v_f = 10.3\hat i - 5.72\hat j`

Part a)

now magnitude of initial momentum is given as

`P = mv`

`P = 0.256(11.8)`

`P = 5.72 kg m/s`

Part b)

Change in momentum is given as

`\Delta P = m(v_f - v_i)`

`\Delta P = 0.256(5.72 + 5.72)`

`\Delta P = 2.93 kg m/s`

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

`F = (\Delta P)/(\Delta t)`

`F = (2.93)/(0.066)`

`F = 44.4 N`

Part d)

Magnitude of change in momentum is given as

`\Delta P = m(v_f - v_i)`

`\Delta P = 0.256(7.8 + 11.8)`

`\Delta P = 5.02 kg m/s`

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

`(\Delta P)/(\Delta t) = F`

`(5.02)/(\Delta t) = 44.4`

`\Delta t = 0.113 s`

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

`\Delta K = 0`