# Analyzing Racquet Ball Collision: Momentum, Time & Kinetic Energy

• ctwokay
In summary: It is t = 0.07 s. Your calculation using ft=Δp is also correct, but it is giving you the average force, not the time. So, the time is already given and cannot be changed.
ctwokay

## Homework Statement

A racquet ball with mass m = 0.223 kg is moving toward the wall at v = 16.3 m/s and at an angle of θ = 35° 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 = 0.07 s.

Now the racquet ball is moving straight toward the wall at a velocity of vi = 16.3 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.6 m/s. The ball 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?

Δp=m(vf-vi)
ft=Δp
KE=1/2mv^2

## The Attempt at a Solution

The first question I get 6.00kg-m/s by using m(vf-vi) which is 0.223*(-10.6-16.3). Is that correct?

Second question I use ft=Δp which gives me t=Δp/ave force; which i have the values are 6.00/85.1=0.0704s but it seems to me that the time should be lesser cause when travel without any angles it cuts down the time. Is the caluation correct?

The third question is ΔKE=1/2m(vf^2-vi^2) which i get 1/2*0.223*(10.6^2-16.3^2)
which i get -17.1J.

All 3 question just want to check am I doing the correct way,thanks for helping out.

I didn't check the numbers but your reasoning on every question is correct so assuming you made no calculation errors your answers should be correct.

In regards to your second question, the time the ball was in contact with the wall is given in the problem statement.

## 1. What is the purpose of analyzing racquet ball collision?

The purpose of analyzing racquet ball collision is to understand the principles of momentum, time, and kinetic energy and how they apply to the game of racquetball. By studying these concepts, we can better understand the physics behind racquetball and potentially improve our performance in the game.

## 2. How is momentum involved in racquet ball collision?

Momentum is involved in racquet ball collision as it is a measure of the quantity of motion of an object. When two racquet balls collide, their momenta are transferred to each other, resulting in a change in their velocities. This change in velocity is what causes the balls to move and bounce off each other.

## 3. Why is time an important factor in racquet ball collision?

Time is an important factor in racquet ball collision because it determines the duration of the collision and the amount of force that is exerted on the balls. A shorter collision time means a greater force is applied, resulting in a more powerful bounce. This can affect the trajectory and speed of the balls after the collision.

## 4. How does kinetic energy play a role in racquet ball collision?

Kinetic energy is the energy an object possesses due to its motion. In racquet ball collision, the kinetic energy of the balls is transferred between them during the collision. This transfer of energy affects the speed and direction of the balls after the collision, determining the outcome of the game.

## 5. What are some applications of studying racquet ball collision?

Studying racquet ball collision has various applications, including improving gameplay by understanding the physics behind it, designing better racquetball equipment, and developing safety measures to prevent potential injuries during gameplay. It also has broader applications in other sports and industries that involve collisions, such as car crashes and ballistics.

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