Momentum of ball bouncing off wall

[aszymans]

Homework Statement

A racquet ball with mass m = 0.247 kg is moving toward the wall at v = 12.4 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 = 0.063 s.

Homework Equations

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

The Attempt at a Solution

I found the initial P already and am not sure if I should use the (delta)P= F(net)* (delta)t and use 1-d kinematics to find acceleration or how exactly to begin.

 

[grzz]

Since the collision is assumed to be perfectly elastic, try using conservation of KE.

 

[grzz]

Then find the change in momentum.

 

[HallsofIvy]

Momentum is a vector quantity. The component of momentum parallel to the way will not change. The component of momentum perpendicular to the way is multiplied by -1.

 

[rwisz]

I would approach this problem by first defining the physical principles at play. In this case, we are dealing with the conservation of momentum and the concept of elastic collisions.

The initial momentum of the ball can be calculated using the formula P = mv, where m is the mass of the ball and v is its velocity. In this case, the initial momentum would be P = (0.247 kg)(12.4 m/s) = 3.0628 kg*m/s.

Since the collision with the wall is perfectly elastic, we know that the total momentum of the system (ball + wall) before and after the collision should be the same. This means that the change in momentum of the ball (delta P) should be equal to the negative of the initial momentum (since it is rebounding in the opposite direction).

Therefore, the magnitude of the change in momentum would be delta P = |-3.0628 kg*m/s| = 3.0628 kg*m/s.

To find the acceleration of the ball during its contact with the wall, we can use the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time of contact. In this case, we know that vf = -vi (since it rebounds with the same speed in the opposite direction), so we can simplify the equation to a = (-2vi)/t.

Using the given values, we can calculate the acceleration to be a = (-2)(12.4 m/s)/0.063 s = -393.65 m/s^2. Note that the negative sign indicates that the acceleration is in the opposite direction of the initial velocity, as expected.

Overall, the magnitude of the change in momentum of the ball and its acceleration during the collision can be calculated using the principles of momentum and elastic collisions.