The Impulse and Average Force of a Bouncing Handball

[maniacp08]

A 60g handball moving with a speed of 5m/s strikes the wall at a 40 degree angle with the norm, and then it bounces off with the same speed at the same angle with the norm. It is in contact with the for 2.0ms.

a)What is the average force exerted by the ball on the wall?
b)The rebounding ball is caught by a player who brings it to rest. In the process her hand moves back .50m. what is the impulse received by the player?
c)what average force was exerted on the player by the ball?

Impulse = change in momentum = m * change in velocity
Avg force = Impulse / collision time

Isn't velocity the same as speed but with a direction?
So can vi = 5m/s and vf = -5m/s?

The problem gave me an angle so I know I have to do something with it but I am not sure what.
I try doing 60g * -10m/s as the impulse then divide it by 2ms for average force but it is wrong.
Can someone help me.

 

[LowlyPion]

A 60g handball moving with a speed of 5m/s strikes the wall at a 40 degree angle with the norm, and then it bounces off with the same speed at the same angle with the norm. It is in contact with the for 2.0ms.

a)What is the average force exerted by the ball on the wall?
b)The rebounding ball is caught by a player who brings it to rest. In the process her hand moves back .50m. what is the impulse received by the player?
c)what average force was exerted on the player by the ball?

Impulse = change in momentum = m * change in velocity
Avg force = Impulse / collision time

Isn't velocity the same as speed but with a direction?
So can vi = 5m/s and vf = -5m/s?

The problem gave me an angle so I know I have to do something with it but I am not sure what.
I try doing 60g * -10m/s as the impulse then divide it by 2ms for average force but it is wrong.
Can someone help me.

You mostly have it.

Velocity is a vector. So resolve the velocity into the x,y components. The vertical component before/after remains in the same direction. It's apparently only the horizontal component that is reversed.

 

[maniacp08]

Ahh. Thanks it is -5cos40-5cos40 * 60g then divide by 2ms.

b)The rebounding ball is caught by a player who brings it to rest. In the process her hand moves back .50m. what is the impulse received by the player?
So Would the vi of the ball be -5cos40 and vf = 0 since it is caught and goes to rest?
so is -5cos40 * 60g?

c)what average force was exerted on the player by the ball?
Is this just answer to B/2ms?

 

[LowlyPion]

Ahh. Thanks it is -5cos40-5cos40 * 60g then divide by 2ms.

b)The rebounding ball is caught by a player who brings it to rest. In the process her hand moves back .50m. what is the impulse received by the player?
So Would the vi of the ball be -5cos40 and vf = 0 since it is caught and goes to rest?
so is -5cos40 * 60g?

c)what average force was exerted on the player by the ball?
Is this just answer to B/2ms?

The ball was caught - both components of it - presumably at the same height.

So that's V² = 2*a*(.5m)
Use the "a" to yield the time of the impulse.

 

[maniacp08]

Im kind of confuse.
Wouldn't the rebounding ball vi be -5cos40?
bringing it to rest wouldn't that be vf = 0?

If I am wrong then
What would V^2 be?
The sum of the 2 components of the speed?

a yields the time of the impulse is that the collision time?

 

[LowlyPion]

Im kind of confuse.
Wouldn't the rebounding ball vi be -5cos40?
bringing it to rest wouldn't that be vf = 0?

If I am wrong then
What would V^2 be?
The sum of the 2 components of the speed?

a yields the time of the impulse is that the collision time?

The vertical component is not affected.

and then it bounces off with the same speed at the same angle with the norm.

So when the ball arrives at the glove it is still traveling at 5 m/s - if it is caught at the same height it was thrown.