# What average force was exerted on the player by the ball ?

1. Dec 24, 2008

### nns91

1. The problem statement, all variables and given/known data

1. A 300g handball moving with a speed of 5 m/s strikes the wall at an angle of 40 degree and then bounces off with the samw speed at the same angle. It is in contact with the wall for 2ms. What is the average force exerted by the ball on the wall ??

2. A handball of mass 300g is thrown straight against a wall with a speed of 8 m/s. It rebounds with the same speed. (a) What impulse is delivered to the wall ? (b) If the ball is in contact with the wall of 0.003 s, what average force is exerted on the wall by the ball ? (c) The ball is caught by a player who brings it to rest. In the process, her hand moves back 0.5m. What is the impulse received by the player ? (d) What average force was exerted on the player by the ball ??

2. Relevant equations

p=mv, I=$$\Delta$$p

3. The attempt at a solution

I don't know if the bouncing will affect anything ???

1. I am clueless since the ball hit at 40 degree angle.

2. For (a) I did I=$$\Delta$$p and get I=0.3 * 8=2.4 N. Am I right ??

2. Dec 24, 2008

### jgens

Re: Impulse

Clarifying question on number 1: Assuming the ball hits the wall at 40 degrees below the horizontal, does it rebound at 40 degrees above or below the horizontal?

To get you started on number one, since momentum is a vector quantity, resolve the ball's intial and final momentum into its respective x and y components. From there determine the impulse.

Also, your answer to 2a looks right to me.

Last edited: Dec 24, 2008
3. Dec 24, 2008

### nns91

Re: Impulse

They don't say above or below the horizontal. What role does the angle play ? It affects velocity right ??

Thanks. How about part (c) and ( d) of 2 ??

4. Dec 24, 2008

### jgens

Re: Impulse

Well, if you resolve the momentum into its components you'll see it actually plays a fairly significant role in the total change in momentum.

Alright, for 2c we'll label the intial momentum p_i and the final momentum p_f. Using the data given earlier in the problem it's easy to determine the initial momentum. Now, since she brings the ball to rest what does that suggest about p_f and what does that suggest about the impulse exerted on the player?

For 2d, think Newton's second law.

5. Dec 24, 2008

### nns91

Re: Impulse

For 1, P_i = m*v_x + m*v_y= m ( v*sin40 + v*cos40). Is that right ?? How about P_f ??

so p_f = 0 so I= -2.4 N ??

6. Dec 24, 2008

### jgens

Re: Impulse

Should be right. Curious though, which direction did you take the ball's momentum to be? (positive or negative)

7. Dec 24, 2008

### nns91

Re: Impulse

positive ??

So what can be the final ??

In problem 2, do you think the acceleration is constant so that I can use vf^2= vi^2 +2ax ??

8. Dec 24, 2008

### jgens

Re: Impulse

Well, if you chose the direction of the balls momentum to be positive the answer for 2c isn't quite right. While it's true that the impulse exerted on the ball was -2.4 Ns, what was the impulse exerted on the player?

You don't know whether the acceleration is constant or not; however, in this instance the average acceleration will suffice.

9. Dec 24, 2008

### nns91

Re: Impulse

2.4 N ???

so I will use vf^2 = vi^2 +2ax to find a then use F=ma ?

10. Dec 24, 2008

### jgens

Re: Impulse

Yes, if you choose the ball's momentum to be the positive direction than the player must have recieved 2.4 Ns of impulse. Does that make sense?

That should method should yield the average force exerted. Good job!

As a side note, be careful with units. You tend to leave the s off on Ns.

11. Dec 24, 2008

### nns91

Re: Impulse

It does since we talk about different objects. Thanks

which one is Ns ?

12. Dec 24, 2008

### nns91

Re: Impulse

I still have not figured number 1 yet.

13. Dec 24, 2008

### jgens

Re: Impulse

Units of momentum are newton seconds, hence Ns.

For number one, assume that if the ball collides with the wall at 40 degrees below the horizontal that it rebounds at 40 degrees below the horizontal. Resolve the components of the initial and final momentums and go from there.

14. Jan 4, 2009

### nns91

Re: Impulse

So Initially: P= mvcos40 + mvsin40.

Will it be the same in Final momentum ?? since the angle is 40 degree

15. Jan 4, 2009

### rl.bhat

Re: Impulse

If the ball collides with the wall at 40 degrees below the horizontal, it component along +x axis is mvcos40 and along +y axis is mvsin40. When it rebounds it will be moving along 40 degree above the horizontal. Then mvcos40 will be along -x axis and mvsin40 will be along =y axis. Find the change in momentum. that will be impulse.

Last edited: Jan 5, 2009
16. Jan 5, 2009

### nns91

Re: Impulse

Thus, change in y component is 0 and in x component is -2mvcos40 ??

17. Jan 5, 2009

### rl.bhat

Re: Impulse

That is right.