# Homework Help: What is the change of momentum of the ball?

1. Dec 11, 2012

### underduck

1. A 0.095 kg tennis ball is travelling 40 m/s, hits a wall and travels in the opposite site direction it came from. The bounced leaving the wall with a speed of 30 m/s.

a. What is the change of momentum of the ball?

b. If the time contact is 20ms, find the impulse force.

m1= 0.095 v1= 40 m/s
m2= 0.095 v2= -30 m/s
My equation for:

a. change of momentum= (m1 x v1) - (m2 x v2)
(0.095 x 40) - (0.095 x -30) = 6.65 kg m/s
is this equation right?

b. i have no idea. i need some help here please :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 11, 2012

### CWatters

a) A change in momentum would be = momentum after - momentum before. It's also a vector quantity so is the ball going in the same direction afterwards?

b) Impulse = Change in momentum = force * time

3. Dec 11, 2012

### Redbelly98

Staff Emeritus
Yes, correct.

For (b), see the post by CWatters.

p.s. Welcome to Physics Forums.

4. Dec 11, 2012

### underduck

B. F= (mv - mu)/t
= (0.095 x -30) - (0.095 x 40) / 20
= -0.33 N

is that correct ?
hehe my first time doing physics. thanks for the help guys!

5. Dec 12, 2012

### CWatters

I dissagree with the answer you give for a)

The change in momentum = MV2 - MV1

m1= 0.095 v1= 40 m/s
m2= 0.095 v2= -30 m/s

Therefore the change

= (0.095 * -30) - (0.095 * 40) = -2.85 - 3.8 = -6.65 kg.s-1

Note the minus sign.

Last edited: Dec 12, 2012
6. Dec 12, 2012

### CWatters

For b)

The impact force = change of momentum / time (in seconds)

= -6.65/(20 x 10-3)
= -332.5 N

Negative because the applied force is in the opposite direction to the initial direction the ball was travelling in.

7. Dec 12, 2012

### underduck

I got it for the first question.

for the second one there is one thing i'm confused there:
(20 x 10-3)

why do i need to add 10-3 ?

thank you. i really appreciate your help, making things easier for me.

8. Dec 12, 2012

### underduck

I have another question on this Impulse and Momentum topic.

Two identical blocks, each mass 10kg, are used in an experiment. The First is held at rest on 20° incline plane 10m from the second, which at rest at the foot of the plane. the one descends the incline, slams into and sticks to the second, and the sail off together horizontally. given μk= 0.35, g 9.8 m/s2

1. Find the velocity of the first block just before it hit the second block

2. Find the velocity after the impact

3. Find the change of kinetic energy

9. Dec 12, 2012

### CWatters

The contact time is 20ms not 20s.

20ms = 20 * 10-3seconds.

10. Dec 12, 2012

### CWatters

Which bit are you stuck on? Have you drawn a diagram showing the forces on the first block?

11. Dec 12, 2012

### underduck

I'm stuck from the first question. getting a lil bit confused on how to use the equation. maybe you can explain.

Here i attach the free body diagram for the question. thank you

#### Attached Files:

• ###### scan0002.jpg
File size:
98.8 KB
Views:
174
12. Dec 13, 2012

### Redbelly98

Staff Emeritus
There are problems with your free body diagram:

1. There is no applied force Fapp. (There would have been a force holding the block in place initially, but then it is released and allowed to slide down the incline.)

2. There are other forces involved that you did not draw for the top block. The "μk" and "g" suggest what two of those forces are.