# Change in Momentum Problem

Tags:
1. Oct 20, 2014

### mememe1245

Question
A 0.0420-kg hollow racquetball with an initial speed of 12.0 m/s collides with a backboard. It rebounds with a speed of 6.0 m/s.
a. Calculate the total impulse on the ball.
b. If the contact time lasts for 0.040 s, calculate the average force on the ball.

What I did on the test:
(A)Impulse=F x change in T=change in momentum(mass x velocity)

F= unknown, so use the formula, F=M x A

_________________________

Momentum=

P(momentum)= 0.0420kg x 6m/s

P= .252N

Therefore, the impulse on the ball is, .252kg. m/s.

****************************************************************

(B) F= Mass x Acceleration

change in Velocity= Acceleration x Time

Acceleration= change in Velocity/ Time

A= Velocity final - Velocity initial/ Time

A= 6m/s -12m/s/0.040s

A= -6m/s/0.040s

A= - 150 m/s^2

_____________

F= M x A

F= 0.042kg x (-150m/s^2)

F= -6.3 N

Last edited by a moderator: Oct 20, 2014
2. Oct 20, 2014

### Staff: Mentor

(-6) - 12 = -18

3. Oct 22, 2014

### mememe1245

Why is the "6" negative?

4. Oct 22, 2014

### Staff: Mentor

Because after it bounces off the wall it moves in the opposite direction.

5. Oct 23, 2014

### mememe1245

Ohhh....Okay.
so, 18 x 0.0420kg
P = .756N?

6. Oct 24, 2014

### Staff: Mentor

Are you sure about the units of P (if you indeed mean that P is the impulse of the force)?

Chet

7. Oct 24, 2014

### mememe1245

Nm

8. Oct 24, 2014

### Staff: Mentor

No. You're just guessing.

What are the units of mass? What are the units of velocity? What are the fundamental units of a Newton?

Chet