# Very simple momentum question

1. Jan 2, 2009

### thomas49th

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

When an pea bounces off the surface, it exeriences a change of momentum of 0.0051.

Show that the rebound velocity of this pea is about 5m/s

Mass of pea = 5x10^-4

2. Relevant equations

m1v1 = m2v2

3. The attempt at a solution

So a change in momentum of 0.0051 means change in m1v1 = 0.0051

as we know the mass of the pea we can do

v2 = (m1v1)/m2
= 0.0051 / 5x10^-4

but my answer is a factor of 2 too big? What have I done wrong?

Thanks :)

2. Jan 2, 2009

### Staff: Mentor

No, the change in momentum = 0.0051, not the momentum.

Assume that the pea bounces off with the same speed as it landed. Careful: While the speed hasn't changed, the velocity sure has. What's the change in momentum?

Hint: mv1 ≠ mv2

3. Jan 3, 2009

mv1 = -mv2

4. Jan 3, 2009

### Staff: Mentor

Exactly. So what's the change in momentum?

5. Jan 3, 2009

### thomas49th

mv1 -- mv2 = mv1 + mv2

mv1 + mv2 = 0.0051

m(v1 + v2) = 0.0051
v1 + v2 = 10.2
if say the speed before = speed after then v1 = v2:

2v1 = 10.2
v1 = 5.1m/s

also is it true that in the change of momentum only speed will change as mass is constant?

THANKS :)

6. Jan 3, 2009

### Staff: Mentor

Good. Here's how I'd do it.

If we call the initial velocity V, then the final velocity will be -V (since the direction reverses). Thus the initial momentum is mV and the final momentum is -mV. The change in momentum is -mV -mV = -2mV. (The magnitude of the change is just 2mV.)

The mass is constant but the velocity changes (not the speed).

7. Jan 3, 2009

### thomas49th

sorted! Thanks alot :)