What is the Rebound Velocity of a Pea Bouncing off a Surface?

[thomas49th]

Homework Statement

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

Homework Equations

m1v1 = m2v2

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 :)

 

[Doc Al]

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

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

 

[thomas49th]

mv1 = -mv2

 

[Doc Al]

Exactly. So what's the change in momentum?

 

[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

TRADA!

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

 

[Doc Al]

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

TRADA!

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.)

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

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

 

[thomas49th]

sorted! Thanks a lot :)