Question

If % decrease in k.E energy is 10% what is % decrease in momentum?

Answer 1

The momentum decreased to `94.9% "of " p`.
.

Explanation:

I will assume that the 10% decrease is due to speed decrease (rather than throwing 10% of the mass overboard).

We have `KE_i = 1/2*m*v_i^2` and `KE_f = 1/2*m*v_f^2`

`0.9*KE_i = KE_f`

`0.9*1/2*m*v_i^2 = 1/2*m*v_f^2`

Cancelling where possible, `0.9*cancel(1/2*m)*v_i^2 = cancel(1/2*m)*v_f^2`

`0.9*v_i^2 = v_f^2`

`v_f^2 = 0.9*v_i^2`

`v_f = sqrt(0.9)*v_i = 0.949*v_i`

The velocity decreased to `0.949 *v_i`, therefore the momentum decreased to `0.949*p`.

I hope this helps,
Steve