# Homework Help: Impulse Question

1. Nov 19, 2005

### amcavoy

Here is a link to an image of my problem:

http://img54.imageshack.us/img54/7268/prob31iq.jpg [Broken]

For part (a), I said that the impulse is the area under the graph and that is equal to Mv (since the initial velocity is zero):

$$J=\int\limits_{0}^{0.4}F\,dt=.36\implies v=.072\text{m}/\text{s}$$

Part (b) is where I'm having trouble. Using g=10 N/kg, I can find the coefficient of kinetic friction:

$$\mu_kmg=F_{0}\implies \mu=.48$$

This is what I did to find the initial impulse, although I'm not sure if it's right:

$$F_M=\left(.48\right)F_m\implies F_m=5N$$

Now I have the initial force; do I just multiply 5 by 0.4 seconds?

Thank you very much.

Last edited by a moderator: May 2, 2017
2. Nov 20, 2005

### lightgrav

This is a "slow, soft collision".

How fast is the little mass moving at the end of the 4 seconds?
What was the total momentum of the system at t=0?
Which mass was it that was given momentum, by t=0?

Was that Work done on the little mass initially, on the big mass initially,
or on the pair of them finally?

Since the Friction Force varies greatly (looks like proportional to v!)
your coefficient of dry friction isn't going to be valid.

3. Nov 20, 2005

### amcavoy

OK. Here is a second thought then:

The change in momentum (impulse) on block m is .072(.5)=.036. The change in momentum on block M is .072(5)=.36. Adding these together gives a total impulse of .396 N-s. Is this correct? Thank you.