Elastic collision of block on block

[Beblak2]

There is 4 parts to this one question.
A 2.0 Kg block is traveling to the right with a velocity of 3.0 m/s. The block collides with a stationary 1.0 kg block and the blocks bounce apart (elastic collision).
1. Homework Statement
(a). if the velocity of the 1 kg block is +4 m/s after the collision, what is the velocity of the 2.0 kg block post collision? use only conservation of momentum.
(b). what impulse does the 2.0 kg block experience?
(c). with no further calculations explain/justify whether the magnitude of the impulse experienced by the 1.0 kg block is greater then, the same as or less than the 2.0 kg block. be sure to use words in response.
(d). if during the collision, the force applied to the 1.0 kg block is as shown, how long did the collision take?

Homework Equations

Fnet=ma
p=mv
m1v1+m2v2=m1v1'+m2v2

The Attempt at a Solution

I used m1v1 + m2v2 = m1v1 +m2v2 and got -0.5 m/s?

 

[Simon Bridge]

Welcome to PF.
To get the most out of these forums, it is best practise to post your reasoning as well as the equations you use... how did you get from the equation to the answer?

For (a) you used the conservation of momentum equation ... well done.
How did you use it to get that answer (we do not check arithmetic or answers here, just reasoning)?

Have you checked your working?

It can help to be more formal about it ... like this:
before: $p_b = 3\times 2 + 1\times 0 = 6$kg.m/s
after: $p_a = 4+3v$
conservation of momentum: $p_a = p_b \implies 4+3v = 6 \implies \cdots$
... check my working and complete.

What about the others?

 

[haruspex]

if the velocity of the 1 kg block is +4 m/s

This is unnecessary information. It can be deduced from the facts already given.

we do not check arithmetic or answers here

I do, and it is wrong.

 

[Orodruin]

This is unnecessary information. It can be deduced from the facts already given.

Not if you are to use only momentum conservation as asked for in (a). Of course, you can use this information to deduce that the collision was in fact elastic or you could use the fact that the collision was elastic to deduce the 4 m/s.

 

[haruspex]

Not if you are to use only momentum conservation as asked for in (a)

Yes, it often helps to read the whole question.