Question

Ball 1 initially travels with a velocity of u m/s.Ball 2 is stationery and has a mass of 0.2kg and this collision lasts for 0.1 s.Afterwards both balls move in the direction of ball 1's initial velocity. Each ball has a different final velocity.

During the collision a force of -6N is exerted on ball 1 by ball 2 calculate the velocity of ball 2 after the collision.

Answer 1

Hmmm. If -6N was exerted on ball 1, then 6N was exerted on ball2

6*.1=.2w
w=3m/s

Answer 2

Since p=mv, the momentum for the ball2 initially is 0.2 * 0 = 0 kg m/s.
It is stated that -6N of force was exerted by ball2 on ball1, so because of Newton's third law stating that there is always a force acting in the equal and opposite direction, we know that 6N has been acted by ball1 on ball2.
Since F=(mv-mu)/t, we can substitute some values in:
6=(mv-mu)/0.1
0.6=mv-mu
mv-mu is the change in momentum, so 0.6 is the change in momentum.
Using the formula p=mv again, we can calculate the final velocity of ball2 now.
p=mv
0.6=0.2v
0.6/0.2=v
v=3 m/s