Question

2 spheres, of unequal masses, hang on strings and are touching. Sphere a is pulled back to the left, string taught, and released, hitting sphere B at a known velocity. Sphere B's post impact velocity is known, and to the right.

i have to find A's post collision velocity.

Intuitively, I know it will be the difference between the impact velocity of A and the final velocity of B.

I can demonstrate this using conservation of momentum:

m1u1+m2u2 = m1v1 + m2v2

Should i do any other calculations?

Answer 1

No. As I read the question, u2 is zero.

Answer 2

Yes, it's zero. That seems to make the second term (m2u2) zero, but I still get what i think is the correct answer.

What should be my next approach?

Answer 3

(ma ua + mb ub)before = (ma va + mb vb)after

ua is known
ub is 0
so
ma ua = ma va + mb vb
but vb is known
so
ma va = ma ua - mb vb
so
va = (maua -mb vb)/ma
everything on the right is known

Answer 4

Thanks both, for the answers.