Physics; Momentum - Not homework - hardest question?

Question

Three particles A, B and C have masses 0.2kg, mkg and 0.3kg respectively. the particles are free to move on a smooth horizonatal table. initially the particles lie in a straight line, with B and C at rest, and A is moving directly towords B with speed 'u'ms-1. After A's collision with B, the speed of A is '1/2u'ms-1 in the same direction as before. When B collides with C, the particles B and C coalesce and begin to move with speed '1/4u'ms-1.

a) Find the value of m
b) Find in terms of u, the speed of B before it collides with C

the answer should come to 0.1, u ms-1, but i need the working out.

Answer 1

a) Using conservation of momentum:

First collision:

m1u1 + m2u2 = m1v1 + m2v2
0.2u = 0.2 x u/2 + mvB
mvB = 0.1u

Second collision:

m1u1 + m2u2 = m1v1 + m2v2
mvB = vBC(mB + mC)
0.1u = u/4(m + 0.3)
0.4u = u(m + 0.3)
0.4 = m + 0.3
m = 0.1

b) Substitute m = 0.1 back into mvB = 0.1u from the first collision.
mvB = 0.1u
0.1vB = 0.1 u
vB = u

I tried to use the conventional formulae/notation, but couldn't subscript stuff like the 1 on m1. Hope it makes sense. The answers come out right, though :)

Answer 2

Well, the way to this question is to do a and b at the same time. To find B, you have to use the info on the collision of B and C. C is intiallly at rest. Therefore, m(1/2)=(m+0.3)(1/4)
0.5m= 0.25m+0.075
0.25m= 0.075
m=0.3kg

b) Let's say that the speed of B is x.
0.2(u)+m(0)= 0.2(1/2)+m(x)
0.2u=0.1+0.3x
2u=1+3x
(2u-1)/3=x

The speed of B is therefore:(2u-1)/3 ms^-1

I hope this helps. :)

Answer 3

is A moving at a constant velocity or acceleration and are you looking for the 'muzzle' velocity as B moves off?