# Elastic Collisions

1. Oct 30, 2005

### ViewtifulBeau

Two particles of masses M and 3M are moving towards each other along the +x and -x directions with the same initial speed 3.29 m/s. They undergo a 'head-on' elastic collision and each rebounds along the same line as it approached. What is the final speed of the particle of mass M in m/s?

so i used mv + mv = mv + mv so
3.29M -3.29(3M) = M(v) + 3M(v):
-6.58 = v + 3v
v = -1.645

now im confused, i know this isn't the right answer. I don't know what to do next. thanks.

2. Oct 31, 2005

### daniel_i_l

The final speeds of M and 3M aren't equal.

3. Oct 31, 2005

### mathmike

ok so we have m1v1_ + m2v2_i = m1v1_f + m2v2_f

and

.5m1v1_i^2 + .5m2v2_i^2 = .5m1v1_f + .5m2v2_f

now lets manipulate it some and we get

v1_i + v1_f = v2_f + v2_i

and

v1_i - v2_i = -(v1_f - v2_f)

since we know the mass and initial velocity

v1_f = [(m1 - m2) / (m1 + m2)) * v1_i + [(2 * m2) / ( m1 + m2)) * v2_i

and

v2_f = [(2 * m1) / (m1 + m2)] * v1_i + [( m2 - m1) / (m1 + m2)] * v2_i

so lets see what you get

4. Oct 31, 2005

### ViewtifulBeau

i got v1_f is -6.58 m/s
and v2_f is 0 m/s

is the sign on v1_f correct?

5. Oct 31, 2005

### mathmike

does this sound correct to you??

i think that v2 should have a final vel since it had an initial vel, but i havent calc'd it out

and as far as the sign goes, it depends which way it was going initially