Question

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 250 g, remains at rest. (a) What is the mass (in g) of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is 4.5 m/s?

Answer 1

For two masses m and M,
initial momentum p = mv - Mv = (m - M)v
where m is presumed headed to the right (+ve) and M to the left (-ve).
final momentum p = MV = (m - M)v
where V is the velocity of the particle still moving post-collision.
So V = v(m - M)/M
For an elastic head-on collision, we know that the
relative velocity of approach = relative velocity of separation, or in this case,
2v = V
So 2v = v(m - M)/M
and 2M = m - M
and m = 3M
and, since M was moving after the collision, m = 250g and
M = 250/3 = 83.33 g

b) 250g * 4.5m/s - 83.33g * 4.5m/s = 166.67g * 4.5m/s = p
v = p / (m + M) = 166.67g * 4.5m/s / 333.33 g = 2.25 m/s