Two titanium spheres approach each other headon with the sam

Two titanium spheres approach each other head-on with the same speed and collide elastically. Initially m₁ is moving to the right while m₂ is moving to the left. After the collision, m₂ remains at rest. Given: m₂ = 0.30 kg.

a) What is the mass of the other sphere?

b) What is the speed of the two-sphere center of mass if the initial speed of each was 2 m/s?

Solution

initial momentum p = mv - Mv = (m - M)v

i am using m1 = m and m2 as M

final momentum p = MV = (m - M)v

V = v(m - M)/M

For an elastic head-on collision,

relative velocity of approach = relative velocity of separation, or in this case,

2v = V

So 2v = v(m - M)/M

and 2M = m - M

m = 3M

a)

given m= 0.30 and

M = 0.1 kg

b) 0.3 *2 - 0.1*2 = 0.4*V

=> V = 1 m/s