Question

A block of mass 200 g, sliding with a speed of 12 cm/sec on a smooth level surface, makes a head‑on, elastic collision with a block of unknown mass, initially at rest. After the collision the velocity of the 200 g block is 4 cm/sec in the same direction as its initial velocity. Determine the mass of the 2nd block and its speed after the collision.

Answer 1

This deals with conservation of momentum.
The sum of the momentum before the collision must equal the sum of the momentum after the collision. Since the collision is elastic, no energy is lost.
Initial momentum = mv = .2 kg x .12 m/s = .024 kg x m/s
Final momentum = .2 kg x .04 m/s + M kg x V m/s = .008 + MV kg x m/s
So, we know that the momentum of the second body is .024 - .008 = .016 kg x m/s

Up to this point, we don't know either the mass or velocity of the second object. To solve this, we need a system of equations.

Answer 2

To create this system, let's use Kinetic Energy. Kinetic Energy = v^2 x m/2. Since the collision is elastic, the final kinetic energy will equal the initial kinetic energy.
The initial kinetic energy is (.2*.012*.012)/2 = .0000144 J.
The final kinetic energy is the sum of the individual kinetic energies. The first block has a kinetic energy of (.2*.004*.004)/2 = .0000016 J.
The difference, the kinetic energy of the unknown block, is .0000128 J = .5 MV^2
So, MV^2 = .0000256 J.

We now have two equations:
MV^2 = .0000256 and
MV = .016

Can you solve the system from here?

Answer 3

Thank very much, I will give it a try.