Question

On a frictionless surface, a block of mass m1 and velocity v1 = vi is moving to the right when it collides with a spring, of spring constant k, attached to the left side of a stationary block of mass m2. Both blocks are free to slide on the surface.

(a) Argue either from physical principles or with equations that the velocities of the two blocks are the same when the spring is at its maximum compression.

(b) What is the maximum compression of the spring, xmax?
(c) In terms of m1,m2, and v, what are the final velocities of the two blocks after

they finish interacting?

Answer 1

a) Let us now analyse the matter. The first block starts compressing the spring. The velocity of the 2nd block slowly starts increasing form 0. In the beginning the velocity of the 1st block is greater than the 2nd. So, the spring keeps on getting compressed. The velocity of the 2nd keeps on increasing and the 1st keeps on decreasing. The compression takes place as long as the velocity of 1st is more than that of the 2nd. After sometime their velocities becomes equal. Then after the velocity of 2nd become greater than that of 1st and so the spring length will starts becoming longer. So, compression takes place till their velocities are equal and hence

the velocities of the two blocks are the same when the spring is at its maximum compression.

b) Let v' be the common velocity at maximum compression.

By, conservation of momentum,

m₁ v' + m₂ v' = m₁ v
=> v' =  m₁ v / (m₁ + m₂)

c) Let v₁₁ and v₂₂ be the final velocities.

This is the case of elastic collision where momentum and kinetic energy both are conserved(before and after interaction)

Now,

This can be also be solved as follows:

By conservation of momentum,
m₁ v₁₁ + m₂ v₂₂ = m₁ v
=>  m₂ v₂₂ = m₁ v - m₁ v11
=> m₂ v₂₂ = m₁ (v - v₁₁) -------- (1)

By conservation of energy,

Dividing (2) by (1) gives


Putting this in (2)

Now,