A block of mass M1 is attached by string to a support. The block is raised to a height h and released. It then strikes a block of mass M2 on a frictionless surface. Find the velocity of the block M2, assuming a totally elastic collision.

Given Information:

Mass of block 1 = m₁

Mass of block 2 = m₂

Block 1 raised to height = h

Required Information:

Velocity of block 2 = v₂ = ?

Velocity of block 2 = v₂ = √2m₁gh/m₂

Explanation:

Please refer to the attached diagram, as you can see the block 1 is attached to a support and it is released from a height and it strikes the block 2. Applying the principle of conservation of mechanical energy, the potential energy of block 1 before the collision is transformed into kinetic energy of block 2 after the collision,

PE₁ = KE₂   eq. 1

We know that potential energy is given by

PE₁ = m₁gh

Where m₁ is mass of block 1 and h is the height the block 1 is raised to

We know that kinetic energy is given by

KE₂ = ½m₂v₂²

Where m₂ is the mass of block 2 and v₂ is the velocity of block 2

Therefore, the equation 1 becomes

m₁gh = ½m₂v₂²

We want an expression for the velocity of block 2

½m₂v₂² = m₁gh

v₂² = 2m₁gh/m₂

v₂ = √2m₁gh/m₂

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.