Question of Physics

Question

A block of mass of 2 kg comes with initial velocity v₁ = 4i

m/s and collides with another block of mass of 1 kg coming with initial velocity v₂ = -5i m/s on a frictionless horizontal table. Determine the velocities of the blocks after the collision if the collision is

i) elastic, and

ii) completely (perfectly) inelastic.

Answer 1

Let

m₁ = 2kg be the mass of the first block,

m₂ = 1kg be the mass of the second block,

v₁ = 4m/s be the velocity of the first block before the collision,

v₂ = -5m/s be the velocity of the second block before the collision,

w₁ (to be calculated) be the velocity of the first block after the collision,

w₂ (to be calculated) be the velocity of the second block after the collision.

From conservation of momentum:

m₁w₁ + m₂w₂ = m₁v₁ + m₂v₂

Substituting actual numbers and simplifying

w₂ = 3 - 2w₁ (1)

i) The kinetic energy of the blocks after the collision equals their initial kinetic energy.

m₁w₁²/2 + m₂w₂²/2 = m₁v₁²/2 + m₂v₂²/2

Substituting actual numbers and simplifying

2w₁²/2 + w₂²/2 = 28.5

Using (1) and simplifying we get the quadratic equation

w₁² - 2w₁ - 8 = 0

It has two solutions

w₁ = 4, w₂ = -5

w₁ = -2, w₂ = 7

The first solution is not physically possible because it corresponds to the

situation of the two blocks passing through each other and continuing undisturbed.

Only the second solution is consistent with the statement that a collision occurred.

ii) The collision causes the blocks to stick together.

Therefore

w₁ = w₂

In combination with (1)

w₁ = w₂ = 1m/s