Question

A body of mass 300g kept at rest breaks into two part due to internal force. One part of mass200g is found to move at a speed of 12m*s-1 . Towards the east. What would be the velocity of other part

Answer 1

Initially the object was at rest. So its velocity v = 0 m/s so its momentum p = 0 kg m/s

Finally

Now, After it breaks into two parts

For part 1 i.e. of 200g = 0.2kg,

Momentum = m₁v₁ = 0.2 * 12

For part 2 i.e. of 100g = 0.1 kg,

Momentum = m₂v₂ = 0.1 * v₂

Now.

As per the concept of law of conservation of momentum.

Final momentum = Initial momentum

Final momentum = m₁v₁ + m₂v₂

Initial Momentum = 0

So,

m₁v₁ + m₂v₂ = 0

(0.2 * 12) + (0.1 * v₂) = 0

2.4 = -0.1v₂

∴ v₂ = 2.4/(-0.1)

∴ v₂ = -0.24 m/s

Answer :- Velocity of another part is -0.24 m/s