Question

A 60 kg man skating with a speed of 10 m/s collides with a 40 kg skater at rest and they cling to each other. Find the loss of kinetic energy during the collision.

Answer 1

It is given that:
Mass of the skater who is skating, m₁ = 60 kg
Initial speed of this man, v₁ = 10 m/s
Mass of the skater at rest, m₂ = 40 kg 
Initial speedof this man, v₂ = 0

Let the velocity of both men after collision be v.

Using the law of conservation of momentum, we can write:

\[m_1 v_1 + m_2 v_2 = ( m_1 + m_2 )v\]

\[ \Rightarrow 60 \times 10 + 0 = 100 \times v\]

\[ \Rightarrow v = 6 \text{ m/s }\]

\[\text{  The loss in kinetic energy during collision is given as, }\]

\[ ∆ KE = \frac{1}{2} m_1 v_1^2 - \frac{1}{2}( m_1 + m_2 ) v^2 \]

\[ \Rightarrow ∆ KE = \frac{1}{2} \times 60 \times (10 )^2 - \frac{1}{2} \times 100 \times 36\]

\[ \Rightarrow ∆ KE = 1200 J\]