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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.
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Solution
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:
\[m_1 v_1 + m_2 v_2 = ( m_1 + m_2 )v\]
\[ \Rightarrow 60 \times 10 + 0 = 100 \times v\]
\[ \Rightarrow v = 6 \text{ m/s }\]
\[\text{ The loss in kinetic energy during collision is given as, }\]
\[ ∆ KE = \frac{1}{2} m_1 v_1^2 - \frac{1}{2}( m_1 + m_2 ) v^2 \]
\[ \Rightarrow ∆ KE = \frac{1}{2} \times 60 \times (10 )^2 - \frac{1}{2} \times 100 \times 36\]
\[ \Rightarrow ∆ KE = 1200 J\]
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