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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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प्रश्न

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.

योग
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उत्तर

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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Momentum Conservation and Centre of Mass Motion
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Centre of Mass, Linear Momentum, Collision - Exercise [पृष्ठ १६२]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 9 Centre of Mass, Linear Momentum, Collision
Exercise | Q 32 | पृष्ठ १६२

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