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

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Solution

Momentum should be conserved due to the absence of external force on the system (skater + man).

So from conservation of momentum, we have
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) V$
where $V$ is common velocity after the cling.

we get $60 \times 10 + 0 = 100 V$ or $V = 6 m / s$

Initial Kinetic Energy of the man and skater was:
$K_{1} = \frac{1}{2} m_{1} v_{1}^{2} + 0$
$= 30 \times 100 = 3000 J o u l e$

Final Kinetic Energy of the man and skater is
$K_{2} = \frac{1}{2} (m_{1} + m_{2}) V^{2}$
$= \frac{1}{2} 100 \times 36 = 1800 J o u l e$

Loss in the kinetic energy will be
$Δ K E = 3000 - 1800 = 1200 J o u l e$

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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.

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