A block of mass $1.2 k g$ moving at a speed of $20 c m / s$ collides head-on with a similar block kept at rest. The coefficient of restitution is $3 / 5$ , then the loss of kinetic energy during collision.

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Solution

Here, $m_{1} = 1.2 k g, u_{1} = 20 c m / s, m_{2} = 1.2 k g, u_{2} = 0$
If $v_{1}$ and $v_{2}$ are velocities of the two blocks after collision, then according to principle of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} = 1.2 \times 20 + 0 = 1.2 v_{1} + 1.2 v_{2}$ ... (i)

velocity of approach $= u_{1} - u_{2} = 20 c m / s$
velocity of separation $= v_{2} - v_{1}$
By definition, $e = v_{2} - v_{1} / u_{1} - u_{2} = 3 / 5 = v_{2} - v_{1} / u_{1} - u_{2}$
Therefore, $v_{2} - v_{1} = 20 \times 3 / 5 = 12$ --- (ii)

From (i) and (ii),
v_1 $= 4 c m / s, v_{2} = 16 c m / s,$

Loss in K.E. $= 1 / 2 m_{1} u_{1}^{2} - 1 / 2 (m_{1}) v_{1}^{2} - 1 / 2 m_{2} v_{2}^{2} = 1 / 2 \times 1.2 (20 / 100)^{2} - 1 / 2 (1.2) (4 / 100)^{2} - 1 / 2 \times 1.2 (16 / 100)^{2}$
$= 2.4 \times 10^{- 2} - 0.096 \times 10^{- 2} - 1.536 \times 10^{- 2}$

Loss in K.E. $= 0.768 \times 10^{-} 2 = 7.7 \times 10^{- 3} J$

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A block of mass 1.2 kg moving at a speed of 20 cm/s collides head-on with a similar block kept at rest. The coefficient of restitution is 3/5, then the loss of kinetic energy during collision.

A block of mass $1.2 k g$ moving at a speed of $20 c m / s$ collides head-on with a similar block kept at rest. The coefficient of restitution is $3 / 5$ , then the loss of kinetic energy during collision.