If the coefficient of restitution between A and B (shown in figure) is e=12, then select the correct statement(s)

Loss in kinetic energy during the collision is 3 16 m v 2 , Impulse between the bodies during collision is 3 4 m v , Loss in kinetic energy during the collision is 1 16 m v 2

If the coefficient of restitution between A and B (shown in figure) is e=12, then select the correct statement(s)

Since, v 1 = ( m 1 − e m 2 m 1 + m 2 ) μ 1 + [ ( 1 + e ) m 2 m 1 + m 2 ] μ 2 ⇒ v 1 = ( m 1 − m / 2 m + m ) v + 0 = v 4 A l s o , v 2 = [ ( 1 + e ) m 1 m 1 + m 2 ] u 1 + ( m 2 − e m 1 m 1 + m 2 ) u 2 ⇒ v 2 = ( 1 + 1 / 2 ) m 2 m = 3 v 4 | Im p u l s e | = | Δ p 1 | = | Δ p 2 | ⇒ | Im p u l s e | = m ( 3 4 v ) = 3 m v 4 Loss in kinetic energy is K i − K f = − Δ K ⇒ − Δ K = 1 2 m v 2 − [ 1 2 m ( 3 v 4 ) 2 + 1 2 m ( v 4 ) 2 ] ⇒ − Δ K = 3 16 m v 2

Loss in kinetic energy during the collision is 3 16 m v 2

Impulse between the bodies during collision is 3 4 m v

Loss in kinetic energy during the collision is 1 16 m v 2

Velocity of B after collision is 3 v 4

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If the coefficient of restitution between A and B (shown in figure) is $e = \frac{1}{2}$ , then select the correct statement(s)

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Loss in kinetic energy during the collision is $\frac{3}{16} m v^{2}$

Impulse between the bodies during collision is $\frac{3}{4} m v$

Loss in kinetic energy during the collision is $\frac{1}{16} m v^{2}$

Velocity of B after collision is $\frac{3 v}{4}$

answer is A, B, C.

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Detailed Solution

Since, $v_{1} = (\frac{m_{1} - e m_{2}}{m_{1} + m_{2}}) μ_{1} + [\frac{(1 + e) m_{2}}{m_{1} + m_{2}}] μ_{2}$

$\Rightarrow v_{1} = (\frac{m_{1} - m / 2}{m + m}) v + 0 = \frac{v}{4}$

$A l s o, v_{2} = [\frac{(1 + e) m_{1}}{m_{1} + m_{2}}] u_{1} + (\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}) u_{2}$

$\Rightarrow v_{2} = \frac{(1 + 1 / 2) m}{2 m} = \frac{3 v}{4}$

$| Im p u l s e | = | Δ p_{1} | = | Δ p_{2} | \Rightarrow | Im p u l s e | = m (\frac{3}{4} v) = \frac{3 m v}{4}$

Loss in kinetic energy is $K_{i} - K_{f} = - Δ K$

$\Rightarrow - Δ K = \frac{1}{2} m v^{2} - [\frac{1}{2} m {(\frac{3 v}{4})}^{2} + \frac{1}{2} m {(\frac{v}{4})}^{2}] \Rightarrow - Δ K = \frac{3}{16} m v^{2}$