A body of mass 2 ,kg collides head-on with an | vedclass

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(B) Given: $m_1 = 2 \,kg$, $m_2 = 4 \,kg$, relative velocity before collision $u_{rel} = u_1 - u_2 = 10 \,ms^{-1}$, relative velocity after collision

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(B) Given: $m_1 = 2 \,kg$, $m_2 = 4 \,kg$, relative velocity before collision $u_{rel} = u_1 - u_2 = 10 \,ms^{-1}$, relative velocity after collision $v_{rel} = v_2 - v_1 = 4 \,ms^{-1}$.Coefficient of restitution $e = \frac{v_{rel}}{u_{rel}} = \frac{4}{10} = 0.4$.The loss in kinetic energy during a collision is given by $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} ight) (u_{rel})^2 (1 - e^2)$.Substituting the values: $\Delta K = \frac{1}{2} \left( \frac{2 	imes 4}{2 + 4} ight) (10)^2 (1 - (0.4)^2)$.$\Delta K = \frac{1}{2} \left( \frac{8}{6} ight) (100) (1 - 0.16)$.$\Delta K = \frac{1}{2} 	imes \frac{4}{3} 	imes 100 	imes 0.84$.$\Delta K = \frac{2}{3} 	imes 84 = 56 \,J$.

$A$ body of mass $2 \,kg$ collides head-on with another body of mass $4 \,kg$. If the relative velocities of the bodies before and after collision are $10 \,ms^{-1}$ and $4 \,ms^{-1}$ respectively, the loss of kinetic energy of the system due to the collision is (in $\,J$)

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