A body of mass 40,kg having velocity 4,m/s co | vedclass

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(C) For a perfectly inelastic collision, the loss in kinetic energy $(\Delta K)$ is given by the formula:$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) For a perfectly inelastic collision, the loss in kinetic energy $(\Delta K)$ is given by the formula:$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} ight) (u_1 - u_2)^2$Given:$m_1 = 40\,kg$, $u_1 = 4\,m/s$$m_2 = 60\,kg$, $u_2 = 2\,m/s$Substituting the values into the formula:$\Delta K = \frac{1}{2} \left( \frac{40 	imes 60}{40 + 60} ight) (4 - 2)^2$$\Delta K = \frac{1}{2} \left( \frac{2400}{100} ight) (2)^2$$\Delta K = \frac{1}{2} 	imes 24 	imes 4$$\Delta K = 12 	imes 4 = 48\,J$Therefore, the loss in kinetic energy is $48\,J$.

$A$ body of mass $40\,kg$ having velocity $4\,m/s$ collides with another body of mass $60\,kg$ having velocity $2\,m/s$. If the collision is perfectly inelastic, then the loss in kinetic energy will be .............. $J$.

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