A body of mass 5 times 10^{3} , kg moving wit | vedclass

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(A) Given: $m_1 = 5 	imes 10^{3} \, kg$,$u_1 = 2 \, m/s$,$m_2 = 15 	imes 10^{3} \, kg$,$u_2 = 0 \, m/s$.Since the collision is perfectly inelastic $

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$7.5$

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(A) Given: $m_1 = 5 	imes 10^{3} \, kg$,$u_1 = 2 \, m/s$,$m_2 = 15 	imes 10^{3} \, kg$,$u_2 = 0 \, m/s$.Since the collision is perfectly inelastic $(e = 0)$,the bodies stick together.The loss in kinetic energy $(\Delta K.E.)$ for a perfectly inelastic collision is given by:$\Delta K.E. = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2$Substituting the values:$\Delta K.E. = \frac{1}{2} 	imes \frac{(5 	imes 10^{3}) 	imes (15 	imes 10^{3})}{5 	imes 10^{3} + 15 	imes 10^{3}} 	imes (2 - 0)^2$$\Delta K.E. = \frac{1}{2} 	imes \frac{75 	imes 10^{6}}{20 	imes 10^{3}} 	imes 4$$\Delta K.E. = \frac{1}{2} 	imes 3.75 	imes 10^{3} 	imes 4$$\Delta K.E. = 7.5 	imes 10^{3} \, J = 7.5 \, kJ$.

$A$ body of mass $5 \times 10^{3} \, kg$ moving with speed $2 \, m/s$ collides with a body of mass $15 \times 10^{3} \, kg$ inelastically and sticks to it. The loss in $K.E.$ of the system will be (in $kJ$):

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