A body of mass 4, kg moving with velocity 12, | vedclass

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(C) Initial kinetic energy of the system $(K_i)$ = $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} 	imes 4 	imes (12)^2 + 0 = 288\, J$.

Vedclass · Accepted Answer

$172.8$

Vedclass · Answer

(C) Initial kinetic energy of the system $(K_i)$ = $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} 	imes 4 	imes (12)^2 + 0 = 288\, J$. Since the bodies stick together,it is a perfectly inelastic collision. By the law of conservation of linear momentum,$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$. $4 	imes 12 + 6 	imes 0 = (4 + 6) v \implies 48 = 10v \implies v = 4.8\, m/s$. Final kinetic energy of the system $(K_f)$ = $\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} 	imes 10 	imes (4.8)^2 = 5 	imes 23.04 = 115.2\, J$. Loss in kinetic energy = $K_i - K_f = 288 - 115.2 = 172.8\, J$.

$A$ body of mass $4\, kg$ moving with velocity $12\, m/s$ collides with another body of mass $6\, kg$ at rest. If two bodies stick together after collision,then the loss of kinetic energy of the system is .............. $J$.

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