A 50, g bullet moving with velocity 10, m/s s | vedclass

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(B) Initial kinetic energy of the system $(K_i)$ = Kinetic energy of the bullet = $\frac{1}{2} m_B v_B^2 = \frac{1}{2} 	imes 0.05 \, kg 	imes (10 \,

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(B) Initial kinetic energy of the system $(K_i)$ = Kinetic energy of the bullet = $\frac{1}{2} m_B v_B^2 = \frac{1}{2} 	imes 0.05 \, kg 	imes (10 \, m/s)^2 = 2.5 \, J$.By the law of conservation of linear momentum,$m_B v_B = (m_B + M) v_{sys}$.$0.05 	imes 10 = (0.05 + 0.95) 	imes v_{sys} \implies 0.5 = 1 	imes v_{sys} \implies v_{sys} = 0.5 \, m/s$.Final kinetic energy of the system $(K_f)$ = $\frac{1}{2} (m_B + M) v_{sys}^2 = \frac{1}{2} 	imes 1 \, kg 	imes (0.5 \, m/s)^2 = 0.125 \, J$.Loss in kinetic energy = $K_i - K_f = 2.5 - 0.125 = 2.375 \, J$.Percentage loss = $\frac{K_i - K_f}{K_i} 	imes 100 = \frac{2.375}{2.5} 	imes 100 = 95\%$.

$A$ $50\, g$ bullet moving with velocity $10\, m/s$ strikes a block of mass $950 \,g$ at rest and gets embedded in it. The loss in kinetic energy will be ............... $\%$

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