A bullet of mass 10 g pierces through a plat | vedclass

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(C) Let $m = 10 \ g = 0.01 \ kg$ be the mass of the bullet,$M_A = 500 \ g = 0.5 \ kg$ be the mass of plate $A$,and $M_B = 1.49 \ kg$ be the mass of pl

Vedclass · Accepted Answer

$43.75$

Vedclass · Answer

(C) Let $m = 10 \ g = 0.01 \ kg$ be the mass of the bullet,$M_A = 500 \ g = 0.5 \ kg$ be the mass of plate $A$,and $M_B = 1.49 \ kg$ be the mass of plate $B$.Let $v_1$ be the initial velocity of the bullet,$v_3$ be the velocity of the bullet after passing through plate $A$,and $v_2$ be the final velocity of both plates $A$ and $B$ after the bullet embeds into $B$.Applying the law of conservation of momentum for the entire system:$m v_1 = M_A v_2 + (M_B + m) v_2$$0.01 v_1 = (0.5 + 1.49 + 0.01) v_2 = 2 v_2$$v_1 = 200 v_2$  --- $(i)$Now,considering the collision between the bullet and plate $B$ only:$m v_3 = (M_B + m) v_2$$0.01 v_3 = (1.49 + 0.01) v_2 = 1.5 v_2$$v_3 = 150 v_2$  --- $(ii)$The percentage loss in kinetic energy of the bullet after passing through plate $A$ is:$	ext{Loss} \% = \frac{\frac{1}{2} m v_1^2 - \frac{1}{2} m v_3^2}{\frac{1}{2} m v_1^2} 	imes 100 = \left( 1 - \frac{v_3^2}{v_1^2} ight) 	imes 100$Substituting the values from $(i)$ and $(ii)$:$	ext{Loss} \% = \left( 1 - \left( \frac{150 v_2}{200 v_2} ight)^2 ight) 	imes 100 = \left( 1 - \left( \frac{3}{4} ight)^2 ight) 	imes 100$$	ext{Loss} \% = \left( 1 - \frac{9}{16} ight) 	imes 100 = \frac{7}{16} 	imes 100 = 43.75 \%$

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