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

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

Vedclass · Accepted Answer

$43.75$

Vedclass · Answer

(C) Let the mass of the bullet be $m = 10 \ g = 0.01 \ kg$,mass of plate $A$ be $M_A = 500 \ g = 0.5 \ kg$,and mass of plate $B$ be $M_B = 1.49 \ kg$.Let the initial velocity of the bullet be $u$ and its velocity after passing through plate $A$ be $v_1$.Let the velocity of plate $A$ after the bullet passes through it be $v_A$ and the velocity of plate $B$ after the bullet gets embedded in it be $v_B$.According to the problem,$v_A = v_B = v$.For plate $A$,by conservation of linear momentum:$m u = m v_1 + M_A v \Rightarrow m(u - v_1) = M_A v \quad ... (1)$For plate $B$,by conservation of linear momentum:$m v_1 = (m + M_B) v \Rightarrow v = \frac{m v_1}{m + M_B} \quad ... (2)$Substituting $v$ from $(2)$ into $(1)$:$m(u - v_1) = M_A \left( \frac{m v_1}{m + M_B} ight)$$u - v_1 = v_1 \left( \frac{M_A}{m + M_B} ight)$$u = v_1 \left( 1 + \frac{M_A}{m + M_B} ight) = v_1 \left( \frac{m + M_B + M_A}{m + M_B} ight)$$v_1 = u \left( \frac{m + M_B}{m + M_B + M_A} ight) = u \left( \frac{0.01 + 1.49}{0.01 + 1.49 + 0.5} ight) = u \left( \frac{1.5}{2.0} ight) = 0.75 u = \frac{3}{4} u$.The initial kinetic energy of the bullet is $K_i = \frac{1}{2} m u^2$.The kinetic energy of the bullet after passing through plate $A$ is $K_f = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left( \frac{3}{4} u ight)^2 = \frac{9}{16} K_i$.The percentage loss in kinetic energy is:$	ext{Percentage Loss} = \frac{K_i - K_f}{K_i} 	imes 100 = \left( 1 - \frac{9}{16} ight) 	imes 100 = \frac{7}{16} 	imes 100 = 43.75 \%$.

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