A bullet of mass 4.2 times 10^{-2} text{ kg}, | vedclass

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(B) Let the mass of the bullet be $m = 4.2 	imes 10^{-2} 	ext{ kg}$.Mass of the block $M = 9m = 9 	imes 4.2 	imes 10^{-2} 	ext{ kg}$.Initial velo

Vedclass · Accepted Answer

$405$

Vedclass · Answer

(B) Let the mass of the bullet be $m = 4.2 	imes 10^{-2} 	ext{ kg}$.Mass of the block $M = 9m = 9 	imes 4.2 	imes 10^{-2} 	ext{ kg}$.Initial velocity of the bullet $v = 300 	ext{ m/s}$.By the law of conservation of linear momentum, the momentum before collision equals the momentum after collision:$mv = (m + M)V$$m(300) = (m + 9m)V$$300m = 10mV$$V = 30 	ext{ m/s}$.Heat generated is equal to the loss in kinetic energy:$\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)V^2$$\Delta K = \frac{1}{2} 	imes (4.2 	imes 10^{-2}) 	imes (300)^2 - \frac{1}{2} 	imes (10 	imes 4.2 	imes 10^{-2}) 	imes (30)^2$$\Delta K = \frac{1}{2} 	imes 4.2 	imes 10^{-2} 	imes (90000 - 10 	imes 900)$$\Delta K = 2.1 	imes 10^{-2} 	imes (90000 - 9000) = 2.1 	imes 10^{-2} 	imes 81000 = 1701 	ext{ J}$.Since $1 	ext{ cal} = 4.2 	ext{ J}$, the heat in calories is:$	ext{Heat} = \frac{1701}{4.2} = 405 	ext{ cal}$.

$A$ bullet of mass $4.2 \times 10^{-2} \text{ kg}$, moving at a speed of $300 \text{ m/s}$, gets stuck into a block with a mass $9$ times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be (in $\text{ cal}$)

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