Two identical blocks of ice move in opposite | vedclass

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(A) The total kinetic energy of the two blocks before the collision is $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Since the collision is perfec

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(A) The total kinetic energy of the two blocks before the collision is $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Since the collision is perfectly inelastic (the blocks melt), the maximum loss in kinetic energy is equal to the total initial kinetic energy, which is $mv^2$.This energy is used to raise the temperature of the ice from $-8^{\circ}C$ to $0^{\circ}C$ and then to melt it.The energy required for one block is $Q = ms\Delta	heta + mL$.For two blocks, the total energy required is $2(ms\Delta	heta + mL)$.Equating the energy loss to the heat required: $mv^2 = 2m(s\Delta	heta + L)$.Canceling $m$ from both sides: $v^2 = 2(s\Delta	heta + L)$.Given: $s = 2100 \ Jkg^{-1}K^{-1}$, $\Delta	heta = 8^{\circ}C$, $L = 3.36 	imes 10^5 \ Jkg^{-1}$.$v^2 = 2(2100 	imes 8 + 3.36 	imes 10^5) = 2(16800 + 336000) = 2(352800) = 705600$.$v = \sqrt{705600} = 840 \ ms^{-1}$.

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