A bomb projected from the ground at an angle | vedclass

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Question

(C) At the topmost point,the velocity of the bomb of mass $M$ is $v = u \cos 	heta$ in the horizontal direction. The momentum is $P = M u \cos 	heta

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) At the topmost point,the velocity of the bomb of mass $M$ is $v = u \cos 	heta$ in the horizontal direction. The momentum is $P = M u \cos 	heta$.After the explosion,the bomb splits into two fragments of mass $m = M/2$. Let the velocity of the first fragment (which returns to the point of projection) be $v_1$ and the second fragment be $v_2$.Since the first fragment returns to the point of projection,it must have a horizontal velocity of $-u \cos 	heta$ (as it travels the same horizontal distance back in the same time).By conservation of linear momentum in the horizontal direction:$M u \cos 	heta = (M/2)(-u \cos 	heta) + (M/2) v_2$$u \cos 	heta = -0.5 u \cos 	heta + 0.5 v_2$$1.5 u \cos 	heta = 0.5 v_2 \implies v_2 = 3 u \cos 	heta$.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.For the bomb just before explosion: $\lambda_{bomb} = \frac{h}{M u \cos 	heta}$.For the second fragment just after explosion: $\lambda_2 = \frac{h}{(M/2) v_2} = \frac{h}{(M/2) (3 u \cos 	heta)} = \frac{2h}{3 M u \cos 	heta}$.The ratio of the de Broglie wavelength of the second fragment to that of the bomb is:$\frac{\lambda_2}{\lambda_{bomb}} = \frac{\frac{2h}{3 M u \cos 	heta}}{\frac{h}{M u \cos 	heta}} = \frac{2}{3}$.

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