A body of mass M moving at speed


A body of mass $M$ moving at speed $V_{0}$ collides elastically with a mass ' $m$ ' at rest. After the collision, the two masses move at angles $\theta_{1}$ and $\theta_{2}$ with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio $\mathrm{M} / \mathrm{m}$, for which the angles $\theta_{1}$ and $\theta_{2}$ will be equal, is :

  1. 4

  2. 1

  3. 3

  4. 2

Correct Option: , 3


given $\theta_{1}=\theta_{2}=\theta$

from momentum conservation

in $\mathrm{x}$-direction $\mathrm{MV}_{0}=\mathrm{MV}_{1} \cos \theta+\mathrm{mV}_{2} \cos \theta$

in $\mathrm{y}$-direction $0=M \mathrm{MV}_{1} \sin \theta-\mathrm{mV}_{2} \sin \theta$

Solving above equations

$\mathrm{V}_{2}=\frac{\mathrm{MV}}{\mathrm{m}}, \mathrm{V}_{0}=2 \mathrm{~V}_{1} \cos \theta$

From energy conservation

$\frac{1}{2} \mathrm{MV}_{0}^{2}=\frac{1}{2} \mathrm{MV}_{1}^{2}+\frac{1}{2} \mathrm{MV}_{2}^{2}$

Substituting value of $V_{2} \& V_{0}$, we will get

$\frac{\mathrm{M}}{\mathrm{m}}+1=4 \cos ^{2} \theta \leq 4$

$\frac{\mathrm{M}}{\mathrm{m}} \leq 3$

Option (3)

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