Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

122 of 130

Question 122 of 130

Two particles AA and BB of equal mass MM are moving with the same speed vv as shown in figure. They collide completely inelastic and move as a single particle CC. The angle θ\theta that the path of CC makes with the XX-axis is given by-

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Answer: (D) tan⁡⁡θ=3+21−2\tan ⁡\theta =\dfrac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}

After inelastic collision mass of the combination is 2m\text{2m}
v′v' be the velocity of the combined mass
Using conservation of momentum along XX and YY-axis

2mv′sin⁡⁡θ=mv2+mv322m{v}^{'}\sin ⁡\theta =\dfrac{mv}{\sqrt{2}}+\dfrac{mv\sqrt{3}}{2}

2mv′cos⁡⁡θ=mv2−mv22m{v}^{'}\cos ⁡\theta =\dfrac{mv}{2}-\dfrac{mv}{\sqrt{2}}

2sinθ2cosθ=12+3212−12=2+31−2\dfrac{2sin\theta }{2cos\theta }=\dfrac{\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}}=\dfrac{\sqrt{2}+\sqrt{3}}{1-\sqrt{2}}

tan⁡θ=2+31−2\tan \theta =\dfrac{\sqrt{2}+\sqrt{3}}{1-\sqrt{2}}

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