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MCQ · Q5

Q.A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. Ratio of their respective speeds after collision is (A) 2/3 (B) 3/2 (C) 2 (D) 1/2

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Let mass 2m2m have initial velocity u1=+su_1=+s and mass mm, approaching it from the opposite direction at double the speed, have u2=−2su_2=-2s (taking 2m's direction of motion as positive). Momentum conservation: 2m(s)+m(−2s)=2m v1+m v2⇒2ms−2ms=2mv1+mv2⇒2v1+v2=0⇒v2=−2v12m(s)+m(-2s)=2m\,v_1+m\,v_2\Rightarrow 2ms-2ms=2mv_1+mv_2\Rightarrow 2v_1+v_2=0\Rightarrow v_2=-2v_1. Coefficient of restitution (using the book's definition e=−v2−v1u2−u1e=-\frac{v_2-v_1}{u_2-u_1}): e=−v2−v1−2s−s=v2−v13s=0.5⇒v2−v1=1.5se=-\frac{v_2-v_1}{-2s-s}=\frac{v_2-v_1}{3s}=0.5\Rightarrow v_2-v_1=1.5s. Substituting v2=−2v1v_2=-2v_1: $-2v_1-v_1=1.5s\Rightarrow -3 …

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