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Numericals · Q42

Q.A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double the speed. After collision, the heavier one travels with the average of the initial speeds of the two. Calculate the coefficient of restitution.

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Let the heavier marble (mass 2m) have initial speed u1=6u_1=6 cm/s and the lighter marble (mass m), following behind with double speed, have u2=12u_2=12 cm/s (both moving in the same direction, m catching up to 2m from behind). Momentum conservation: 2m(6)+m(12)=2m v1+m v2⇒12+12=2v1+v2⇒2v1+v2=242m(6)+m(12)=2m\,v_1+m\,v_2\Rightarrow 12+12=2v_1+v_2\Rightarrow 2v_1+v_2=24 --- (I). Given: the heavier marble's final speed equals the AVERAGE of the two initial speeds, v1=6+122=9v_1=\frac{6+12}{2}=9 cm/s. Substituting into (I): 2(9)+v2=24⇒v2=24−18=62(9)+v_2=24\Rightarrow v_2=24-18=6 cm/s. Coefficient of restitution: e=v2−v1u1−u2e=\frac{v_2-v_1}{u_1-u_2} -- using signed values consistently with the standard convention (e=relative speed of separationrelative speed of approache=\frac{\text{relative speed of separation}}{\text{relative speed of approach}}, both taken as positive magnitudes since m approaches 2m from behind), relative speed of approach =u2−u1=12−6=6=u_2-u_1=12-6=6 cm/s (m catching up to 2m), relative speed of separation =v1−v2=9−6=3=v_1-v_2=9-6=3 cm/s …

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