A sphere of mass m moving with velocity v col | vedclass

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(C) According to the law of conservation of linear momentum:$m v + m(0) = m v_1 + m v_2$$v_1 + v_2 = v \quad \dots (i)$Using the definition of the coe

Vedclass · Accepted Answer

$\frac{e+1}{2}$

Vedclass · Answer

(C) According to the law of conservation of linear momentum:$m v + m(0) = m v_1 + m v_2$$v_1 + v_2 = v \quad \dots (i)$Using the definition of the coefficient of restitution $(e)$:$e = \frac{v_2 - v_1}{v - 0}$$e v = v_2 - v_1 \quad \dots (ii)$Adding equation $(i)$ and equation $(ii)$:$(v_1 + v_2) + (v_2 - v_1) = v + e v$$2 v_2 = v(1 + e)$$v_2 = \frac{v(e + 1)}{2}$Therefore,the ratio of the final velocity of the second sphere $(v_2)$ to the initial velocity of the first sphere $(v)$ is:$\frac{v_2}{v} = \frac{e + 1}{2}$

$A$ sphere of mass $m$ moving with velocity $v$ collides head-on with another sphere of the same mass which is at rest. The ratio of the final velocity of the second sphere to the initial velocity of the first sphere is ($e$ is the coefficient of restitution and the collision is inelastic).

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