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A ball of mass m moving with a speed `2v_0` collides head-on with an identical ball at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?A. `(1-e)/(1+e)`B. `(1+e)/(1-e)`C. `(e-1)/(e+1)`D. `(e+1)/(e-1)` |
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Answer» Correct Answer - A According to law of conservation of linear momentum , we get `m(2v_0)+mxx0=mv_1 + mv_2` where `v_1` and `v_2` are the velocities of the balls after collision. `v_1+v_2 =2v_0`..(i) By definition of coefficient of restitution `e=(v_2-v_1)/(u_1-u_2)=(v_2-v_1)/(2v_0)" " (because u_1=2v_0, u_2=0)` `v_2-v_1=2ev_0`...(ii) Adding (i) and (ii), we get `2v_2-2v_0+2ev_0 , v_2=(1+e)v_0` Subtract (ii) from (i), we get `2v_1 =2v_0 - 2ev_0 , v_1=v_0 (1-e)` Their corresponding ratio is `v_1/v_2=(1-e)/(1+e)` |
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