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A ball of mass m moving with a speed v makes a head on collision with an identical ball at rest. The kinetic energy after collision of the balls is three fourth of the original kinetic energy. The coefficient of restitution is |
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Answer» `1/2` Applying the principle of conservation of linear momentum , we get `mv = mv_1 + mv_2 " or " v = v_1 + v_2 "" ….(i)` By DEFINATION of coefficient of restitution `e = (v_2 - v_1)/(u_1 - u_2) = (v_2 - v_1)/(v - 0)" or " v_2 - v_1 = ev "" .....(II)` As KE after COLLISION = `3/4 KE` before collision `:. 1/2 m (v_1^2 + v_2^2) = 3/4 xx 1/2 mv^2 " or " v_1^2 + v_2^2 = 3/4 v^2 "".....(iii)` Squaring eq. (i), we get `v_1^2 + v_2^2 + 2v_1 v_2 = v^2 "".........(IV)` Subtracting (iiii), from EQUATION (iv), we get `2v_1 v_2 = 1/4 v^2"".....(v)` Squaring (ii), we get `v_2^2 + v_1^2 - 2v_1 v_2 = e^2 v^2 "" ....(vi)` Using (iii) and (v) in (vi), we get `3/4 v^2 - 1/4 v^2 = e^2 v^2 , 1/2 v^2 = e^2 v^2 or e = 1/(sqrt2)`. |
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