Three identical blocks A , B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m s - 1 . The coefficient of restitution for all collisions is 0.5 . The speed of the block C just after the collision is

Question

Three identical blocks A , B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m s - 1 . The coefficient of restitution for all collisions is 0.5 . The speed of the block C just after the collision is, 5.6 m s - 1, 6 m s - 1, 8 m s - 1, 10 m s - 1

MARKS App · Accepted Answer

For collision between blocks A and B , e = v B - v A u A - u B = v B - v A 10 - 0 = v B - v A 10 ∴ v B - v A = 10 e = 10 × 0 .5 = 5 … . ( i ) From principle of momentum conservation, m A u A + m B u B = m A v A + m B v B m × 10 + 0 = mv A + mv B ∴ v A + v B = 10 … . ( ii ) Adding eqs. (i) and (ii), we get v B = 7.5 m s - 1 … ( iii ) Similarly for collision between B and C , v C - v B = 7 .5 e = 7 .5 × 0 .5 = 3 .75 ∴ v C - v B = 3 . 75 m s - 1 ...(iv) Adding Eqs. (iii) and (iv) we get 2 v C = 11.25 ∴ v C = 11 .25 2 = 5.6 m s - 1

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