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 ms -1 . The coefficient of restitution for all collisions is 0.5. The speed of the block C just after the collision is
Text Solution
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For collision between blocks A and B,
…. (i)
From principle of momentum conservation,
m A u A + m B u B = m A v A + m B v B
m x 10 + 0 = mv A +mv B
v a + v b = 10 …. (ii)
Adding eqs. (i) and (ii), we get
v B = 7.5m s -1 . .. (iii)
Similarly, for collision between B and C,
v C -v b = 7 .5 e = 7 .5 x0.5= 3.75
v c -v b = 3.75 m s -1 ...(iv)
Adding Eqs. (iii) and (iv) we get
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