Three identical block 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 `10ms^(-1)`. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after the collision is

Options

A. `5.6 ms^(-1)`
B. `6.4 ms^(-1)`
C. `3.2 ms^(-1)`
D. `4.6 ms^(-1)`

Answer

A

Explanation

Between A and B

`m xx 10 = mv_(A) + mv_(B)` ................(i)
`e = 1/2 = (v_(B)-v_(A))/10`
or `v_(B) - v_(A) = 5` ..............(ii)
Solving Eqs. (i) and (ii), we get, `v_(B) = 7.5 ms^(-1)`
Hence A has given 75% of its speed to B and will also transfer its 75% speed of C.
`therefore v_(C) = (75)/(100) xx 7.5 = 5.625 ms^(-1)`