First slide

Coefficient of restitution

Question

Three identical blocks A, B and C are placed on 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 collision is approximately

Difficult

A

$5.6 {ms}^{- 1}$
B

$6.4 {ms}^{- 1}$
C

$3.2 {ms}^{- 1}$
D

$4.6 {ms}^{- 1}$

Solution

For collision between A and B,
$m \times 10 = {mv}_{A} + {mv}_{B} \Rightarrow v_{B} + v_{A} = 10$ …(i)
$e = \frac{1}{2} = \frac{v_{B} - v_{A}}{10}$ 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 B will also transfer its 75% speed to C.
$∴$ $v_{C} = \frac{75}{100} \times 7.5 = 5.625 {ms}^{- 1} ≃ 5.6 {ms}^{- 1}$

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