First slide

Coefficient of restitution

Question

Three identical blocks ,4, B and C are placed on horizontal friction less surface as shown in fig. (+). The
blocks B and C are at rest. The block A is approaching towards B with a speed t0 m,/s. The coefficient of restitution for all collisions is 0 . 5. The speed of block C just after collision is

Moderate

A

5.6 m/s
B

6 m/s
C

8 m/s
D

10 m/s

Solution

For collision between A and B
$e = \frac{v_{B} - v_{A}}{u_{A} - u_{B}} = \frac{v_{B} - v_{A}}{10 - 0} = \frac{v_{B} - v_{A}}{10} ∴ V_{B} - V_{A} = 10 e = 10 \times 0 \cdot 5 = 5 Applying the] aw of conservation of momentum m_{A} u_{A} + m_{B} u_{B} = m_{A} v_{A} + m_{B} v_{B} m \times 10 + 0 = {mv}_{A} + {mv}_{B} ∴ v_{A} + v_{B} = 10 From eqs . (1) and (2), we get V_{B} = 7 \cdot 5 m / s For collision between B and C, we have v_{C} - v_{B} = 7 \cdot 5 e = 7 \cdot 5 \times 0 \cdot 5 = 3 \cdot 75 V_{C} + V_{B} = 7 \cdot 5 Solving eqs . (3) and (4), we get v_{C} = 5 \cdot 6 m / s$

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