Coefficient of restitution

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Question

Three identical blocks A, B and C are at rest. But A is approaching towards B with a speed 10m/s. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is approximately

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Solution

When A collide with B,
$\begin{array}{l} m \times 10 = m v_{A} + m v_{B} ..... (i) \\ e = \frac{1}{2} = \frac{v_{B} - v_{A}}{10} \\ v_{B} - v_{A} = 5 ........ (i i) \end{array}$
Solving equation (I) and (ii), we get
$v_{B} = 7.5 m / s$
i.e., A has transferred 75% of its speed to B. Similarly, B will also transfer 75% of its speed to C.
$v_{c} = \frac{75}{100} \times 7.5 = 5.625 m / s$

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Coefficient of restitution

Remember concepts with our Masterclasses.

Three identical blocks A, B and C are at rest. But A is approaching towards B with a speed 10m/s. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is approximately

When A collide with B,m×10=mvA+mvB .....(i)e=12=vB−vA10vB−vA=5 ........(ii)Solving equation (I) and (ii), we getvB=7.5 m/si.e., A has transferred 75% of its speed to B. Similarly, B will also transfer 75% of its speed to C.vc=75100×7.5=5.625 m/s

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