Question

Hey All!

QUESTION:

Figure shows three identical discs with discs B and C placed on a smooth horizontal plane and another disc A is moving symmetrically hits the two discs and stops. Find the coefficient of restitution between discs.


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Answer 1

Given :

▪ Discs B and C are Initially at rest.

▪ Final velocity of Disc A = zero

To Find :

▪ Co-efficient of restitution.

Concept :

→ Here, no external force acts on the whole system, we can easily apply concept of momentum conservation to solve this type of questions.

Diagram :

→ Please, see the attachment for better understanding.

Assumption :

☞ Initial velocity of disc A = u

☞ Final velocity of disc B/C = v

☞ Mass of each disc = m

☞ Radius of each disc = R

Calculation :

$\implies\sf\:\sin\theta=\dfrac{R}{2R}\\ \\ \implies\bf\:\sin\theta=\dfrac{1}{2}$

Momentum conservation along x-axis :

$\mapsto\sf\:mu=mv\cos\theta+mv\cos\theta\\ \\ \mapsto\sf\:mu=2mv\cos\theta\\ \\ \mapsto\sf\:u=v\cos\theta\\ \\ \mapsto\bf\:v=\dfrac{u}{\cos\theta}$

Co-efficient of restitution :

$\dashrightarrow\sf\:e=\dfrac{-(v-0)}{(0-u\cos\theta)}\\ \\ \dashrightarrow\sf\:e=\dfrac{v}{u\cos\theta}\\ \\ \dashrightarrow\sf\:e=\dfrac{u}{2u\cos^2\theta}\\ \\ \dashrightarrow\sf\:e=\dfrac{1}{2\cos^2\theta}=\dfrac{1}{2(1-\sin^2\theta)}\\ \\ \dashrightarrow\sf\:e=\dfrac{1}{2(1-\frac{1}{4})}\\ \\ \dashrightarrow\sf\:e=\dfrac{4}{2\times 3}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{e=\dfrac{2}{3}}}}}\:\red{\bigstar}\gray{\bigstar}$