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Question: A body X with a momentum p collides with another identical stationary body Y one dimensionally. Duri...

A body X with a momentum p collides with another identical stationary body Y one dimensionally. During the collision, Y gives an impulse J to body X. Then, coefficient of restitution is,
(a)2Jp1\left( a \right)\dfrac{2J}{p}-1
(b)Jp+1\left( b \right)\dfrac{J}{p}+1
(c)Jp1\left( c \right)\dfrac{J}{p}-1
(d)J2p1\left( d \right)\dfrac{J}{2p}-1

Explanation

Solution

We will use diagrams to understand the given condition better. Also, we will use the formula of impulse which will give the value of impulse created by X on Y . This impulse is a result of momentum of X. After collision we see impulse created in both X and Y objects. We will conclude the solution by using the formula of coefficient of restitution.

Formula used:
J=ΔpJ=\Delta p where J is an impulse and p is momentum, e=momentum of both objects after collisionmomentum-e\,\text{=}\dfrac{\text{momentum of both objects after collision}}{\text{momentum}} where e is restitution coefficient.

Complete answer:
Impulse: When a force is exerted on any object for any interval of time then, that sudden force that helps in quick movement of an object is known as impulse. If impulse is measured with force so, the impulse will be this sudden force which makes an effect on an object. Here, we will denote impulse by J and numerically, we will write J=ΔpJ=\Delta p.

Coefficient of restitution: It is the ratio of those frequencies which has two values in the form of initial and final velocities of an object before and after collision of two objects. Its value never goes outside 0 and 1 rather it lies between them. In case the value of it is 1 then, there will be a case of elastic collision.

We will take the help of a diagram which is drawn according to the question below.

For the object X the difference between the momentums will be Δp=momentum of X after colliding Y- momentum of X\Delta p=\text{momentum of X after colliding Y- momentum of X}. Since, after colliding Y, J will get negative for X so, this can be written as Δp=pxp-\Delta p={{p}_{x}}-p. Thus, by formula J=ΔpJ=\Delta p we get
J=pxp-J={{p}_{x}}-p
px=pJ\Rightarrow {{p}_{x}}=p-J ….(i)

Similarly, For the object Y the difference between the momentums will be Δp=momentum of Y after colliding X- momentum of Y\Delta p=\text{momentum of Y after colliding X- momentum of Y}. Since, the momentum of Y was 0 so, this can be written as Δp=py0\Delta p={{p}_{y}}-0. Thus, by formula J=ΔpJ=\Delta p we get J=pyJ={{p}_{y}}….(ii).

By the coefficient of restitution formula, we get
e=pypx0p ep=pypx \begin{aligned} & -e=\dfrac{{{p}_{y}}-{{p}_{x}}}{0-p} \\\ & \Rightarrow ep={{p}_{y}}-{{p}_{x}} \\\ \end{aligned}
Now, we will use (i) and (ii) to get,
ep=J(pJ) ep=Jp+J ep=2Jp e=2Jpp e=2Jp1 \begin{aligned} & ep=J-\left( p-J \right) \\\ & \Rightarrow ep=J-p+J \\\ & \Rightarrow ep=2J-p \\\ & \Rightarrow e=\dfrac{2J-p}{p} \\\ & \Rightarrow e=\dfrac{2J}{p}-1 \\\ \end{aligned}

So, the correct answer is “Option A”.

Note:
We have to make a correct diagram to start the question because if we don’t form a diagram correctly then, it means the question is misunderstood by us or the value will get wrong. The negative sign of coefficient restitution is important here. As its value should never be negative this is why the use of negative is applied otherwise, the solution will get into a negative value. As after colliding Y, the impulse will be opposite in the same direction because Y was at rest and X was coming with impulse so, this is why after colliding, J will become negative for X.