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Question: A ball of mass m moving with speed u undergoes a head – on elastic collision with a ball of mass m i...

A ball of mass m moving with speed u undergoes a head – on elastic collision with a ball of mass m initially at rest. The fraction of the Incident energy transferred to the second ball is
A. n1+n\dfrac{n}{1+n}
B. (n1+n)2{{\left( \dfrac{n}{1+n} \right)}^{2}}
C. 2n1+n2\dfrac{2n}{1+{{n}^{2}}}
D. 4n(1+n)2\dfrac{4n}{{{\left( 1+n \right)}^{2}}}

Explanation

Solution

Hint: During the head on elastic collision coefficient of restitution (e) will be equal to 1 (one). In this question the energy transferred by a small ball is equal to the final energy of a heavier ball.

Complete step-by-step answer:
As given, the Ball of Mass ‘m’ moving with speed ‘u’ has a head on collision with a ball of mass ‘nm’ which is initially at rest.
For elastic collision e=Ie=I
e=Velocity of separation Velocity of approache=\dfrac{\text{Velocity of separation}}{\text{ Velocity of approach}}
⇒e=V2−V1u1−u2\Rightarrow e=\dfrac{{{V}_{2}}-{{V}_{1}}}{{{u}_{1}}-{{u}_{2}}}
⇒I=V2−V1u−0\Rightarrow I=\dfrac{{{V}_{2}}-{{V}_{1}}}{u-0}
⇒V2−V1 ........ 1\Rightarrow {{V}_{2}}-{{V}_{1}}\text{ }........\text{ 1}
By Momentum conservation
Momentum before collision momentum after collision
⇒mu+nm×0\Rightarrow mu+nm\times 0
⇒u=V1×nV2 ........... 2\Rightarrow u={{V}_{1}}\times n{{V}_{2}}\text{ }...........\text{ 2}
By equation 1 and 2
V1+nV2=V2−V1{{V}_{1}}+n{{V}_{2}}={{V}_{2}}-{{V}_{1}}
⇒2V1=V2−nV1\Rightarrow 2{{V}_{1}}={{V}_{2}}-n{{V}_{1}}
⇒V1=V2(1−h)2 ........ 3\Rightarrow {{V}_{1}}=\dfrac{{{V}_{2}}\left( 1-h \right)}{2}\text{ }........\text{ 3}
By 1 and 3
u=V2−V2(1−h)2⇒V2=2u1+uu={{V}_{2}}-\dfrac{{{V}_{2}}\left( 1-h \right)}{2}\Rightarrow {{V}_{2}}=\dfrac{2u}{1+u}
⇒V2=(1−n1+n)u\Rightarrow {{V}_{2}}=\left( \dfrac{1-n}{1+n} \right)u
Initially incident energy =12mu2=\dfrac{1}{2}m{{u}^{2}}
Final Energy of heavier Ball = Energy transferred
=12mn.V22=\dfrac{1}{2}mn.{{V}_{2}}^{2}
=12mn.4u2(1+n)2=\dfrac{1}{2}mn.\dfrac{4{{u}^{2}}}{{{\left( 1+n \right)}^{2}}}
=2u2.mn(1+n)2=\dfrac{2{{u}^{2}}.mn}{{{\left( 1+n \right)}^{2}}}
Required fraction of Incident energy
=2u2.mn(1+n)212mu2=4u(1+n)2=\dfrac{\dfrac{2{{u}^{2}}.mn}{{{\left( 1+n \right)}^{2}}}}{\dfrac{1}{2}m{{u}^{2}}}=\dfrac{4u}{{{\left( 1+n \right)}^{2}}}
Therefore, the correct choice is: (D) 4n(1+n)2\dfrac{4n}{{{\left( 1+n \right)}^{2}}}

Note: The collision problem required the coefficient of restitution (e) and the value of 'e’ depends on the type or collision. Also it requires the conservation of momentum and conservation of energy.
Coefficient of restitution ranges from 0 to 1 where 1 would be a perfectly elastic collision.