Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

125 of 130

Question 125 of 130

This question has Statement−IStatement - I and Statement−IIStatement - II of the four choices given after the Statements, choose the one that best describes the two Statements.

Statement−I:Statement - I: A point particle of mass mm moving with speed ν\nu collides with stationary point particle of mass MM. If the maximum energy loss possible is given as
f(12mν2)f\left(\dfrac{1}{2}m{\nu }^{2}\right) then f=(mM+m)f=\left(\dfrac{m}{M+m}\right).

Statement−II:Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Show answer and solution

Answer: (B) Statement-II is false, Statement-IIII is true.

Maximum energy loss will take place in perfectly inelastic collision, and is given by, ∆K=P22m−P22(m+M)∆K=\dfrac{{P}^{2}}{2m}-\dfrac{{P}^{2}}{2\left(m+M\right)}
⇒∆K=P22m[1−m(m+M)]\Rightarrow ∆K=\dfrac{{P}^{2}}{2m}\left[1-\dfrac{m}{\left(m+M\right)}\right]
⇒∆K=P22m[M(m+M)]\Rightarrow ∆K=\dfrac{{P}^{2}}{2m}\left[\dfrac{M}{\left(m+M\right)}\right]
⇒∆K=12mv2[M(m+M)]\Rightarrow ∆K=\dfrac{1}{2}m{v}^{2}\left[\dfrac{M}{\left(m+M\right)}\right]
On comparing with the given form of equation, we have, (f=Mm+M)\left(f=\dfrac{M}{m+M}\right)

Hence Statement-II is false, Statement-IIII is true. ( As it is perfect Inelastic collision)

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