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JEE Main 2013 Tes Paper

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Question : 3 of 8

Marks: +1, -0

This question has statement I and Statement II. Of the four choice given after the statements, choose the one that best describes the two statements.
Statement-I: A Point particle of mass m moving with speed v collides with stationary point particle of mass M . If the maximum energy loss possible is given as

\mathrm{f}\left(\frac{1}{2} \mathrm{mv}^2\right) \text { then } \mathrm{f}=\left(\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}\right) .

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

Statement - I is true, Statement - II is true, statement - II is a correct explanation of Statement - I
Statement - I is true, Statement - II is true, statement - II is not a correct explanation of Statement -I
Statement - I is true, Statement - II is false
Statement-I is false, Statement-II is true

Solution:

Maximum energy loss when inelastic collision takes place

\mathrm{mv}=(\mathrm{m}+\mathrm{M}) \mathrm{v}^{\prime}

\mathrm{v}^{\prime}=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}} \mathrm{v}

\mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2

\mathrm{k}_{\mathrm{f}}=\frac{1}{2}(\mathrm{~m}+\mathrm{M}) \mathrm{v}^{\prime 2}=\frac{1}{2}(\mathrm{~m}+\mathrm{M}) \frac{\mathrm{m}^2 \mathrm{v}^2}{(\mathrm{~m}+\mathrm{M})^2}=\frac{1}{2} \mathrm{mv}^2\left(\frac{\mathrm{~m}}{\mathrm{M}+\mathrm{m}}\right)

Loss of energy

=k_i-k_f=\frac{1}{2} m v^2\left[1-\frac{m}{M+m}\right]=\frac{M}{M+m} \times \frac{1}{2} m v^2

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