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Units And Measurement

Question
CBSEENPH11020286

This question has Statement I and Statement II. Of the four choices 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 f open parentheses 1 half mv squared close parentheses comma space then space straight f space equals space open parentheses fraction numerator straight m over denominator straight M plus straight m end fraction close parentheses
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

D.

Statement – I is false, Statement – II is true 

Before collision, the mass is m and after collision, the mass is m+M
therefore, Maximum energy loss
fraction numerator straight p squared over denominator 2 straight m end fraction minus fraction numerator straight p squared over denominator 2 left parenthesis straight m plus straight M right parenthesis end fraction
space equals space fraction numerator straight p squared over denominator 2 straight m end fraction open square brackets fraction numerator begin display style straight m end style over denominator straight m plus straight M end fraction close square brackets space space space space
space space space open square brackets because KE space equals space fraction numerator straight p squared over denominator 2 straight m end fraction close square brackets
equals space 1 half mv squared open curly brackets fraction numerator straight m over denominator straight m plus straight M end fraction close curly brackets
open square brackets straight f space equals space fraction numerator straight m over denominator straight m plus straight M end fraction close square brackets