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Laws Of Motion

Question
CBSEENPH11020686

A ball moving with velocity 2 ms-1 collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms-1) after collision will be

  • 0,1

  • 1,1

  • 1,0.5

  • 0,2

Solution

A.

0,1

If two bodies collide head on with coefficient of restitution
straight e space equals space fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
From space the space law space of space conservation space of space linear space momentum

straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus straight m subscript 2 straight v subscript 2
rightwards double arrow space straight v subscript 1 space equals space open square brackets fraction numerator straight m subscript 1 minus em subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 plus open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
Substituting space straight u subscript 1 space equals space 2 space ms to the power of negative 1 end exponent comma space straight u subscript 2 space equals 0 comma space straight m subscript 1 equals straight m subscript 2 space and space straight m subscript 2 space equals space 2 straight m comma space straight e space equals space 0.5
we space get space space straight v subscript 1 space equals space open square brackets fraction numerator straight m minus straight m over denominator straight m plus 2 straight m end fraction close square brackets space straight x 2
similarly comma
straight v subscript 2 space equals open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 space plus open square brackets fraction numerator straight m subscript 2 minus em subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
equals space open square brackets fraction numerator 1.5 space straight x space straight m over denominator 3 straight m end fraction close square brackets space straight x space 2
space equals space 1 space ms to the power of negative 1 end exponent