Laws of Motion

Question
CBSEENPH11020350

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 over 3
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 half
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 third
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2

Solution

D.

straight lambda subscript straight A over straight lambda subscript straight B space equals space 2
By conservation of linear momentum
mv space equals space mv subscript 1 space plus straight m over 2 straight v subscript 2
2 straight v space equals space 2 straight v subscript 1 space plus straight v subscript 2 space... space left parenthesis 1 right parenthesis
By space law space of space collision
straight e space equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
straight u space equals space straight v subscript 2 minus straight v subscript 1 space space... space left parenthesis 2 right parenthesis
by space equ. space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
straight v subscript 1 space equals space straight v over 3 semicolon space straight v subscript 2 space equals space fraction numerator 4 straight v over denominator 3 end fraction
straight lambda subscript 1 space equals space straight h over straight p subscript 1 semicolon space straight lambda subscript 2 space equals space straight h over straight o subscript 2
straight lambda subscript 1 over straight lambda subscript 2 space equals space 2 over 1

Question
CBSEENPH11020376

A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as

  • 17.6 kg ms−1 

  • 17.565 kg ms−1

  • 17.56 kg ms−1

  • 17.57 kg ms−1

Solution

A.

17.6 kg ms−1 

P = mv = 3.513 × 5.00 ≈ 17.6

Question
CBSEENPH11020382

A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

  • 0.16 J

  • 1.00 J

  • 0.67 J

  • 0.34 J

Solution

C.

0.67 J

m1u1 + m2u2 = (m1 + m2)v
v = 2/3 m/s
Energy space loss space equals space 1 half space left parenthesis 0.5 right parenthesis space straight x space left parenthesis 2 right parenthesis squared space minus space 1 half space left parenthesis 1.5 right parenthesis space straight x space open parentheses 2 over 3 close parentheses squared
space equals space 0.67 space straight J

Question
CBSEENPH11020392

A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (mass less) of spring constant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is stretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’.

  • mF/M

  • (M+m)F/m

  • mF/(m+ M)

  • MF (m+M)

Solution

C.

mF/(m+ M)