Laws of Motion

Question 521
CBSEENPH11020580

A car of mass 1000 kg negotiates a banked curve of radius 90m on a frictionless road. If the banking angle is 45o, the speed of the car is 

  • 20 ms-1

  • 30ms-1

  • 5 ms-1

  • 10 ms-1

Solution

B.

30ms-1

The angle of banking
tan space straight theta space equals space straight v squared over rg
Given comma space straight theta space equals space 45 to the power of straight o comma space straight r equals space 90 space straight m
space and space straight g space equals space 10 space straight m divided by straight s squared
tan space 45 to the power of straight o space equals space fraction numerator straight v squared over denominator 90 space straight x space 10 end fraction
straight v space equals space square root of 90 space straight x 10 space straight x space tan space 45 to the power of straight o end root
speed space of space car space straight v space equals space 30 space straight m divided by straight s

Question 522
CBSEENPH11020597

Two sphere A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along the x -axis. After the collision, B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction

  • same as that of B

  • opposite to that of B

  • straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x space minus axis
  • straight theta space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses space to space the space straight x space minus axis

Solution

C.

straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x space minus axis

Here, Pi = m2vi +m2 x 0

straight P subscript straight f space equals space straight m subscript 2 straight v over 2 straight j space plus space straight m subscript 1 space straight x space straight v subscript 2
Law space of space conservation space of space momentum
straight P subscript straight i space equals space straight P subscript straight f
straight m subscript 2 straight v subscript straight i space equals straight m subscript 2 space straight v over 2 straight j space plus straight m subscript 1 straight x space straight v subscript 1
straight v subscript 1 space equals space fraction numerator straight m subscript 2 over denominator straight m subscript 1 space end fraction vi space plus straight m subscript 2 over straight m subscript 1 straight v over 2 straight j
From space this space equation space we space can space find
tan space straight theta space equals space 1 half
straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x minus space axis

Question 523
CBSEENPH11020605

The upper half of an inclined plane of the inclination straight theta is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by,

  • straight mu space equals space fraction numerator 1 over denominator tan space straight theta end fraction
  • straight mu space equals space fraction numerator 2 over denominator tan space straight theta end fraction
  • straight mu space equals space 2 space tan space straight theta
  • straight mu space equals space tan space straight theta

Solution

C.

straight mu space equals space 2 space tan space straight theta
The block will be stationary when,

mg sin straight theta . L = straight alpha mg cos straight theta space. space straight L over 2
straight mu space equals space fraction numerator mg space sin space straight theta space straight L over denominator mg space cos space straight theta space begin display style bevelled straight L over 2 end style space end fraction

space space equals space 2 space fraction numerator sin space straight theta over denominator cos space straight theta end fraction

space space equals space 2 space tan space straight theta

straight mu equals space 2 space tan space straight theta
Question 524
CBSEENPH11020607

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m/s and the second part of mass 2 kg moves with second part of mass 2 kg moves with 8 m/s speed. If the third part flies off with 4 m/s speed, then its mass is,

  • 3 kg

  • 5 kg

  • 7 kg

  • 17 kg

Solution

B.

5 kg

We have,
p1 + p2 + p3 = 0 [ because p = mv]
Therefore,
1 x 12 i  + 2 x 8 j + p3 = 0
rightwards double arrow space 12 space straight i space plus space 16 space straight j space plus space straight p subscript 3 space equals space 0

rightwards double arrow space straight p subscript 3 space equals space minus left parenthesis 12 space straight i space space plus space 16 space straight j right parenthesis

therefore space straight p subscript 3 space equals space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root

space space space space space space space space space space equals space square root of 144 space plus space 256 space end root

space space space space space space space space space space equals space 20 space kg minus straight m divided by straight s
Now. p3 = m3 v3
This implies,


straight m subscript 3 space equals space straight p subscript 3 over straight v subscript 3 space
space space space space space equals space 20 over 4 space
space space space space space equals space 5 space kg

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