Laws of Motion

More Topic from Physics

Question 1

A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2 )
  • 2.0

  • 4.0

  • 1.6 

  • 2.4

Solution

A.

2.0

Let the mass of block be m.
Frictional force in rest position
F = mg sin 30

10 space equals space straight m space straight x space 10 space straight x space 1 half
space straight m space equals space fraction numerator 2 space straight x space 10 over denominator 10 end fraction space equals space 2 space kg

Question 2

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

  • h/9 metres from the ground

  • 7h/9 metres from the ground

  • 8h/9 metres from the ground

  • 17h/18 metres from the ground.

Solution

C.

8h/9 metres from the ground


second law of motion
straight s space equals ut space plus space 1 half space plus gT squared
or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis
therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root
At space straight t space equals space straight T over 3 straight s comma
straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared
rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9
rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses
therefore s = h/9 m
Hence, the position of ball from the ground= h- h/9 = 8h/9 m
Question 3

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2

  • 22 N

  • 4 N

  • 20 N

  • 30 N

Solution

C.

20 N

mgh = Fs
F = 20 N

Question 4

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)

Question 5

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