# Topic: Laws of Motion (Test 1)

Topic: Laws of Motion
Q.1
In which of the following cases is the net force zero
A. a high-speed electron in space far from all material objects, and free of electric and magnetic fields
B. a coin falling under gravity
C. a rocket just before lift off
D. a feather falling under gravity
Explaination / Solution:

As no external fields like, electric,magnetic and gravitational field acting on electron the net force is zero

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Q.2
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, during its upward motion, Ignore air resistance
A. 0
B. 0.05 N vertically down
C. 0.5 N vertically down
D. 0.5 N vertically up
Explaination / Solution:

F = mg = 0.05 x 10 = 0.5 N

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Q.3
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, during its downward motion, Ignore air resistance
A. 0.5 N vertically up
B. 0
C. 0.5 N vertically down
D. 0.05 N vertically down
Explaination / Solution:

F = mg = 0.05 x 10 = 0.5 N downwards

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Q.4
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, if the pebble was thrown at an angle of 45° with the horizontal direction, Ignore air resistance
A. 0
B. 0.5 N vertically down
C. 0.5 N vertically up
D. 0.05 N vertically down
Explaination / Solution:

F = mg = 0.05 x 10 = 0.5 N. The horizontal component of the velocity remain constant.

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Q.5
Neglecting air resistance, in which of the following cases the net force acting on a stone of mass 0.1 kg is not acting downward?
A.
just after it is dropped from the window of a train running at a constant velocity of 36 km/h

B.
just after it is dropped from the window of a train accelerating with 1 m
C. lying on the floor of a train which is accelerating with 1 m S-2 , the stone being at rest relative to the train
D. just after it is dropped from the window of a stationary train
Explaination / Solution:

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms-2

force in horizontal direction

F = ma = 0.1 x 1 = 0.1N

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Q.6

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 m , the stone being at rest relative to the train. Neglect air resistance.

A. 1.0 N in the direction of motion
B. 0.1 N along the direction of motion
C. 0.2 opposite to the direction of motion
D. 0.1 N opposite to the direction of motion
Explaination / Solution:

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms-2

force in horizontal direction

F = ma = 0.1 x 1 = 0.1N

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Q.7
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, T is the tension in the string, the net force on the particle directed towards the centre is
A. T
B. T+mv2l
C. Tmv2l
D. 0
Explaination / Solution:

According to Newton's third law of motion, the net force on a rotating particle is equal to Tension in the String. As the action (ie force towards the centre) and reaction (tension in the string) are equal in magnitude and opposite in direction.

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Q.8

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m . How long does the body take to stop?

A. 4.0 s
B. 6.0 s
C. 4.0 s
D. 5.0 s
Explaination / Solution:

F=mdvdtdt=mdvF=20×1550=6s
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Q.9
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m  to 3.5 m  in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
A. 0.07 m , 0.16 N in the direction of motion
B. 0.08 m , 0.14 N in
C. 0.18 N along the direction of motion
D. 0.09 m , 0.12 N in the direction of motion
Explaination / Solution:

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Q.10
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the net force acting on the body.
A. 14 N at an angle of 37 to the 8 N force
B. 16 N at an angle of 46 to the 8 N force
C. 12 N at an angle of 3652 towards the 8 N force
D. 12 N at an angle of 38 to the 8 N force
Explaination / Solution:

Let F1 = 8N & F2 = 6 N.

then the resultant force is

=

= 10N

The direction is determined as

= 6/8

therefore

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