When a car is taking a circular turn on a horizontal road, the centripetal force is the force of:
- AFriction.
- BWeight of the car.
- CWeight of the tyres.
Question types
Laws of Motion question types
506 questions across 6 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.
506
Questions
6
Question groups
5
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M.C.Q (1 Marks)
80 Q→021 Marks Question
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98 Q→054 Marks Question
8 Q→065 Marks Questions
109 Q→Sample Questions
Laws of Motion questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
A book is lying on the table. What is the angle between the action of the book on the table and reaction of the table on the book?
- A0°
- B30°
- C45°
- D180°
A trolley is carrying a box on its surface having coefficient of static friction equal to 0.3. Now the trolley starts moving with increasing acceleration. Find the maximum acceleration of the trolley so that the box does not slide back on the trolley.
- A2ms-2
- B3ms-2
- C4ms-2
- D5ms-2
A particle of man 2kg is moving on a circular path of radius 10m with a speed of 5ms and its speed is increasing at rate of 3ms-1. Find the force acting on the particle.
- A5N
- B10N
- C12N
- D14N
A body of mass 2kg travels according to the law x(t) = pt + qt2 rt3 where p = 3m s-1, q = 4m s-2 and r = 5m s-3. The force acting on the body at t = 2 seconds is.
- A136N
- B134N
- C158N
- D68N
If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
- The stone moves radially outwards.
- The stone flies off tangentially from the instant the string breaks.
- The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Give the magnitude and direction of the net force acting on.
A kite skillfully held stationary in the sky.
A kite skillfully held stationary in the sky.
A monkey of mass 40kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey, 
- Climbs up with an acceleration of 6ms-2.
- Climbs down with an acceleration of 4ms-2.
- Climbs up with a uniform speed of 5ms-1.
- Falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).
Give the magnitude and direction of the net force acting on.
A drop of rain falling down with a constant speed.
A drop of rain falling down with a constant speed.
Give the magnitude and direction of the net force acting on.
A car moving with a constant velocity of 30km/h on a rough road.
A car moving with a constant velocity of 30km/h on a rough road.
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.)
Just after it is dropped from the window of a train accelerating with 1ms-2.
Just after it is dropped from the window of a train accelerating with 1ms-2.
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.)
Just after it is dropped from the window of a train running at a constant velocity of 36km/h.
Just after it is dropped from the window of a train running at a constant velocity of 36km/h.
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.)
Lying on the floor of a train which is accelerating with 1ms-2, the stone being at rest relative to the train.
Lying on the floor of a train which is accelerating with 1ms-2, the stone being at rest relative to the train.
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.)
Just after it is dropped from the window of a stationary train.
Just after it is dropped from the window of a stationary train.
The motion of a particle of mass $m$ is described by $y=u t+\frac{1}{2} g t^2$. Find the force acting on the particle.
View full solution →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 the net force on the particle (directed towards the centre) is:
View full solution →- $\text{T}.$
- $\text{T}-\frac{\text{m}\nu^2}{\text{l}}$
- $\text{T}+\frac{\text{m}\nu^2}{\text{l}}$
- $0$
A bob of mass 0.1kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of the bob at its mean position is 1ms-1 . What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15ms-1. How long does the body take to stop?
A man of mass 70kg stands on a weighing scale in a lift which is moving.
Downwards with a uniform acceleration of 5ms-2.
Downwards with a uniform acceleration of 5ms-2.
A man of mass 70kg stands on a weighing scale in a lift which is moving.
Upwards with a uniform speed of 10 ms-1.
Upwards with a uniform speed of 10 ms-1.
Read the passage given below and answer the following questions from (i) to (v). This principle is a consequence of Newton’s second and third laws of motion. In an isolated system (i.e. a system having no external force), mutual forces (called internal forces) between pairs of particles in the system causes momentum change in individual particles. Let a bomb be at rest, then its momentum will be zero. If the bomb explodes into two equal parts, then the parts fly off in exactly opposite directions with same speed, so that the total momentum is still zero. Here, no external force is applied on the system of particles (bomb).
View full solution →- A bullet of mass 10g is fired from a gun of mass 1kg with recoil velocity of gun 5m/ s. The muzzle velocity will be:
- 30km/ min
- 60km/ min
- 30m/ s
- 500m/ s
- A shell of mass 10 kg is moving with a velocity of 10 ms" 1 when it blasts and forms two parts of mass 9 kg and 1 kg respectively. If the first mass is stationary, the velocity of the second is:
- 1ms-1
- 10ms-1
- 100ms-1
- 1000ms-1
- A bullet of mass 0.1kg is fired with a speed of 100ms’ 1 . The mass of gun being 50kg, then the velocity of recoil becomes:
- 0.05ms-1
- 0.5ms-1
- 0.lms-1
- 0.2ms-1
- A unidirectional force F varying with time Tas shown in the figure acts on a body initially at rest for a short duration:
2.T. Then, the velocity acquired by the body is
- $\frac{\pi\text{F}_\text{o}\text{T}}{4\text{m}}$
- $\frac{\pi\text{F}_\text{o}\text{T}}{2\text{m}}$
- $\frac{\text{F}_\text{o}\text{T}}{4\text{m}}$
- $\text{zero}$
- Two masses ofM and 4Af are moving with equal kinetic energy. The ratio of their linear momenta is
- 1 : 8
- 1 : 4
- 1 : 2
- 4 : 1
Read the passage given below and answer the following questions from 1 to 5.
The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is net external force acting on the body. It relates the net external force to the acceleration of the body.
These qualitative observations lead to the second law of motion expressed by Newton as follow:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval $\triangle\text{t},$ the velocity of a body of mass m changes from v to $\text{v}+\triangle\text{v},$ i.e. its initial momentum
p = m v changes by $\triangle\text{p}=\text{m}\triangle\text{v}.$ According to the Second Law
$\text{F}\alpha\frac{\triangle\text{p}}{\triangle\text{t}}$ or $\text{F}=\text{k}\frac{\triangle\text{P}}{\triangle\text{t}}$
Where k is a constant of proportionality. Mathematically,
F = ma, the unit of force is kg-m/s2 or Newton, which has the symbol N. Let us note at this stage some important points about the second law:
View full solution →The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is net external force acting on the body. It relates the net external force to the acceleration of the body.
These qualitative observations lead to the second law of motion expressed by Newton as follow:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval $\triangle\text{t},$ the velocity of a body of mass m changes from v to $\text{v}+\triangle\text{v},$ i.e. its initial momentum
p = m v changes by $\triangle\text{p}=\text{m}\triangle\text{v}.$ According to the Second Law
$\text{F}\alpha\frac{\triangle\text{p}}{\triangle\text{t}}$ or $\text{F}=\text{k}\frac{\triangle\text{P}}{\triangle\text{t}}$
Where k is a constant of proportionality. Mathematically,
F = ma, the unit of force is kg-m/s2 or Newton, which has the symbol N. Let us note at this stage some important points about the second law:
- In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law.
- The second law of motion is a vector law.
- The second law of motion given by is applicable to a single point particle as well as to the rigid body but internal forces is not considered in F.
- The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant.
- SI unit of force is:
- Newton
- Pascal
- m/s
- None of the above
- According to second law of motion The rate of change of momentum of a body is directly proportional to
- Velocity of body
- Applied force
- Only mass of body
- None of the above.
- The second law of motion is:
- Vector law
- Scalar law
- State second law of motion.
- Write a note on 2nd law of motion. Enlist some deductions from 2nd law.
Read the passage given below and answer the following questions from (i) to (v).
Momentum and Newton’s Second Law of Motion Momentum of a body is the quantity of motion possessed by the body. It depends on the mass of the body and the velocity with which it moves. When a bullet is fired by a gun, it can easily pierce human tissue before coming to rest resulting in casualty. The same bullet fired with moderate speed will not cause much damage. The greater the change in momentum in a given time, the greater is the force that needs to be applied. The second law of motion refers to the general situation, where there is a net external force rating on the body.
View full solution →Momentum and Newton’s Second Law of Motion Momentum of a body is the quantity of motion possessed by the body. It depends on the mass of the body and the velocity with which it moves. When a bullet is fired by a gun, it can easily pierce human tissue before coming to rest resulting in casualty. The same bullet fired with moderate speed will not cause much damage. The greater the change in momentum in a given time, the greater is the force that needs to be applied. The second law of motion refers to the general situation, where there is a net external force rating on the body.
- A satellite in force-free space sweeps stationary interplanetary dust at a rate $\frac{\text{dM}}{\text{dt}}=\text{dv,}$ where M is the mass, v is the dt velocity of satellite and a is a constant. What is the deceleration of the satellite?
- $\frac{-2\text{av}^2}{\text{M}}$
- $\frac{-\text{av}^2}{\text{M}}$
- $\text{-av}^2$
- $\frac{\text{av}^2}{\text{M}}$
- A body of mass 5 kg is moving with velocity of $\text{v} = (2\hat{\text{i}} + 6\hat{\text{j}})$ ms-1 at t = 0s. After time t = 2s, velocity of body is $(10\hat{\text{i}} + 6\hat{\text{j}})$ ms-1, then change in momentum of body is:
- $40\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
- $20\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
- $30\hat{\text{i}} \text{kg}-\text{ms}^{-1}$
- $(50\hat{\text{i}}+30\hat{\text{j}})\text{ Kg - ms}^{-1}$
- A cricket ball of mass 0.25kg with speed 10m/ s collides with a bat and returns with same speed with in 0.01s. The force acted on bat is:
- 25N
- 50N
- 250N
- 500N
- A stationary bomb explodes into three pieces. One piece of 2 kg mass moves with a velocity of 8 ms” 1 at right angles to the other piece of mass 1 kg moving with a velocity of 12 ms -1 . If the mass of the third piece is 0.5 kg, then its velocity is:
- 10ms-1
- 20ms-1
- 30ms-1
- AOms-1
- A force of 10 N acts on a body of mass 0.5kg for 0.25s starting from rest. What is its momentum now?
- 0.25 N/ s
- 2.5 N/ s
- 0.5 N/ s
- 0.75 N/ s
Read the passage given below and answer the following questions from (i) to (v).
Force of Friction on Connected Bodies When bodies are in contact, there are mutual contact forces satisfying the third law of motion. The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction
In the above figure, 8 kg and 6 kg are hanging stationary from a rough pulley and are about to move. They are stationary due to roughness of the pulley.
View full solution →Force of Friction on Connected Bodies When bodies are in contact, there are mutual contact forces satisfying the third law of motion. The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction
In the above figure, 8 kg and 6 kg are hanging stationary from a rough pulley and are about to move. They are stationary due to roughness of the pulley.
- Which force is acting between pulley and rope?
- Gravitational force
- Tension force
- Frictional force
- Buoyant force
- The normal reaction acting on the system is
- 8g
- 6g
- 2g
- 4g
- The tension is more on side having mass of:
- 8kg
- 6kg
- Same on both
- Nothing can be said
- The force of friction acting on the rope is:
- 20N
- 30N
- 40N
- 50N
- Coefficient of friction of the pulley is
- $\frac{1}{6}$
- $\frac{1}{7}$
- $\frac{1}{5}$
- $\frac{1}{4}$
Read the passage given below and answer the following questions from 1 to 5.
Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p:
p = m v
Momentum is clearly a vector quantity. SI unit is kg m/s. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) is parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.
Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p:
p = m v
Momentum is clearly a vector quantity. SI unit is kg m/s. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) is parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.
- If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
- Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.
- SI unit of momentum is:
- Kgm/s
- Kgm/s2
- m/s2
- None of these
- Momentum is:
- Scalar quantity
- Vector quantity
- Define momentum. Give its SI unit.
- Explain with example how mass of body is important for determining effect of force on its motion?
- Explain with example how speed is important for determining effect of force on its motion?
A truck starts from rest and accelerates uniformly at 2.0ms-2. At t = 10s, a stone is dropped by a person standing on the top of the truck (6m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of:
- The force on the 7th coin (counted from the bottom) due to all the coins on its top.
- The force on the 7th coin by the eighth coin.
- The reaction of the 6th coin on the 7th coin.
A block of mass 15kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5ms-2 for 20s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Two bodies A and B of masses 5kg and 10kg in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between µs and µk.
A pebble of mass 0.05kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
- During its upward motion
- During its downward motion.
- At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
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