MCQS Questions
3.1 When we kick a stone, we get hurt. This is due to
(a) inertia
(b) velocity
(c) momentum
(d) reaction
Correct Option (a) inertia
Reason Inertia is the property of matter to resist changes in motion. When we kick a stone, its large inertia resists motion, and the reaction force hurts our foot.
3.2 An object will continue its motion with constant acceleration until
(a) the resultant force on it begins to decrease.
(b) the resultant force on it is zero.
(c) the resultant force on it begins to increase.
(d) the resultant force is at right angle to its tangential velocity.
Correct Option (b) the resultant force on it is zero.
Reason According to Newton’s First Law of Motion, an object in motion will remain in motion with constant velocity unless acted upon by a net force.
3.3 Which of the following is a non-contact force?
(a) Friction
(b) Air resistance
(c) Electrostatic force
(d) Tension in the string
Correct Option (c) Electrostatic force
Reason Electrostatic force acts at a distance and does not require physical contact, unlike the other forces mentioned.
3.4 A ball with initial momentum \( p \) hits a solid wall and bounces back with the same velocity. Its momentum \( p’ \) after collision will be:
(a) \( p’ = p \)
(b) \( p’ = -p \)
(c) \( p’ = 2p \)
(d) \( p’ = -2p \)
Correct Option (b) \( p’ = -p \)
Reason The ball bounces back with the same velocity but in the opposite direction, reversing the momentum (negative sign).
3.5 A particle of mass \( m \) moving with a velocity \( v \) collides with another particle of the same mass at rest. The velocity of the first particle after collision is:
(a) \( v \)
(b) \( -v \)
(c) \( 0 \)
(d) \( -1/2 \)
Correct Option (c) \( 0 \)
Reason After a perfectly elastic collision between two particles of equal mass, the moving particle transfers all its velocity to the stationary particle and comes to rest.
3.6 Conservation of linear momentum is equivalent to:
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) None of these
Correct Option (c) Newton’s third law of motion
Reason Conservation of momentum is a direct consequence of Newton’s Third Law of Motion, which states that every action has an equal and opposite reaction.
3.7 An object with a mass of 5 kg moves at constant velocity of 10 m/s. A constant force then acts for 5 seconds on the object and gives it a velocity of 2 m/s in the opposite direction. The force acting on the object is:
(a) \( 5 \, \text{N} \)
(b) \( -10 \, \text{N} \)
(c) \( -12 \, \text{N} \)
(d) \( -15 \, \text{N} \)
Correct Option (c) \( -12 \, \text{N} \)
Reason Using the equation \( F = \frac{\Delta p}{\Delta t} \):
Initial momentum = \( 5 \times 10 = 50 \, \text{kg·m/s} \)
Final momentum = \( 5 \times (-2) = -10 \, \text{kg·m/s} \)
Change in momentum = \( -10 – 50 = -60 \, \text{kg·m/s} \)
Force = \( \frac{-60}{5} = -12 \, \text{N} \).
3.8 A large force acts on an object for a very short interval of time. In this case, it is easy to determine:
(a) magnitude of force
(b) time interval
(c) product of force and time
(d) none of these
Correct Option (c) product of force and time
Reason The product of force and time is called impulse, which can be determined even if the force and time interval are not known individually.
3.9 A lubricant is usually introduced between two surfaces to decrease friction. The lubricant:
(a) decreases temperature
(b) acts as ball bearings
(c) prevents direct contact of the surfaces
(d) provides rolling friction
Correct Option (c) prevents direct contact of the surfaces
Reason Lubricants form a thin film between surfaces, reducing direct contact and thereby minimizing friction.
Short Questions
3.1 What kind of changes in motion may be produced by a force?
A force can produce the following changes in motion:
It can start or stop motion.
It can change the speed (increase or decrease).
It can change the direction of motion.
It can change the shape of an object.
3.2 Give 5 examples of contact forces.
Friction
Normal force
Air resistance
Tension in a string
Applied force (e.g., pushing or pulling an object)
3.3 An object moves with constant velocity in free space. How long will the object continue to move with this velocity?
The object will continue to move with constant velocity indefinitely unless acted upon by an external force (Newton’s First Law of Motion).
3.4 Define impulse of force.
Impulse is the product of force and the time duration for which it acts. It is given by
\[\text{Impulse} = F \times t\]
It is equal to the change in momentum of an object.
3.5 Why has not Newton’s first law been proved on the Earth?
Newton’s First Law cannot be directly observed on Earth because of external forces like friction and air resistance, which always act on moving objects and prevent them from moving indefinitely.
3.6 When sitting in a car which suddenly accelerates from rest, you are pushed back into the seat. Why?
When the car accelerates, your body tends to remain in its state of rest due to inertia. The seat pushes you forward, making it feel like you are being pushed back.
3.7 The force expressed in Newton’s second law is a net force. Why is it so?
Newton’s Second Law states that acceleration is caused by the net force acting on an object. The net force is the vector sum of all forces acting on the object.
3.8 How can you show that rolling friction is less than sliding friction?
By observing that it is easier to roll a heavy object (like a barrel) than to slide it on the ground. Rolling objects require less effort because rolling friction is smaller than sliding friction.
3.9 Define terminal velocity of an object.
Terminal velocity is the maximum velocity an object reaches when the downward force of gravity is balanced by the upward force of air resistance.
3.10 An astronaut walking in space wants to return to his spaceship by firing a hand rocket. In what direction does he fire the rocket?
The astronaut should fire the hand rocket in the direction opposite to the spaceship. By Newton’s Third Law, the reaction force will push him toward the spaceship.
Constructed Response Questions
3.1 Two ice skaters weighing 60 kg and 80 kg push off against each other on a frictionless ice track. The 60 kg skater gains a velocity of 4 m/s. Considering all the relevant calculations involved, explain how Newton’s third law applies to this situation.
By Newton’s Third Law, the force exerted by the 60 kg skater on the 80 kg skater is equal and opposite to the force exerted by the 80 kg skater on the 60 kg skater.
Momentum of 60 kg skater = \( m \times v = 60 \times 4 = 240 \, \text{kg·m/s} \)
By conservation of momentum, the momentum of the 80 kg skater will also be 240 kg·m/s but in the opposite direction.
Velocity of 80 kg skater = \( \frac{\text{Momentum}}{\text{Mass}} = \frac{240}{80} = 3 \, \text{m/s} \).
This shows equal and opposite reactions.
3.2 Inflatable air bags are installed in vehicles as safety equipment. In terms of momentum, what is the advantage of air bags over seatbelts?
Airbags increase the time of impact during a collision, reducing the force experienced by the person. By Newton’s Second Law, \( F = \frac{\Delta p}{\Delta t} \), increasing the time \( \Delta t \) reduces the force \( F \), making the collision less harmful.
3.3 A horse refuses to pull a cart. The horse argues, “according to Newton’s third law, whatever force I exert on the cart, the cart will exert an equal and opposite force on me. Since the net force will be zero, therefore, I have no chance of accelerating (pulling) the cart.” What is wrong with this reasoning?
The horse overlooks the fact that the cart moves due to the force exerted by the horse on the ground. The ground reacts with an equal and opposite force that helps the horse move forward, overcoming the force exerted by the cart.
3.4 When a cricket ball hits high, a fielder tries to catch it. While holding the ball, he/she draws hands backward. Why?
The fielder draws hands backward to increase the time of impact. This reduces the force acting on the hands, as \( F = \frac{\Delta p}{\Delta t} \). It prevents injury.
3.5 When someone jumps from a small boat onto the river bank, why does the jumper often fall into the water?
When the jumper pushes the boat backward, by Newton’s Third Law, the boat exerts an equal and opposite force. This backward movement of the boat causes instability, making the jumper likely to fall into the water.
3.6 Imagine that if friction vanishes suddenly from everything, then what could be the scenario of daily life activities?
If friction vanishes:
Vehicles will not move or stop.
Walking will become impossible.
Objects will slide indefinitely.
Machines will fail as their parts will slip.
Friction is essential for almost all activities in daily life.
Comprehensive Questions
3.1 Explain the concept of force by practical examples.
Force is a push or pull that changes an object’s state of motion or shape. Examples include:
Pushing a car A force is applied to move it forward.
Lifting a box A force is applied upward to overcome gravity.
Kicking a football A force is applied by the foot to change its motion.
Stretching a rubber band A force is applied to change its shape.
3.2 Describe Newton’s laws of motion.
First Law (Law of Inertia)
An object remains at rest or in uniform motion unless acted upon by an external force.
Example
A book stays on a table until pushed.
Second Law
The rate of change of momentum of an object is proportional to the net force acting on it. Mathematically, \( F = ma \).
Example
Pushing a lighter object results in greater acceleration than a heavier object with the same force.
Third Law
For every action, there is an equal and opposite reaction.
Example
When you jump, your legs push the ground backward, and the ground pushes you upward.
3.3 Define momentum and express Newton’s 2nd law of motion in terms of change in momentum.
Momentum Momentum is the product of an object’s mass and velocity. It is given by:
\[p = m \times v\]
Newton’s Second Law in terms of momentum
The net force is equal to the rate of change of momentum:
\[F = \frac{\Delta p}{\Delta t}\]
Where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval.
3.4 State and explain the principle of conservation of momentum.
Principle of Conservation of Momentum
The total momentum of an isolated system remains constant if no external force acts on it.
Explanation
In an isolated system, the momentum before an interaction is equal to the momentum after the interaction.
\[m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\]
Here:
\( m_1, m_2 \): masses of the objects
\( u_1, u_2 \): initial velocities
\( v_1, v_2 \): final velocities
Example: When two ice skaters push off each other, their total momentum remains constant.
3.5 Describe the motion of a block on a table taking into account the friction between the two surfaces. What is the static friction and kinetic friction?
Static Friction
The force that opposes the initiation of motion. It acts when the block is at rest.
Example
A block remains stationary on a table until a certain force is applied to overcome static friction.
Kinetic Friction
The force that opposes the motion of the block when it starts moving.
Example
Once the block starts sliding, kinetic friction acts to slow it down.
Motion on a Table
Initially, the applied force is balanced by static friction, so the block does not move.
When the applied force exceeds the maximum static friction, the block starts to slide.
Kinetic friction comes into play, which is less than static friction, making the block easier to move.
3.6 Explain the effect of friction on the motion of vehicles in the context of tire surface and braking force.
Friction plays a crucial role in the motion of vehicles
Tire Surface
The rough surface of tires increases friction with the road, providing better grip for motion and turning.
Smooth or worn-out tires reduce friction, causing slipping and loss of control.
Braking Force
Friction between brake pads and wheels helps slow down or stop a vehicle.
Without sufficient friction, brakes would fail to stop the vehicle effectively.
Thus, friction ensures safe driving and effective braking.
Numerical Problems
3.1 A 10 kg block is placed on a smooth horizontal surface. A horizontal force of 5 N is applied to the block. Find:
(a) The acceleration produced in the block.
(b) The velocity of the block after 5 seconds.
To Find
(a) Acceleration \( a \)
(b) Velocity \( v \) after \( t = 5 \, \text{s} \).
Solution
Given
Mass \( m = 10 \, \text{kg} \)
Force \( F = 5 \, \text{N} \)
Time \( t = 5 \, \text{s} \).
(a) Acceleration
Using Newton’s second law:
\[a = \frac{F}{m} = \frac{5}{10} = 0.5 \, \text{m/s}^2\]
(b) Velocity after 5 seconds
\[v = u + at\]
Initial velocity \( u = 0 \)
\[v = 0 + (0.5 \times 5) = 2.5 \, \text{m/s}\]
(a) \( a = 0.5 \, \text{m/s}^2 \)
(b) \( v = 2.5 \, \text{m/s} \)
3.2 The mass of a person is 80 kg. What will be his weight on the Earth? What will be his weight on the Moon? The value of acceleration due to gravity on the Moon is 1.6 m/s².
To Find
(a) Weight on Earth.
(b) Weight on Moon.
Solution
Given
Mass \( m = 80 \, \text{kg} \)
Acceleration due to gravity on Earth \( g = 9.8 \, \text{m/s}^2 \)
Acceleration due to gravity on Moon \( g_{\text{moon}} = 1.6 \, \text{m/s}^2 \).
(a) Weight on Earth
\[W = m \times g = 80 \times 9.8 = 784 \, \text{N} \, (\text{rounded to } 800 \, \text{N})\]
(b) Weight on Moon
\[W_{\text{moon}} = m \times g_{\text{moon}} = 80 \times 1.6 = 128 \, \text{N}\]
(a) \( W = 800 \, \text{N} \)
(b) \( W_{\text{moon}} = 128 \, \text{N} \)
3.3 What force is required to increase the velocity of 800 kg car from 10 m/s to 30 m/s in 10 seconds?
To Find
Force \( F \).
Solution
Given
Mass \( m = 800 \, \text{kg} \)
Initial velocity \( u = 10 \, \text{m/s} \)
Final velocity \( v = 30 \, \text{m/s} \)
Time \( t = 10 \, \text{s} \).
Using Newton’s second law
\[F = m \cdot a\]
First, find acceleration \( a \)
\[a = \frac{v – u}{t} = \frac{30 – 10}{10} = 2 \, \text{m/s}^2\]
Now calculate force
\[F = 800 \cdot 2 = 1600 \, \text{N}\]
\( F = 1600 \, \text{N} \)
3.4 A 5 g bullet is fired by a gun. The bullet moves with a velocity of 300 m/s. If the mass of the gun is 10 kg, find the recoil speed of the gun.
To Find
Recoil speed of the gun \( v_{\text{gun}} \).
Solution
Given
Mass of bullet \( m_{\text{bullet}} = 5 \, \text{g} = 0.005 \, \text{kg} \)
Velocity of bullet \( v_{\text{bullet}} = 300 \, \text{m/s} \)
Mass of gun \( m_{\text{gun}} = 10 \, \text{kg} \).
Using conservation of momentum:
\[m_{\text{bullet}} v_{\text{bullet}} = m_{\text{gun}} v_{\text{gun}}\]
Rearrange for \( v_{\text{gun}} \)
\[v_{\text{gun}} = \frac{m_{\text{bullet}} v_{\text{bullet}}}{m_{\text{gun}}} = \frac{0.005 \cdot 300}{10} = 0.15 \, \text{m/s}\]
Answer \( v_{\text{gun}} = -0.15 \, \text{m/s} \) (negative sign indicates opposite
direction).
3.5 An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m/s. Determine:
(a) The speed of the astronaut as he recoils away from the wrench.
(b) The distance covered by the astronaut in 30 minutes.
Solution
Given
Mass of astronaut \( m_{\text{astro}} = 70 \, \text{kg} \)
Mass of wrench \( m_{\text{wrench}} = 0.3 \, \text{kg} \)
Velocity of wrench \( v_{\text{wrench}} = 3.5 \, \text{m/s} \).
(a) Recoil Speed of Astronaut:
Using conservation of momentum
\[m_{\text{astro}} v_{\text{astro}} = -m_{\text{wrench}} v_{\text{wrench}}\]
Rearrange for \( v_{\text{astro}} \)
\[v_{\text{astro}} = -\frac{m_{\text{wrench}} v_{\text{wrench}}}{m_{\text{astro}}} = -\frac{0.3 \cdot 3.5}{70}\]
\[v_{\text{astro}} = -\frac{1.05}{70} = -0.015 \, \text{m/s}\]
(b) Distance Covered by the Astronaut in 30 Minutes
Time \( t = 30 \, \text{minutes} = 30 \cdot 60 = 1800 \, \text{s} \).
\[\text{Distance} = v_{\text{astro}} \cdot t = -0.015 \cdot 1800 = -27 \, \text{m}\]
(Note: The negative sign shows the direction opposite to the wrench.)
Correct Answer
(a) \( v_{\text{astro}} = -1.5 \times 10^{-2} \, \text{m/s} \)
(b) \( \text{Distance} = 27 \, \text{m} \)
3.6 A \( 6.5 \times 10^3 \, \text{kg} \) bogie of a goods train is moving with a velocity of 0.8 m/s. Another bogie of mass \( 9.2 \times 10^3 \, \text{kg} \) coming from behind with a velocity of 1.2 m/s collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.
To Find
The common velocity \( v \) after coupling.
Solution
Given
Mass of first bogie \( m_1 = 6.5 \times 10^3 \, \text{kg} = 6500 \, \text{kg} \)
Velocity of first bogie \( u_1 = 0.8 \, \text{m/s} \)
Mass of second bogie \( m_2 = 9.2 \times 10^3 \, \text{kg} = 9200 \, \text{kg} \)
Velocity of second bogie \( u_2 = 1.2 \, \text{m/s} \).
Using conservation of momentum
\[m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\]
Rearranging for \( v \):
\[v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}\]
Substitute the values
\[v = \frac{(6500 \cdot 0.8) + (9200 \cdot 1.2)}{6500 + 9200}\]
\[v = \frac{5200 + 11,040}{15,700} = \frac{16,240}{15,700} \approx 1.03 \, \text{m/s}\]
Answer \( v = 1.03 \, \text{m/s} \)
3.7 A cyclist weighing 55 kg rides a bicycle of mass 5 kg. He starts from rest and applies a force of 90 N for 8 seconds. Then he continues at a constant speed for another 8 seconds. Calculate the total distance travelled by the cyclist.
To Find
The total distance travelled by the cyclist.
Solution
Given
Total mass \( m = 55 + 5 = 60 \, \text{kg} \)
Force \( F = 90 \, \text{N} \)
Time for acceleration \( t = 8 \, \text{s} \).
Find acceleration \( a \)
\[a = \frac{F}{m} = \frac{90}{60} = 1.5 \, \text{m/s}^2\]
Find distance during acceleration
Use \( s = ut + \frac{1}{2}at^2 \)
\[s_1 = 0 \cdot 8 + \frac{1}{2} \cdot 1.5 \cdot (8)^2 = 0 + 0.75 \cdot 64 = 48 \, \text{m}\]
Find speed after 8 seconds
Use \( v = u + at \)
\[v = 0 + 1.5 \cdot 8 = 12 \, \text{m/s}\]
Find distance at constant speed for the next 8 seconds
\[s_2 = v \cdot t = 12 \cdot 8 = 96 \, \text{m}\]
Total distance travelled
\[s_{\text{total}} = s_1 + s_2 = 48 + 96 = 144 \, \text{m}\]
Answer \( s_{\text{total}} = 144 \, \text{m} \)
3.8 A ball of mass 0.4 kg is dropped on the floor from a height of 1.8 m. The ball rebounds straight upward to a height of 0.8 m. What is the magnitude and direction of the impulse applied to the ball by the floor?
To Find
Impulse magnitude and direction.
Solution
Given
Mass \( m = 0.4 \, \text{kg} \)
Initial height \( h_1 = 1.8 \, \text{m} \), rebound height \( h_2 = 0.8 \, \text{m} \).
Find velocity just before hitting the ground
Using \( v = \sqrt{2gh} \)
\[v_1 = \sqrt{2 \cdot 9.8 \cdot 1.8} = \sqrt{35.28} \approx 5.94 \, \text{m/s}\]
Find velocity just after rebounding
\[v_2 = \sqrt{2 \cdot 9.8 \cdot 0.8} = \sqrt{15.68} \approx 3.96 \, \text{m/s}\]
Find change in momentum (impulse)
Impulse \( I = \Delta p = m (v_2 – (-v_1)) = 0.4 \cdot (3.96 + 5.94) \)
\[I = 0.4 \cdot 9.9 = 3.96 \, \text{N·s}\]
Direction
The impulse acts **upward**, as the floor exerts the force to change the ball’s direction.
Answer \( I = 4 \, \text{N·s}, \text{upward} \)
3.9 Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities 20 m/s and 5 m/s respectively. After collision, the velocity of 0.2 kg ball becomes 6 m/s. What will be the velocity of the 0.4 kg ball?
Solution
Given
Mass \( m_1 = 0.2 \, \text{kg}, m_2 = 0.4 \, \text{kg} \).
Initial velocities: \( u_1 = 20 \, \text{m/s}, u_2 = -5 \, \text{m/s} \).
Final velocity of ball 1: \( v_1′ = 6 \, \text{m/s} \).
Using conservation of momentum
\[m_1 u_1 + m_2 u_2 = m_1 v_1′ + m_2 v_2’\]
Substitute values
\[(0.2 \cdot 20) + (0.4 \cdot -5) = (0.2 \cdot 6) + (0.4 \cdot v_2′)\]
\[4 – 2 = 1.2 + 0.4v_2’\]
\[2 = 1.2 + 0.4v_2’\]
\[0.8 = 0.4v_2′ \]
\[ v_2′ = \frac{0.8}{0.4} = 2 \, \text{m/s}\]
Correct Answer \( v_2′ = 2 \, \text{m/s} \)
