Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupee coin and a one-rupee coin?
Inertia is the tendency of an object to resist any change in its state of rest or motion. Inertia depends directly on mass — greater the mass, greater the inertia.
(a) Stone — The stone has greater mass than the rubber ball of the same size (stone is denser), so the stone has greater inertia.
(b) Train — The train has a much greater mass than the bicycle (a train can weigh hundreds of tonnes vs. ~10 kg for a bicycle), so the train has far greater inertia. This is why a train takes a very long distance to stop even after brakes are applied.
(c) Five-rupee coin — The five-rupee coin has a greater mass than the one-rupee coin (it is larger and heavier), so it has greater inertia.
In the following example, identify the number of times the velocity of the ball changes: "A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team." Also identify the agent supplying the force in each case.
The velocity of the football changes 4 times:
(1) First player kicks the ball: The ball changes from rest (velocity = 0) to moving with some velocity. Force agent: First player's foot.
(2) Second player (teammate) kicks the ball: The ball's velocity changes in magnitude (speed increases/decreases) and direction (directed towards goal). Force agent: Second player's foot.
(3) Goalkeeper collects (catches) the ball: The ball decelerates from its velocity to rest (or near rest) in the goalkeeper's hands. Velocity changes from moving to zero. Force agent: Goalkeeper's hands.
(4) Goalkeeper kicks the ball: The ball changes from rest (or slow) to moving in a new direction towards the goalkeeper's teammate. Force agent: Goalkeeper's foot.
In each case, the velocity of the ball changes because an external force is applied to it (Newton's First Law: velocity of an object changes only when a net external force acts on it).
Explain why some of the leaves may get detached from a tree if we vigorously shake its branches.
This phenomenon is due to the inertia of rest (Newton's First Law of Motion).
When the branches of a tree are vigorously shaken, the branches (and the tree) are set into rapid back-and-forth motion. However, the leaves — due to their inertia — tend to remain in their original position at rest and resist the sudden change of motion imposed on them by the shaking branches.
Since the connection between the leaves and the branches (the leaf stalk or petiole) is relatively weak and cannot transmit enough force to accelerate the leaves with the same rapid motion as the branches, the leaves "lag behind" relative to the branch. The relative displacement between the leaf and the branch becomes large enough to snap the petiole, causing the leaf to get detached.
In summary: Branch moves suddenly → Leaf (due to inertia of rest) remains behind → Stalk breaks → Leaf detaches.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backward when it accelerates from rest?
Both situations are explained by Newton's First Law of Motion (Inertia):
Falling forward when bus brakes:
When a moving bus applies brakes and suddenly stops, the bus decelerates rapidly. However, the passengers' bodies have inertia of motion — they tend to continue moving forward at the original speed. Since the feet (in contact with the floor) are stopped along with the bus, but the upper body continues moving forward due to inertia, the passenger leans or falls forward.
Falling backward when bus accelerates:
When the stationary bus suddenly accelerates forward, the bus floor moves forward rapidly. However, the passengers' bodies have inertia of rest — they tend to remain at their original position. Since the feet are pushed forward (by the floor) but the upper body stays back due to inertia, the passenger falls backward relative to the bus.
This is why seat belts are essential — they provide the force needed to accelerate (or decelerate) the passenger along with the vehicle, preventing injury.
If action is always equal to the reaction, explain how a horse can pull a cart.
By Newton's Third Law: for every action, there is an equal and opposite reaction. Action-reaction pairs act on different bodies, not the same body, so they do not cancel each other.
How the horse pulls the cart:
(i) The horse pushes backward on the ground with its hooves (action). The ground exerts an equal and opposite forward reaction force on the horse's hooves (this is friction). This forward force makes the horse move forward.
(ii) The horse pulls the cart forward through the yoke/harness (action on cart). The cart pulls the horse backward through the same connection (reaction on horse).
(iii) The cart can be pulled forward if the forward friction force from the ground on the horse is greater than the rolling friction / resistive force on the cart.
The key point: action and reaction act on different objects (horse and ground; horse and cart). So the net forward force on the horse − resistive force on the cart > 0 allows the system to accelerate forward. If the horse exerts enough force, the cart moves.
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force, (b) the acceleration of the train.
Given: Mass of engine = 8000 kg. Mass of each wagon = 2000 kg. Number of wagons = 5. Force by engine = 40000 N. Friction force = 5000 N.
Total mass of train:
(a) Net accelerating force:
(b) Acceleration — using F = ma:
The net accelerating force is 35000 N and the acceleration of the train is approximately 1.94 m/s2.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s2?
Given: Mass m = 1500 kg. Acceleration a = −1.7 m/s2 (deceleration / retardation).
Using F = ma:
The braking force between the vehicle and the road must be 2550 N, directed opposite to the direction of motion (i.e., the friction force from the road on the tyres acts as the braking force).
What is the momentum of an object of mass m moving with a velocity v?
The momentum (p) of an object of mass m moving with velocity v is defined as the product of its mass and velocity:
Key properties of momentum:
(i) It is a vector quantity — it has both magnitude and direction, and the direction of momentum is the same as the direction of velocity.
(ii) SI unit of momentum is kg·m/s (kilogram-metre per second) or equivalently N·s (Newton-second).
(iii) Momentum is a measure of the "quantity of motion" of an object. A heavy truck moving slowly can have the same momentum as a small bullet moving very fast.
(iv) Newton's Second Law can be stated as: Force = rate of change of momentum = (mv − mu)/t = m(v−u)/t = ma.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Given: Applied horizontal force = 200 N. The cabinet moves at constant velocity (uniform motion, zero acceleration).
Since the cabinet moves at constant velocity, by Newton's First Law, the net force on the cabinet = 0.
The friction force exerted on the cabinet is 200 N, directed opposite to the direction of motion. At constant velocity, the friction force exactly balances the applied force, resulting in zero net force and zero acceleration.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m/s before the collision. Find the total momentum of the two objects (a) before the collision and (b) after the collision if they stick together.
Given: m1 = m2 = 1.5 kg. Velocity of object 1 (forward): +2.5 m/s. Velocity of object 2 (backward): −2.5 m/s.
(a) Total momentum before collision:
(b) Total momentum after collision:
By the Law of Conservation of Momentum, total momentum after collision = total momentum before collision = 0 kg·m/s.
Since the two objects stick together (perfectly inelastic collision) and total momentum = 0, the combined mass (3 kg) moves with velocity = 0 m/s. The two objects come to rest after the collision.
State Newton's second law of motion. Derive the mathematical expression F = ma.
Statement of Newton's Second Law:
The rate of change of momentum of a body is directly proportional to the applied external force, and the change in momentum takes place in the direction of the force.
Mathematical Derivation:
Let a body of mass m have an initial velocity u. A net external force F acts on it for time t, changing its velocity to v.
Initial momentum: pi = mu
Final momentum: pf = mv
Change in momentum: Δp = mv − mu = m(v − u)
Rate of change of momentum:
Since acceleration a = (v − u)/t.
By Newton's Second Law: F ∝ Δp/t = ma
By choosing SI units (1 N is defined as the force that gives 1 kg an acceleration of 1 m/s2), k = 1, giving:
This is the mathematical expression for Newton's Second Law. Force (F) is measured in Newtons (N = kg·m/s2), mass (m) in kilograms, and acceleration (a) in m/s2.
A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?
Given: m = 5 kg. u = 3 m/s. v = 7 m/s. t = 2 s.
Acceleration:
Force — using F = ma:
If force applied for 5 s (same force = 10 N, same acceleration = 2 m/s2):
The applied force is 10 N. If applied for 5 s, the final velocity would be 13 m/s.
Which would require a greater force — accelerating a 2 kg mass at 5 m/s2 or a 4 kg mass at 2 m/s2?
Using F = ma for each case:
Since 10 N > 8 N, accelerating the 2 kg mass at 5 m/s2 requires a greater force (10 N) compared to accelerating the 4 kg mass at 2 m/s2 (8 N).
This shows that force depends on both mass AND acceleration. A smaller mass can require more force than a larger mass if it needs to be accelerated more rapidly.
A bullet of mass 20 g is horizontally fired with a velocity of 150 m/s from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
Given: Mass of bullet m1 = 20 g = 0.02 kg. Velocity of bullet v1 = +150 m/s. Mass of pistol m2 = 2 kg. Initial state: both at rest, total momentum = 0.
By Conservation of Momentum (total momentum before = total momentum after):
The recoil velocity of the pistol is 1.5 m/s in the direction opposite to the bullet's motion (indicated by the negative sign). This backward kick felt when firing a gun is the "recoil."
Two objects of masses 100 g and 200 g are moving along the same line in the same direction with velocities of 2 m/s and 1 m/s respectively. They collide and after collision, the first object moves at 1.67 m/s. Determine the velocity of the second object after collision.
Given: m1 = 100 g = 0.1 kg, u1 = 2 m/s. m2 = 200 g = 0.2 kg, u2 = 1 m/s. After collision: v1 = 1.67 m/s. Find v2.
Total momentum before collision:
Total momentum after collision:
By Conservation of Momentum (pbefore = pafter):
The velocity of the second object after collision is approximately 1.17 m/s in the same direction as before.
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