Force and Laws of Motion — NCERT Solutions, AI Tutor & Practice

Force is a push or a pull acting on an object that tends to change its state of rest or of uniform motion, or to change its shape.

Two effects of force:

1. It can change the speed of an object (make it move faster, slower, or start/stop it).
2. It can change the direction of motion of an object.

(Other valid effects: it can change the shape or size of an object.)

Newton's First Law of Motion: An object continues in its state of rest or of uniform motion in a straight line unless it is acted upon by an external unbalanced force.

Inertia is the natural tendency of a body to resist any change in its state of rest or of uniform motion.

• A body at rest tends to stay at rest.
• A body in motion tends to keep moving with the same speed in the same direction.

The mass of a body is a measure of its inertia — greater mass means greater inertia.

This happens because of inertia of rest.

Step 1: When the bus is stationary, the passenger's whole body is at rest.

Step 2: When the bus suddenly starts, the lower part of the body (in contact with the bus) moves forward along with the bus.

Step 3: The upper part of the body, due to its inertia of rest, tends to remain at rest.

Conclusion: As a result, the upper body lags behind and the passenger appears to fall backward.

Momentum is the product of the mass and velocity of a body. It measures the 'quantity of motion' in a body.

Formula: $p = m \times v$ where $m$ = mass and $v$ = velocity.

SI unit: $\text{kg m s}^{-1}$ (kilogram metre per second).

Nature: Momentum is a vector quantity — it has both magnitude and direction (its direction is the same as that of the velocity).

Newton's Second Law: The rate of change of momentum of a body is directly proportional to the applied unbalanced force and takes place in the direction of the force.

Derivation:

Let a body of mass $m$ have initial velocity $u$ and final velocity $v$ after time $t$.

Initial momentum $= mu$, Final momentum $= mv$.

Rate of change of momentum $= \dfrac{mv - mu}{t} = \dfrac{m(v-u)}{t}$

By the law, force $F \propto \dfrac{m(v-u)}{t}$

Since $\dfrac{v-u}{t} = a$ (acceleration): $F \propto ma \quad \Rightarrow \quad F = kma$

The unit of force (1 newton) is defined so that $k = 1$. Hence: $\boxed{F = ma}$

Given: Force $F = 20\ \text{N}$, mass $m = 5\ \text{kg}$.

Formula: $F = ma \Rightarrow a = \dfrac{F}{m}$

Substitution: $a = \frac{20}{5} = 4\ \text{m s}^{-2}$

Answer: The acceleration produced is $4\ \text{m s}^{-2}$.

Given: mass $m = 1000\ \text{kg}$, initial velocity $u = 20\ \text{m s}^{-1}$, final velocity $v = 0$, time $t = 5\ \text{s}$.

Step 1 — Find acceleration: $a = \frac{v - u}{t} = \frac{0 - 20}{5} = -4\ \text{m s}^{-2}$

Step 2 — Find force: $F = ma = 1000 \times (-4) = -4000\ \text{N}$

Answer: A force of $4000\ \text{N}$ acts opposite to the motion (retarding force). The negative sign shows it opposes the car's motion.

Newton's Third Law: To every action there is an equal and opposite reaction. The two forces act on different bodies.

If body A exerts a force on body B, then B exerts an equal and opposite force on A.

Examples:

1. Walking: We push the ground backward with our feet (action); the ground pushes us forward (reaction), allowing us to move.
2. Recoil of a gun: When a bullet is fired forward (action), the gun recoils backward (reaction).

(Other example: a swimmer pushes water backward and moves forward.)

Law of Conservation of Momentum: In the absence of an external unbalanced force, the total momentum of a system of bodies remains constant. That is, total momentum before an interaction = total momentum after.

Application to gun and bullet:

Before firing, both gun and bullet are at rest, so total momentum $= 0$.

Let gun mass $= M$, recoil velocity $= V$; bullet mass $= m$, velocity $= v$.

By conservation of momentum: $0 = mv + MV$ $MV = -mv \quad \Rightarrow \quad V = -\frac{mv}{m}$ $V = -\frac{mv}{M}$

The negative sign shows the gun moves backward (recoils), opposite to the bullet. Since $M \gg m$, the recoil velocity $V$ is small.

Given: bullet mass $m = 10\ \text{g} = 0.01\ \text{kg}$, bullet velocity $v = 400\ \text{m s}^{-1}$; gun mass $M = 5\ \text{kg}$, recoil velocity $= V$.

By conservation of momentum: total momentum before = total momentum after = 0 $mv + MV = 0$ $V = -\frac{mv}{M} = -\frac{0.01 \times 400}{5}$ $V = -\frac{4}{5} = -0.8\ \text{m s}^{-1}$

Answer: The recoil velocity of the gun is $0.8\ \text{m s}^{-1}$, directed opposite to the bullet.

Given: $m_A = 2\ \text{kg}$, $u_A = 3\ \text{m s}^{-1}$; $m_B = 4\ \text{kg}$, $u_B = 1\ \text{m s}^{-1}$.

By conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)\,v$

Substitution: $(2 \times 3) + (4 \times 1) = (2 + 4)\,v$ $6 + 4 = 6v$ $10 = 6v$ $v = \frac{10}{6} \approx 1.67\ \text{m s}^{-1}$

Answer: The common velocity after collision is about $1.67\ \text{m s}^{-1}$ in the original direction.

Understand: The ball has a large momentum, which must be reduced to zero when caught.

Analyze: From Newton's second law, force is the rate of change of momentum: $F = \frac{\Delta p}{t}$

The change in momentum $\Delta p$ is fixed (the ball must be stopped). By moving his hands backward, the cricketer increases the time $t$ over which the ball is stopped.

Conclude: Since $F$ is inversely proportional to $t$, increasing the time of contact decreases the force exerted on his hands. This prevents injury.