Laws of Motion • Topic 2 of 3

Momentum, Impulse & Conservation

Momentum is the quantity of motion a body carries. A heavy truck rolling slowly and a fast cricket ball can both be hard to stop — what they share is large momentum. Linear momentum is defined as the product of a body's mass and velocity: $\vec{p}=m\vec{v}$. It is a vector pointing in the direction of motion, and its SI unit is $\text{kg}\cdot\text{m/s}$. Newton's second law is most fundamentally written as $\vec{F}=\frac{d\vec{p}}{dt}$ — force is the rate of change of momentum.

Impulse measures the total effect of a force acting over a time interval. When a force $\vec{F}$ acts for a time $\Delta t$, the impulse is $\vec{J}=\vec{F}\Delta t$. From the second law, this equals the change in momentum: $\vec{J}=\vec{F}\Delta t=\Delta \vec{p}=m\vec{v}-m\vec{u}$. This is the impulse–momentum theorem. The SI unit of impulse is the newton-second ($\text{N}\cdot\text{s}$), which is identical to $\text{kg}\cdot\text{m/s}$.

For the same change in momentum, a longer contact time means a smaller force. This is why a cricketer draws the hands back while catching, why cars have crumple zones, and why we bend our knees on landing — all to increase $\Delta t$ and reduce the force.
A large force for a very short time (a hammer blow, a kicked ball) delivers a big impulse even though the time is tiny.

Law of conservation of momentum. If no external force acts on a system, its total linear momentum stays constant. This follows directly from Newton's third law: in a collision the two internal forces are equal and opposite, so they produce equal and opposite changes in momentum that cancel. For two bodies, $m_1\vec{u}_1+m_2\vec{u}_2=m_1\vec{v}_1+m_2\vec{v}_2$.

Applications. In recoil, a gun and bullet start at rest, so $0=m_b v_b + m_g v_g$, giving the gun a backward recoil velocity $v_g=-\frac{m_b v_b}{m_g}$. Rockets work the same way — exhaust gases shoot backward and the rocket moves forward. In collisions (1D), momentum is always conserved; in a perfectly inelastic collision the bodies stick together and move with a common velocity $v=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}$.

Solved Examples

Worked Example

Find the momentum of a 2 kg ball moving at 15 m/s.

Solution

Step 1: Use $p=mv$.
Step 2: Substitute: $p=2 \times 15$.
Step 3: Compute: $p=30\ \text{kg}\cdot\text{m/s}$.

Answer: $p=30\ \text{kg}\cdot\text{m/s}$

Worked Example

A force of 50 N acts on a body for 0.2 s. Find the impulse imparted.

Solution

Step 1: Use $J=F\Delta t$.
Step 2: Substitute: $J=50 \times 0.2$.
Step 3: Compute: $J=10\ \text{N}\cdot\text{s}$.

Answer: $J=10\ \text{N}\cdot\text{s}=10\ \text{kg}\cdot\text{m/s}$

Worked Example

A 0.15 kg cricket ball moving at 20 m/s is struck back along the same line at 30 m/s. If contact lasts 0.01 s, find the average force on the ball.

Solution

Step 1: Take the initial direction as positive: $u=+20\ \text{m/s}$, $v=-30\ \text{m/s}$ (reversed).
Step 2: Change in momentum $\Delta p=m(v-u)=0.15(-30-20)=0.15 \times (-50)=-7.5\ \text{kg}\cdot\text{m/s}$.
Step 3: Average force $F=\frac{\Delta p}{\Delta t}=\frac{-7.5}{0.01}=-750\ \text{N}$.
Step 4: The magnitude is $750\ \text{N}$, directed opposite to the ball's incoming motion.

Answer: $F=750\ \text{N}$ (directed back toward the bowler).

Worked Example

A gun of mass 4 kg fires a bullet of mass 0.02 kg with a velocity of 400 m/s. Find the recoil velocity of the gun.

Solution

Step 1: Conservation of momentum: total momentum before firing is zero.
Step 2: $0=m_b v_b + m_g v_g$, so $v_g=-\frac{m_b v_b}{m_g}$.
Step 3: Substitute: $v_g=-\frac{0.02 \times 400}{4}=-\frac{8}{4}$.
Step 4: Compute: $v_g=-2\ \text{m/s}$ (the minus sign shows it is backward).

Answer: Recoil velocity $=2\ \text{m/s}$ (backward).

Worked Example

A 3 kg trolley moving at 4 m/s collides and sticks to a stationary 1 kg trolley. Find their common velocity after collision.

Solution

Step 1: Perfectly inelastic collision: $v=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}$.
Step 2: Substitute: $v=\frac{3 \times 4 + 1 \times 0}{3+1}=\frac{12}{4}$.
Step 3: Compute: $v=3\ \text{m/s}$.

Answer: Common velocity $=3\ \text{m/s}$ in the original direction.

Worked Example

Two bodies of mass 2 kg and 3 kg move toward each other at 6 m/s and 4 m/s respectively along a straight line. After a head-on collision the 2 kg body comes to rest. Find the velocity of the 3 kg body.

Solution

Step 1: Take the 2 kg body's direction as positive: $u_1=+6$, $u_2=-4\ \text{m/s}$.
Step 2: Conservation of momentum: $m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2$.
Step 3: Substitute with $v_1=0$: $2(6)+3(-4)=2(0)+3 v_2$, so $12-12=3 v_2$.
Step 4: Solve: $0=3 v_2$, hence $v_2=0\ \text{m/s}$.

Answer: The 3 kg body also comes to rest ($v_2=0$).

Key Points

Linear momentum is $\vec{p}=m\vec{v}$ (SI unit $\text{kg}\cdot\text{m/s}$); Newton's second law is $\vec{F}=\frac{d\vec{p}}{dt}$.
Impulse $\vec{J}=\vec{F}\Delta t=\Delta \vec{p}$ — the impulse–momentum theorem; SI unit newton-second.
A longer contact time reduces the force for the same momentum change (catching, crumple zones, bending knees).
Conservation of momentum: with no external force, $m_1\vec{u}_1+m_2\vec{u}_2=m_1\vec{v}_1+m_2\vec{v}_2$.
Recoil of a gun/rocket follows from conservation; in a perfectly inelastic collision the bodies move with a common velocity $v=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}$.

Quick Check — 4 MCQs

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Q1.The SI unit of linear momentum is:

Explanation: $p=mv$ has units of $\text{kg}\cdot\text{m/s}$.

Q2.The impulse–momentum theorem states that impulse equals:

Explanation: $\vec{J}=\vec{F}\Delta t=\Delta \vec{p}$.

Q3.A cricketer pulls the hands back while catching a ball in order to:

Explanation: Increasing $\Delta t$ lowers the average force for the same momentum change.

Q4.The recoil of a gun is a direct consequence of:

Explanation: Total momentum (initially zero) is conserved, so the gun moves opposite to the bullet.

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