Impulse & Momentum – Collision Examples

1. Introduction

Physics studies motion and the forces that cause it. Two of the most powerful tools in mechanics are the concepts of momentum and impulse.

• Momentum measures the “quantity of motion” a body has.
• Impulse measures the effect of a force applied over time.

When objects collide—whether cars on the road, billiard balls on a table, or subatomic particles in an accelerator—the laws of momentum and impulse help us predict the outcome.

In this detailed article, we will cover:

• The basics of momentum
• Newton’s laws and conservation of momentum
• The concept of impulse
• The impulse–momentum theorem
• Types of collisions (elastic, inelastic, perfectly inelastic)
• Derivations and solved examples
• Real-life applications (sports, safety, rockets, etc.)
• Common misconceptions
• Practice problems

By the end, you’ll have a complete understanding of impulse, momentum, and collision problems.

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2. Momentum

Definition

Momentum is defined as the product of mass and velocity: p=mvp = m vp=mv

Where:

• ppp = momentum (kg·m/s)
• mmm = mass (kg)
• vvv = velocity (m/s)

👉 Momentum is a vector quantity, meaning it has both magnitude and direction.

Examples

• A truck moving at 10 m/s has much more momentum than a bicycle at the same speed because its mass is larger.
• A cricket ball hit by a bat changes momentum depending on speed and direction.

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3. Newton’s Second Law in Terms of Momentum

Newton’s second law states: F=dpdtF = \frac{dp}{dt}F=dtdp​

This means that force is the rate of change of momentum.

If mass is constant: F=mdvdt=maF = m \frac{dv}{dt} = m aF=mdtdv​=ma

So, force changes momentum by changing velocity.

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4. Law of Conservation of Momentum

One of the most important principles in mechanics:

“The total momentum of an isolated system remains constant if no external force acts on it.” pinitial=pfinalp_{initial} = p_{final}pinitial​=pfinal​

Applications:

• Rockets conserve momentum when ejecting exhaust gases.
• Recoil of a gun follows conservation of momentum.
• Collisions between vehicles, balls, or atoms obey momentum conservation.

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5. Impulse

Definition

Impulse is defined as the product of force and the time interval during which it acts: J=F⋅ΔtJ = F \cdot \Delta tJ=F⋅Δt

Where:

• JJJ = impulse (N·s or kg·m/s)
• FFF = average force (N)
• Δt\Delta tΔt = time duration (s)

👉 Impulse has the same units as momentum.

Graphical Representation

Impulse = Area under Force–Time graph.

If force is constant: J=F⋅ΔtJ = F \cdot \Delta tJ=F⋅Δt

If force varies: J=∫F dtJ = \int F \, dtJ=∫Fdt

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6. Impulse–Momentum Theorem

The impulse delivered to an object equals the change in its momentum: J=ΔpJ = \Delta pJ=Δp F⋅Δt=mv−muF \cdot \Delta t = m v – m uF⋅Δt=mv−mu

Where:

• uuu = initial velocity
• vvv = final velocity

👉 This theorem connects force, time, and velocity change in collisions.

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7. Types of Collisions

Collisions are situations where two or more bodies exert forces on each other for a short duration.

(a) Elastic Collision

• Both momentum and kinetic energy are conserved.
• Example: Collision of billiard balls.

Equations for 1D elastic collision (masses m1,m2m_1, m_2m1​,m2​): v1=m1−m2m1+m2u1+2m2m1+m2u2v_1 = \frac{m_1 – m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2v1​=m1​+m2​m1​−m2​​u1​+m1​+m2​2m2​​u2​ v2=2m1m1+m2u1+m2−m1m1+m2u2v_2 = \frac{2 m_1}{m_1 + m_2} u_1 + \frac{m_2 – m_1}{m_1 + m_2} u_2v2​=m1​+m2​2m1​​u1​+m1​+m2​m2​−m1​​u2​

(b) Inelastic Collision

• Momentum is conserved, but kinetic energy is not.
• Some energy is converted into heat, sound, or deformation.
• Example: Car crash.

(c) Perfectly Inelastic Collision

• The colliding bodies stick together after collision.
• Maximum loss of kinetic energy.
• Example: Clay balls sticking after collision.

Equation: (m1+m2)v=m1u1+m2u2(m_1 + m_2) v = m_1 u_1 + m_2 u_2(m1​+m2​)v=m1​u1​+m2​u2​

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8. Solved Examples

Example 1: Momentum Conservation

A 100 g bullet is fired at 500 m/s from a 5 kg gun. Find recoil velocity of the gun. mbulletubullet+mgunugun=0m_{bullet} u_{bullet} + m_{gun} u_{gun} = 0mbullet​ubullet​+mgun​ugun​=0 (0.1)(500)+(5)(v)=0(0.1)(500) + (5)(v) = 0(0.1)(500)+(5)(v)=0 50+5v=0⇒v=−10 m/s50 + 5v = 0 \quad \Rightarrow v = -10 \, m/s50+5v=0⇒v=−10m/s

👉 Gun recoils at 10 m/s backward.

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Example 2: Impulse

A cricket ball of mass 0.15 kg moving at 20 m/s is stopped by a bat in 0.01 s. Find impulse and average force. J=Δp=m(v−u)=0.15(0−20)=−3 NsJ = \Delta p = m (v – u) = 0.15 (0 – 20) = -3 \, NsJ=Δp=m(v−u)=0.15(0−20)=−3Ns F=JΔt=−30.01=−300 NF = \frac{J}{\Delta t} = \frac{-3}{0.01} = -300 \, NF=ΔtJ​=0.01−3​=−300N

👉 Negative sign shows force is opposite to motion.

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Example 3: Elastic Collision

Two identical balls collide head-on. One is at rest, the other moving at 10 m/s. After collision, moving ball stops, and the other moves at 10 m/s.

👉 Momentum and kinetic energy both conserved.

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Example 4: Perfectly Inelastic Collision

A 2 kg ball moving at 4 m/s collides with a 3 kg ball at rest. They stick together. (m1+m2)v=m1u1+m2u2(m_1 + m_2) v = m_1 u_1 + m_2 u_2(m1​+m2​)v=m1​u1​+m2​u2​ 5v=85v = 85v=8 v=1.6 m/sv = 1.6 \, m/sv=1.6m/s

👉 Combined velocity = 1.6 m/s.

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9. Real-Life Applications

1. Car Safety:
   • Seatbelts and airbags increase collision time, reducing force.
   • Impulse–momentum theorem explains how longer time reduces impact.
2. Sports:
   • In cricket, batsmen “give” with the ball to reduce impact force.
   • In baseball or tennis, swinging faster increases momentum transfer.
3. Rocket Propulsion:
   • Conservation of momentum explains how rockets move by ejecting exhaust gases.
4. Billiards/Pool:
   • Elastic collisions transfer momentum between balls.
5. Martial Arts:
   • Short contact time (like a punch) increases force delivered.

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10. Common Misconceptions

• Momentum and force are not the same. Force changes momentum.
• Impulse is not a separate force. It is force multiplied by time.
• Kinetic energy is not always conserved—only in elastic collisions.
• Momentum conservation works even when energy is lost.

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11. Key Formulas Recap

1. Momentum:

p=mvp = m vp=mv

2. Newton’s Second Law:

F=dpdtF = \frac{dp}{dt}F=dtdp​

3. Impulse:

J=F⋅Δt=ΔpJ = F \cdot \Delta t = \Delta pJ=F⋅Δt=Δp

4. Conservation of Momentum:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2m1​u1​+m2​u2​=m1​v1​+m2​v2​

5. Elastic Collision (1D):

v1=m1−m2m1+m2u1+2m2m1+m2u2v_1 = \frac{m_1 – m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2v1​=m1​+m2​m1​−m2​​u1​+m1​+m2​2m2​​u2​ v2=2m1m1+m2u1+m2−m1m1+m2u2v_2 = \frac{2 m_1}{m_1 + m_2} u_1 + \frac{m_2 – m_1}{m_1 + m_2} u_2v2​=m1​+m2​2m1​​u1​+m1​+m2​m2​−m1​​u2​

6. Perfectly Inelastic Collision:

(m1+m2)v=m1u1+m2u2(m_1 + m_2) v = m_1 u_1 + m_2 u_2(m1​+m2​)v=m1​u1​+m2​u2​

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12. Practice Problems

1. A 2000 kg car moving at 20 m/s collides with a 1000 kg car at rest. They stick together. Find final velocity.
2. A football of mass 0.5 kg moving at 15 m/s is stopped by a player in 0.05 s. Find average force applied.
3. Two identical balls collide elastically. One moves at 5 m/s, the other at –3 m/s. Find velocities after collision.
4. A 50 g bullet is fired at 400 m/s from a 4 kg gun. Find recoil speed.
5. In a cricket match, a bat exerts an average force of 1000 N on a ball for 0.02 s. If mass of ball = 0.16 kg, find change in velocity.