Example 4.5 NCERT Solutions: Impulse and Reflection of Billiard Balls Explained

  NEET Physics: Impulse and Momentum Wall Collision Solved Step by Step

-Dr.Sanjaykumar Pawar 

Two identical billiard balls reflect from a rigid wall, demonstrating impulse, momentum change, and Newton's Third Law.

Internal Links

Related NCERT Examples

Plain text

Example 4.1 – Momentum of a Particle

Example 4.2 – Impulse and Force Relationship

Example 4.3 – Conservation of Momentum

Example 4.4 – Newton's Second Law Applications

Example 4.6 – Collision Problems

Related Physics Topics

Plain text

What is Momentum in Physics?

Impulse Formula and Numerical Problems

Newton's Laws of Motion Complete Notes

Conservation of Linear Momentum

Elastic and Inelastic Collisions

Projectile Motion Notes

Vector Resolution in Physics

Motion in Two Dimensions

NEET Mechanics Formula Sheet

Common NEET Physics Mistakes

Example 4.5 – Reflection of Two Billiard Balls from a Wall

Question

Two identical billiard balls strike a rigid wall with the same speed but at different angles. After collision, they are reflected without any loss of speed.

Find:

1. Direction of the force on the wall due to each ball.
2. Ratio of the magnitudes of impulses imparted to the balls by the wall.

Concepts Required

1. Momentum

Momentum = Mass × Velocity
p = mv

2. Impulse

Impulse = Change in Momentum
J = Δp = pf − pi

3. Newton's Third Law

If the wall exerts a force on the ball, then the ball exerts an equal and opposite force on the wall.

First find the force on the ball due to the wall, then reverse the direction to get the force on the wall.

Case (a): Ball Strikes Normally

Step 1: Initial Momentum

X-direction:

(px)initial = mu

Y-direction:

(py)initial = 0

Step 2: Final Momentum

After reflection, speed remains same but direction reverses.

(px)final = -mu

(py)final = 0

Step 3: Change in Momentum

Δpx = (-mu) - (mu)
Δpx = -2mu

Δpy = 0

Step 4: Impulse

Jx = -2mu
Jy = 0

Therefore impulse acts completely along the negative x-direction.

Step 5: Direction of Force

• Force on ball due to wall → Negative x-direction.
• Force is normal (perpendicular) to the wall.
• By Newton's Third Law, force on wall due to ball → Positive x-direction.

Case (b): Ball Strikes at 30°

Step 1: Initial Momentum Components

(px)initial = mu cos30°

(py)initial = -mu sin30°

Step 2: Final Momentum Components

After reflection:

• X-component changes sign.
• Y-component remains unchanged.

(px)final = -mu cos30°

(py)final = -mu sin30°

Why does only X-component change?

The wall can exert force only perpendicular (normal) to its surface. It cannot exert force parallel to the surface.

Therefore:

• X-component changes.
• Y-component remains same.

Step 3: Change in Momentum

Δpx = (-mu cos30°) - (mu cos30°)

Δpx = -2mu cos30°

Δpy = 0

Step 4: Impulse

Jx = -2mu cos30°

Jy = 0

Therefore impulse acts only along the negative x-direction.

Step 5: Direction of Force

• Force on ball due to wall → Negative x-direction.
• Normal to the wall.
• Force on wall due to ball → Positive x-direction.
• Also normal to the wall.

Answer to Part (i)

Case (a)

• Force on wall is normal to the wall.
• Direction is positive x-direction.

Case (b)

• Force on wall is also normal to the wall.
• Direction is positive x-direction.

Conclusion:

In both cases, force on the wall is perpendicular (normal) to the wall.

The force is NOT inclined at 30°.

Answer to Part (ii): Ratio of Impulses

Impulse in Case (a)

Ja = 2mu

Impulse in Case (b)

Jb = 2mu cos30°

Ratio

Ja / Jb = 2mu / (2mu cos30°)

Ja / Jb = 1 / cos30°

Since:

cos30° = √3 / 2

Ja / Jb = 2 / √3

≈ 1.15 ≈ 1.2

Final Answers

(i) Direction of Force on Wall

• Case (a): Normal to wall, positive x-direction.
• Case (b): Normal to wall, positive x-direction.

(ii) Ratio of Impulses

Ja : Jb
= 2mu : 2mu cos30°
= 1 : cos30°
= 2/√3 : 1

or

Ja/Jb = 2/√3 ≈ 1.15 ≈ 1.2

NEET Shortcut

• Parallel component of momentum remains unchanged.
• Perpendicular component reverses direction.
• Impulse depends only on perpendicular component.

Impulse = 2m(v⊥)

This shortcut can solve most NEET wall-collision questions within seconds.