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IB DP Physics Collisions Study Notes

IB DP Physics Collisions Study Notes - 2025 Syllabus

IB DP Physics Collisions Study Notes

IB DP Physics Collisions Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with guiding questions of

  • the elastic and inelastic collisions of two bodies
  • explosions
  • energy considerations in elastic collisions, inelastic collisions, and explosions

Standard level and higher level: 10  hours
Additional higher level: There is no additional higher level content

IB DP Physics 2025 -Study Notes -All Topics

IBDP Physics 2025 -Study Notes -All Topics

Elastic and Inelastic Collisions: 

  • Collision: An event in which two or more bodies exert forces on each other for a short time.
  • 1D Collisions: Motion and interaction occur along a single straight line.
  • 2D Collisions: Bodies move in a plane and involve vector components in both x and y directions.

Elastic Collisions

  • Definition: A collision in which both momentum and kinetic energy are conserved.
  • Conditions:
    • Total momentum before = total momentum after
    • Total kinetic energy before = total kinetic energy after
  • Typical examples: Collisions between gas molecules, billiard balls (ideally)

1D Elastic Collision Equations:

  • Total momentum: \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \)
  • Total kinetic energy: \( \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \)

2D Elastic Collision:

  • Break momentum into x- and y-components:
    • Conserve momentum in x: \( \sum p_x^{\text{before}} = \sum p_x^{\text{after}} \)
    • Conserve momentum in y: \( \sum p_y^{\text{before}} = \sum p_y^{\text{after}} \)
  • Also conserve total kinetic energy in system.

Inelastic Collisions

  • Definition: A collision where momentum is conserved, but kinetic energy is not.
  • Perfectly Inelastic Collision: Bodies stick together after impact, moving as one mass.
  • Conditions:
    • Total momentum before = total momentum after
    • Total kinetic energy is not conserved (some energy lost to sound, heat, deformation)
  • Typical examples: Car crashes, clay sticking on collision

1D Perfectly Inelastic Collision Equation:

  • Let both masses stick together after collision and move at velocity \( v \):
  • \( m_1u_1 + m_2u_2 = (m_1 + m_2)v \)

2D Inelastic Collision:

  • Conserve momentum in x- and y-directions separately.
  • Kinetic energy is partially lost — calculate the initial and final energies separately.

Example : 

A 2 kg ball moving at \( 5 \, \text{m/s} \) collides elastically with a 3 kg ball initially at rest. Find their velocities after the collision.

▶️ Answer/Explanation

Given: \( m_1 = 2\, \text{kg}, v_{1i} = 5\, \text{m/s}; m_2 = 3\, \text{kg}, v_{2i} = 0 \)

Use elastic collision formulas:

\( v_{1f} = \frac{m_1 – m_2}{m_1 + m_2} v_{1i} + \frac{2m_2}{m_1 + m_2} v_{2i} \)

\( v_{1f} = \frac{2 – 3}{5} \cdot 5 = -1 \, \text{m/s} \)

\( v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i} + \frac{m_2 – m_1}{m_1 + m_2} v_{2i} \)

\( v_{2f} = \frac{4}{5} \cdot 5 = 4 \, \text{m/s} \)

Final velocities: \( \boxed{v_{1f} = -1 \, \text{m/s},\quad v_{2f} = 4 \, \text{m/s}} \)

Example : 

A 4 kg object moving at \( 3 \, \text{m/s} \) collides and sticks to a 2 kg object at rest. What is their final velocity?

▶️ Answer/Explanation

Total momentum conserved:

\( (4)(3) + (2)(0) = (4+2)v_f \Rightarrow 12 = 6v_f \)

\( \boxed{v_f = 2 \, \text{m/s}} \)

This is a perfectly inelastic collision: kinetic energy is not conserved, but momentum is.

Example : 

A 1 kg puck moving at \( 4 \, \text{m/s} \) on a frictionless surface collides elastically with a 1 kg puck initially at rest. After the collision, the first puck moves at \( 2.8 \, \text{m/s} \) at \( 45^\circ \) above the x-axis. Find the velocity and angle of the second puck.

▶️ Answer/Explanation

Use momentum conservation:

In x-direction: \( 4 = 2.8 \cos 45^\circ + v_2 \cos \theta \)

In y-direction: \( 0 = 2.8 \sin 45^\circ – v_2 \sin \theta \)

\( \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707 \)

Substitute:

\( 4 = 2.8 \cdot 0.707 + v_2 \cos \theta \Rightarrow 4 = 1.98 + v_2 \cos \theta \)

\( v_2 \cos \theta = 2.02 \)

From y-direction: \( 0 = 2.8 \cdot 0.707 – v_2 \sin \theta \Rightarrow v_2 \sin \theta = 1.98 \)

Now use: \( v_2^2 = (v_2 \cos \theta)^2 + (v_2 \sin \theta)^2 \)

\( v_2^2 = 2.02^2 + 1.98^2 = 4.08 + 3.92 = 8 \Rightarrow v_2 = \sqrt{8} = \boxed{2.83 \, \text{m/s}} \)

Angle: \( \theta = \tan^{-1} \left( \frac{1.98}{2.02} \right) \approx \boxed{44.5^\circ} \) below x-axis

Explosions

  • Definition: An explosion is a situation where an object at rest (or in motion) breaks apart into two or more pieces due to an internal force.
  • Key Idea: No external forces act on the system, so total momentum is conserved.
  • Although the objects move away from each other (gain kinetic energy), this energy comes from internal chemical, nuclear, or stored potential energy.

Momentum in Explosions

  • Initial Momentum: If the object was initially at rest, the total initial momentum is 0.
  • Conservation Law: The total momentum after the explosion must also be zero (in the absence of external forces).

If an object of mass \( M \) explodes into two parts, \( m_1 \) and \( m_2 \), moving with velocities \( v_1 \) and \( v_2 \):

  • From momentum conservation: \( m_1v_1 + m_2v_2 = 0 \)
  • This gives: \( m_1v_1 = -m_2v_2 \) → their momenta are equal and opposite

Energy in Explosions

  • Kinetic energy is not conserved in an explosion.
  • In fact, kinetic energy is often gained from the release of stored internal energy (e.g., chemical or nuclear).
  • This is the reverse of an inelastic collision (where kinetic energy is lost); in an explosion, it is gained.

 Real-world Examples of Explosions

  • A firecracker exploding into multiple fragments.
  • Two astronauts pushing away from each other in space – total momentum remains zero.
  • A gun firing a bullet (the gun recoils backward).
  • Nuclear explosion: nucleus breaks into smaller nuclei and releases energy.

Example:

A 4 kg rifle fires a 0.02 kg bullet at 400 m/s. What is the recoil velocity of the rifle?

▶️Answer/Explanation

Initial momentum: System is at rest → Total momentum = 0

After firing: Let \( v_r \) be the recoil velocity of the rifle.

By conservation of momentum:

\( m_b v_b + m_r v_r = 0 \)

\( (0.02)(400) + (4)(v_r) = 0 \)

\( 8 + 4v_r = 0 \Rightarrow v_r = -2 \, \text{m/s} \)

Final Answer: \( \boxed{2 \, \text{m/s} \, \text{(opposite direction)}} \)

Example:

A stationary object explodes into two fragments. One fragment of 3 kg moves left with 6 m/s. Find the velocity of the second fragment if it has a mass of 2 kg.

▶️Answer/Explanation

Initial momentum: Total = 0 (stationary object)

After explosion:

Let \( v \) be the velocity of the second fragment.

By conservation of momentum:

\( (3)(-6) + (2)(v) = 0 \)

\( -18 + 2v = 0 \Rightarrow v = 9 \, \text{m/s} \)

Final Answer: \( \boxed{9 \, \text{m/s to the right}} \)

Example:

A 6 kg object at rest explodes into two fragments. A 2 kg piece moves east at 4 m/s. A 3 kg piece moves north at 2 m/s. Find the velocity (magnitude and direction) of the third piece.

▶️Answer/Explanation

Initial momentum: \( \vec{p}_{\text{total}} = 0 \)

Let:

Piece 1 (2 kg east): \( \vec{p}_1 = (2)(4) = 8 \, \text{kg·m/s} \, \text{east} \)

Piece 2 (3 kg north): \( \vec{p}_2 = (3)(2) = 6 \, \text{kg·m/s} \, \text{north} \)

To conserve momentum: Third piece (1 kg) must balance both components in opposite direction:

\( p_{3x} = -8 \), \( p_{3y} = -6 \)

Magnitude: \( |\vec{p}_3| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{kg·m/s} \)

So velocity = \( \frac{10}{1} = \boxed{10 \, \text{m/s}} \)

Direction: south-west direction

Angle w.r.t. west: \( \theta = \tan^{-1}\left(\frac{6}{8}\right) = 36.9^\circ \)

Final Answer: \( \boxed{10 \, \text{m/s} \text{ at } 36.9^\circ \text{ south of west}} \)

Energy Considerations in Elastic Collisions, Inelastic Collisions, and Explosions

1. Elastic Collisions:

  • Both kinetic energy and momentum are conserved.
  • No energy is transformed into heat, sound, or deformation.
  • Perfectly elastic collisions are rare in real life but well-approximated in particle collisions and billiard balls.

  • Conservation of momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)

    Conservation of kinetic energy: \( \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \)

2. Inelastic Collisions:

  • Momentum is conserved, but kinetic energy is not.
  • Some kinetic energy is transformed into other forms (e.g., heat, sound, deformation).
  • In a perfectly inelastic collision, objects stick together and move with a common velocity after impact.

  • Conservation of momentum: \( m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \)

    Kinetic energy after collision is less than before: \( \text{KE}_{\text{final}} < \text{KE}_{\text{initial}} \)

3. Explosions:

  • Momentum is conserved, assuming the system starts at rest (total initial momentum = 0).
  • Kinetic energy increases after the explosion due to internal energy release (e.g. chemical or nuclear).
  • Energy released often comes from stored potential energy (e.g., chemical bonds, nuclear binding energy).

  • Total momentum: \( \vec{p}_{\text{before}} = \vec{p}_{\text{after}} \)

    But: \( \text{KE}_{\text{after}} > \text{KE}_{\text{before}} \)

Example:

A 2 kg cart moving at 4 m/s collides elastically with a stationary 1 kg cart on a frictionless track. Find their velocities after the collision.

▶️ Answer/Explanation

Given:

  • Before collision: \( m_1 = 2 \, \text{kg}, u_1 = 4 \, \text{m/s} \); \( m_2 = 1 \, \text{kg}, u_2 = 0 \, \text{m/s} \)

Use elastic collision formulas:

\( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} = \frac{(2 – 1) \cdot 4 + 0}{3} = \frac{4}{3} = 1.33 \, \text{m/s} \)

\( v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} = \frac{0 + 2 \cdot 2 \cdot 4}{3} = \frac{16}{3} = 5.33 \, \text{m/s} \)

Final velocities:

  • \( \boxed{v_1 = 1.33 \, \text{m/s}} \)
  • \( \boxed{v_2 = 5.33 \, \text{m/s}} \)

Example:

A 3 kg ball moving at 2 m/s collides inelastically with a 2 kg ball moving at 1 m/s in the same direction. Find their common velocity after they stick together.

▶️ Answer/Explanation

Use conservation of momentum:

\( m_1 = 3 \, \text{kg}, u_1 = 2 \, \text{m/s} \); \( m_2 = 2 \, \text{kg}, u_2 = 1 \, \text{m/s} \)

Total momentum = \( 3 \cdot 2 + 2 \cdot 1 = 6 + 2 = 8 \, \text{kg m/s} \)

Total mass = 3 + 2 = 5 kg

Final velocity \( v = \frac{8}{5} = \boxed{1.6 \, \text{m/s}} \)

Example:

A stationary bomb of mass 4 kg explodes into two fragments. One fragment of 1.5 kg moves at 12 m/s. Find the velocity of the other fragment.

▶️ Answer/Explanation

Initial total momentum = 0 (since it was stationary).

Let \( v \) be the velocity of the second fragment.

\( m_1 = 1.5 \, \text{kg}, v_1 = 12 \, \text{m/s} \)
\( m_2 = 2.5 \, \text{kg}, v_2 = ? \)

By conservation of momentum:

\( 1.5 \cdot 12 + 2.5 \cdot v = 0 \Rightarrow 18 + 2.5v = 0 \Rightarrow v = \boxed{-7.2 \, \text{m/s}} \)

The second fragment moves in the opposite direction.

Example:

A 3 kg ball is moving east at \( 4 \, \text{m/s} \), and a 2 kg ball is moving north at \( 3 \, \text{m/s} \). They collide and stick together. Find the magnitude and direction of their combined velocity after the collision.

▶️ Answer/Explanation

Use conservation of momentum in both x and y directions.

  • Let east be \( +x \), north be \( +y \)
  • \( m_1 = 3 \, \text{kg}, u_{1x} = 4 \, \text{m/s}, u_{1y} = 0 \)
  • \( m_2 = 2 \, \text{kg}, u_{2x} = 0, u_{2y} = 3 \, \text{m/s} \)

 Find total momentum components

\( p_x = 3 \cdot 4 + 2 \cdot 0 = 12 \, \text{kg·m/s} \)
\( p_y = 3 \cdot 0 + 2 \cdot 3 = 6 \, \text{kg·m/s} \)

Total mass after sticking together

\( m = 3 + 2 = 5 \, \text{kg} \)

 Final velocity components

\( v_x = \frac{12}{5} = 2.4 \, \text{m/s} \), \( v_y = \frac{6}{5} = 1.2 \, \text{m/s} \)

 Magnitude and direction of resultant velocity

\( v = \sqrt{(2.4)^2 + (1.2)^2} = \sqrt{5.76 + 1.44} = \sqrt{7.2} \approx \boxed{2.68 \, \text{m/s}} \)

\( \theta = \tan^{-1}\left( \frac{1.2}{2.4} \right) = \tan^{-1}(0.5) \approx \boxed{26.6^\circ} \) north of east

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