Collisions & Explosions in Two-Dimensions (HL) | HL IB Physics Revision Notes 2025

Collisions & Explosions in Two-Dimensions

We know that momentum is always conserved
This doesn't just apply to the motion of colliding objects in one dimension (in one line), but this is true in every direction
Since momentum is a vector, it can be split into its horizontal and vertical component
- This is done by resolving vectors
Consider again the two colliding balls A and B
Before the collision, ball A is moving at speed u_A and hits stationary ball B
- Ball A moves away at speed v_A and angle θ_A
- Ball B moves away at speed v_B and angle θ_B

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This time, they move off in different directions, so we now need to consider their momentum in the x direction and separately, their momentum in the y direction
- This is done by resolving the velocity vector of each ball after the collision

Applying the conservation of momentum along the x direction gives

$m_{A} u_{A} + 0 = m_{A} v_{A} \cos θ_{A} + m_{B} v_{B} \cos θ_{B}$

Applying the conservation of momentum along the y direction gives

$0 + 0 = m_{A} v_{A} \sin θ_{A} - m_{B} v_{B} \sin θ_{B}$

The minus sign now comes from B moving downwards, whilst positive y is considered upwards
The momentum before in the y direction is 0 for both balls A and B because B is stationary and A is only travelling in the x direction, so u_A has no vertical component
Since there are two equations involving sine and cosine, it is helpful to remember the trigonometric identity:

$\tan θ = \frac{\sin θ}{\cos θ}$

When the collision is elastic, the conservation of linear momentum and energy indicates that $θ_{A} + θ_{B} = 90 °$

Worked example

A snooker ball of mass 0.15 kg collides with a stationary snooker ball of mass 0.35 kg. After the collision, the second snooker ball moves away with a speed of 0.48 m s^–1. The paths of the balls make angles of 43° and 47° with the original direction of the first snooker ball.

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Calculate the speed u₁ and v₁ of the first snooker ball before and after the collision.

Answer

Step 1: List the known quantities

Mass of the first snooker ball, m₁ = 0.15 kg
Mass of the second snooker ball, m₂ = 0.35 kg
Velocity of second ball after, v₂ = 0.48 m s^–1
Angle of the first ball, θ₁ = 43°
Angle of the first ball, θ₂ = 47°

Step 2: State the equation for the conservation of momentum in the y (vertical) direction

$0 = m_{1} v_{1} \sin θ_{1} - m_{2} v_{2} \sin θ_{2}$

Step 3: Calculate the speed of the first ball after the collision, v₁

Use the conservation of momentum in the y direction to calculate the speed of the first snooker ball after the collision

$m_{1} v_{1} \sin θ_{1} = m_{2} v_{2} \sin θ_{2}$

$v_{1} = \frac{m_{2} v_{2} \sin θ_{2}}{m_{1} \sin θ_{1}}$

$v_{1} = \frac{0.35 \times 0.48 \times \sin (47)}{0.15 \times \sin (43)} = 1.2 m s^{- 1}$

Step 3: State the equation for the conservation of momentum in the x (horizontal) direction

$m_{1} u_{1} = m_{1} v_{1} \cos θ_{1} + m_{2} v_{2} \cos θ_{2}$

Step 4: Calculate the speed of the first ball before the collision, u₁

$u_{1} = \frac{m_{1} v_{1} \cos θ_{1} + m_{2} v_{2} \cos θ_{2}}{m_{1}}$

$u_{1} = \frac{(0.15 \times 1.2 \times \cos (43)) + (0.35 \times 0.48 \times \cos (47))}{0.15} = 1.6 m s^{- 1}$

Examiner Tip

Make sure you clearly label your diagram or write out the known quantities before you substitute values into the question. It's very easy to substitute in the incorrect velocity or mass. Use subscripts such as '1' '2' or 'A' 'B' depending on the question to help keep track of these.

Although you will get full marks either way, it may be easier in these equations to rearrange first and then substitute instead of the other way around, to keep track of the multiple masses and velocities.

Make sure your calculator is in degree mode if your angles are given in degrees!

If you use the fraction function to input your values, remember that you need to close the brackets on the trig functions or it will give you the wrong answer.

Eg. $\frac{(0.35 \times 0.48 \times \sin (47))}{(0.15 \times \sin (43))} = 1.2 m s^{- 1}$

And if you input the values into your calculator as numerator ÷ denominator, make sure you put brackets around the whole denominator.

Eg. $A n s \div (0.15 \times \sin (43))$

The trig equation for tanθ is also given on your data sheet under 'Mathematical equations', as well as that for resolving forces.

Collisions & Explosions in Two-Dimensions (HL) (HL IB Physics)

Revision Note