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

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Ashika

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Ashika

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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

  • Consider again the two colliding balls A and B

  • Before the collision, ball A is moving at speed uA and hits stationary ball B

    • Ball A moves away at speed vA and angle θA

    • Ball B moves away at speed vB and angle θB

1-2-15-momentum-in-2d-ib-2025-physics

  • 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

1-2-15-momentum-in-2d-x-and-y

m subscript A u subscript A space plus thin space 0 space equals space m subscript A v subscript A cos theta subscript A space plus thin space m subscript B v subscript B cos theta subscript B

 

  • Applying the conservation of momentum along the y direction gives

0 space plus thin space 0 space equals space m subscript A v subscript A sin theta subscript A space minus space m subscript B v subscript B sin theta subscript 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 uA has no vertical component

  • Since there are two equations involving sine and cosine, it is helpful to remember the trigonometric identity:

tan theta space equals fraction numerator space sin theta over denominator cos theta end fraction

 

  • When the collision is elastic, the conservation of linear momentum and energy indicates that theta subscript A space plus thin space theta subscript B space equals space 90 degree

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.

1-2-15-momentum-in-2d-we-ib-2025-physics

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 space space equals space m subscript 1 v subscript 1 sin theta subscript 1 space minus space m subscript 2 v subscript 2 sin theta subscript 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 subscript 1 v subscript 1 sin theta subscript 1 space equals space m subscript 2 v subscript 2 sin theta subscript 2

v subscript 1 space equals space fraction numerator m subscript 2 v subscript 2 sin theta subscript 2 space over denominator m subscript 1 sin theta subscript 1 end fraction

v subscript 1 space equals space fraction numerator 0.35 space cross times space 0.48 space cross times space sin open parentheses 47 close parentheses space over denominator 0.15 space cross times space sin open parentheses 43 close parentheses end fraction space equals space 1.2 space straight m space straight s to the power of negative 1 end exponent  

Step 4: State the equation for the conservation of momentum in the x (horizontal) direction 

m subscript 1 u subscript 1 space equals space m subscript 1 v subscript 1 cos theta subscript 1 space plus space m subscript 2 v subscript 2 cos theta subscript 2

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

u subscript 1 space equals space fraction numerator m subscript 1 v subscript 1 cos theta subscript 1 space plus space m subscript 2 v subscript 2 cos theta subscript 2 over denominator m subscript 1 end fraction

u subscript 1 space equals space fraction numerator open parentheses 0.15 space cross times space 1.2 space cross times space cos open parentheses 43 close parentheses close parentheses space plus space open parentheses 0.35 space cross times space 0.48 space cross times space cos open parentheses 47 close parentheses close parentheses over denominator 0.15 end fraction space equals space 1.6 space straight m space straight s to the power of negative 1 end exponent

Examiner Tips and Tricks

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. space fraction numerator open parentheses 0.35 space cross times space 0.48 space cross times space sin open parentheses 47 close parentheses close parentheses space over denominator open parentheses 0.15 space cross times space sin open parentheses 43 close parentheses close parentheses end fraction space equals space 1.2 space straight m space straight s to the power of negative 1 end exponent

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 space divided by space open parentheses 0.15 space cross times space sin open parentheses 43 close parentheses close parentheses

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

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.