1. To understand collisions in two dimensions (not in AS)

2. What is going on here?
Thankfully you won’t have to answer questions that are this complicated!

3. Conservation of momentum
Linear momentum is always conserved, so we need to use vector triangles and resolving vectors to analyse more complicated problems

4. Looking at collisions A B Head on collision – nice and simple 
Other situations – require a little more thought… A B p

5. Adding momentum A B Before After A moves to the right with momentum p
B stationary
A and B move in different directions with momenta p1 and p2 A B p
p 2
p 1

6. Using vector triangle
The vector sum of the final momenta must equal the initial momentum: p1 p2 p

7. Did you know?
If two objects of the same mass, one moving the other stationary collide perfectly elastically then the angle between the direction of motion of each after the collision will always be 90o.

8. What can we say therefore…
About the mass of an Alpha Particle and a Helium nucleus if when they collide their new paths are at 90o?
This is what was observed in cloud chambers.
(a) if m < M then 0 < 90o and a > 90o (b) if m = M then 0 = 90o and a = 90o (c) if m > M then 0 > 90o and a < 90o

9. Snooker ball
A 200g white ball travelling at 3.0ms-1 hits a stationary black ball of 220g. After impact the balls move apart at approximately 90o to each other, which the white ball travelling at 2.5ms-1. Calculate the magnitude of the final velocity of the black ball.

10. Worked answer: p = mv pwhite before = 0.2 x 3 = 0.6 kgms-1
pwhite after= 0.2 x 2.5 = 0.5 kgms-1
Using pythagoras
pblack after 2= = 0.11 kgms-1
pblack after = 0.33 kgms-1
p = mv therefore v = p/m = 0.33/0.22 = 1.51ms-1

11. Situation m1 initial speed vo collides with stationary object m2.
After the collision m1 travels at angle θ1 to original direction with velocity v1 and m2 travels at angle θ2 with velocity v2.

12. Resolve in x and y directions
Momentum must be conserved in x and y directions therefore:
x: total initial momentum = total final momentum
m1vo = m1v1cos θ1 + m2v2cos θ2
y: total initial momentum = total final momentum
(m1 is only originally moving in x direction therefore no initial y momentum)
0 = m1v1sin θ1 + m2v2sin θ2

13. You Try…
A gas atom of mass m moving with a uniform speed v makes an elastic collision with the wall of the container. What is the magnitude of the change in the momentum of the gas atom?
A 2mv B mv sinθ C mv cosθ
D 2mv sinθ E 2mv cosθ

14. Practice questions P110 qu 3 &4
P113 8 – really good summary of all of momentum, impulse and collisions.