Work Booklet 2.4 page 17 Additional Review Exercise 1. A ball, A, of mass 2.0 kg and moving at 5.0 m/s strikes a glancing blow on a second ball, B, which is initially at rest. After the collision, A moves at right angles to its original direction with a speed of 3.0 m/s.

(a) What is the total momentum to the right before the collision? 10kgm/s note only mass A has velocity P total(horizontally) =2x5kgm/s =10kgm/s (b) What is the total momentum to the right after the collision? Law of Conservation of Momentum : momentum before = momentum after 10kgm/s = 10kgm/s

(c) What is the momentum to the right of A after the collision? Mass A is now moving vertically, hence zero horizontal velocity (component), therefore after the collision, Mass A has momentum to the right = 0kgm/s (d) What is the momentum to the right of B after the collision? Note: Mass B has both Vertical and Horizontal components of Velocity Law of conservation of Momentum still applies. So the total of the momentum before collision (horizontally) must equal the total momentum after collision (horizontally). Since Mass A is moving vertically after collision Mass B assumes all of the original Horizontal momentum from before the collision. Therefore momentum of B after collision to the right = 10kgm/s

(e) What is the total momentum `up the page' before the collision? Mass A is moving to the right before collision, so it has a vertical component of velocity = 0m/s. Mass B is stationary (both horizontal and vertical components of velocity = 0m/s. Therefore the total momentum up the page (vertically) = 0kgm/s After collision both masses are now moving. Mass A has only a vertical component of velocity. Mass B has both vertical and horizontal components of velocity. But we must abide by the Law of Conservation of Momentum. In the vertical direction (before collision) the “net” momentum was = zero! Therefore the net momentum after collision in the vertical direction must also = zero! (f) What is the total momentum `up the page' after the collision?

(g) What is the momentum `up the page' of A after the collision? Mass A is moving vertically at +3m/s, After the collision, momentum A = 6kgm/s (h) What is the momentum `up the page' of B after the collision? As mentioned earlier…and you probably are starting to get the hang of this by now: As Mass B is heading down, it has a negative momentum Momentum of mass B up the page = - 6kgm/s ( here, as usual up is positive)

(i) What is the magnitude and direction of the momentum of B after the collision? Okay we should have enough information now that we have dissected the problem into all its sub components. Mass B has a vertical component of - 6kgm/s Mass B has a horizontal component of +10kgm/s We should draw a right angled triangle (as components are at right angle to each other)

You now need to use Pythagorus to find the magnitude of the Resultant Vector. Momentum B = = 11.66kgm/s All vectors must be stated with direction, We now use trigonometry to determination the direction: