1. Chapter 6 Momentum

2. Let’s start with the following situation: A 70kg box travelling at 5m/s is given an extra push of 140N for 10s. What will be it’s new velocity? F = ma a = F= 140N m 70kg m = 70kga = 2 m/s 2 o = 5m/s F = 140Nv+ at t = 10s = 5m/s + 2m/s 2 (10s) v f ? = 25m/s Force stops after 10s, so the final velocity is 25m/s. Now v is constant because there is no more Force (  F=0)

3. Now let’s combine Force & Kinematics: If: F = ma&a = ∆v = v f – v i ∆t ∆t thenF = m(v f – v i ) ∆t so F.∆t = m(v f – v i ) F.∆t = mv f – mv i Impulse(J)momentum(p) not to be confused with (think Impact!) KE (= ½ mv 2 )

4. Now we can say Impulse(J) = change in momentum(∆p) J = ∆p J = p f – p i F.∆t = mv f – mv i Impulse:J = F.∆t momentum:p = mv Note: ∆t & m are scalar, F, v f & v i are vectors. v f & v i can differ in magnitude, direction, or both.

7. Example: A car at a stop sign starts then stops suddenly. The car behind it hits the car at 5mph and dents the bumper in by 10cm before coming to a halt. The second car has a mass of 2000kg. Calculate the Force of impact.

8. A ball bouncing off the ground looks like this: The Force of Impact can be tracked over time and plotted on a Force-time (F-t) graph. A variable Force is difficult to do calculations with – unless you use Calculus! It would be better to know and work with the Average Force.

9. Previously we have established that the derivative of a graph is the slope of the line and it represents the division of the y-axis and x-axis. m = dy/dxE.g. for a v-t graph, m = a Well, the anti-derivative (integral) is the AREA under the graph and it represents the the product of the y-axis and x-axis. Area = ∫y.dxE.g. for a v-t graph, Area = x But for an F-t graph, Area = J Calculating the area is easier if the graph consists of straight lines and regular shapes.

10. When the area under the graph [Impulse(J)] is determined, it can be divided by the time interval (∆t) to give us the Average Force (F). F avg = J = Area ∆t∆t The area under the graph, and the area under the rectangle are the same, therefore the Impulse is the same.

11. Example: A 100g Tee ball is hit with a bat and the Force of Impact is tracked over time in the graph below. Calculate: a.) The Impulse imparted onto the ball, b.) The average Force by the bat, and c.) The velocity at which the ball takes off.

12. Playing Pool: If you hit the white ball straight onto the black ball – what happens? - The black ball - The white ball bounces back, but with a reduced velocity. - So Energy is transferred from the white ball to the black ball. v Wi v Bi = 0 F W F B Newton’s 3 rd Law applies: “..every action..equal but opposite rn” So

14. F W = - F B } in same ∆t F W.∆t = -F B.∆t} Impulse ∆p W = -∆p B } change in momentum m W (v Wf – v Wi ) = -m B (v Bf – v Bi ) m W v Wf – m W v Wi = -m B v Bf + m B v Bi m W v Wi + m B v Bi = m W v Wf + m B v Bf Total momentum before = total momentum after aka “Conservation of Momentum”

15. Perfectly Elastic: If AND i.e.No loss of Energy through sound, heat, light… e.g.pool balls (near enough!) so m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f applies ANDv 1i – v 2i = -(v 1f – v 2f )} a special case v 1i – v 2i = v 2f – v 1f } formula (see p.173 Eqn. 6.12 for derivation)

16. Inelastic: Iftotal p before = total p after BUT i.e.Energy IS lost through sound, heat, light… & Work Energy of Deformation (damage in a crash) e.g.Almost all examples are Inelastic. so m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f applies

17. Perfectly Inelastic: Iftotal p before = total p after BUT KE before ≠ KE after AND the two objects stick together after the collision i.e. e.g.rail (train) cars, car accidents… so m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f applies but it now becomes m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f combined mass

18. Examples:Elastic Collision. p.179 Q.37A 25.0g ball moving to the right at 20cm/s overtakes and collides elastically with a 10.0g ball moving in the same direction at 15.0cm/s. Find the velocity of each ball after the collision.

19. m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 (v 1i – v 2i + v 1f ) m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 (v 1i – v 2i )+ m 2 v 1f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 1f + m 2 (v 1i – v 2i ) m 1 v 1i + m 2 v 2i - m 2 (v 1i – v 2i ) = v 1f (m 1 + m 2 ) v 1f = m 1 v 1i + m 2 v 2i - m 2 (v 1i – v 2i ) (m 1 + m 2 ) v 1f = 25g(20cm/s) + 10g(15cm/s) – 10g(20cm/s – 15cm/s) (25g + 10g) v 1f = v 2f = v 1i – v 2i + v 1f v 2f = 20cm/s – 15cm/s + 17.1cm/s v 2f =

20. Example:InElastic Collision Jack, a 75kg student is standing on a smooth floor with a pair of roller skates on. He throws the 800g Physics textbook to Jill at a velocity of 12m/s. What will be the velocity of Jack?

21. Example:Perfectly InElatic Collision. Jill (65kg), also on roller skates standing still, now catches the 800g textbook travelling at 12m/s. How fast will she be going?

22. Example:Perfectly Inelastic Collision. A 40 tonne (40,000kg) freight car coasting at a speed of 5m/s along a track strikes a 30 tonne stationary freight car and couples to it. What will be their combined speed after impact?

24. The Ballistic Pendulum A 15.0g bullet is fired at 280m/s horizontally into a 3.00kg block of wood suspended by 2 cords. The bullet sticks in the block. Compute the height the block will rise above the horizontal.

26. A gun fires a 10.0g bullet horizontally at a 1.90kg block of soft wood resting on a frictionless table. The bullet emerges at the other end of the block at a speed of 25.0m/s. The table is 1.20m high and the block lands 0.85m from the edge of the table. Calculate the muzzle velocity of the gun.

29. A 7500kg truck is travelling East at 45mph. A 1500kg car is moving North at 55mph. It approaches the intersection and ignores the red light. The car crashes into the truck and get tangled with it. a.) What speed and direction will the wreckage begin to move? b.) If the coefficient of friction between the tyres and the road is 0.350, then how far will the wreckage travel?