= Momentum after a collision

1. Momentum in a system is always conserved. This means that … = Momentum after a collision Momentum before a collision m1v1+ m2v2 = m1v3+ m2v4 In symbols this is written as Or, if the two masses stick together after the collision … m1v1+ m2v2 = (m1+ m2)v3

2. There is a simple and effective 4-step way to solve all physics problems: • Draw a diagram • Write down what you know (sometimes called what is “given” or “known”) • Write down what you are trying to find out (sometimes called the “unknowns”) • Write a plan or strategy (this often includes the use of mathematical equations that you already know)

3. Let’s see how that works with a real problem …

4. In a 60km/hr zone, a Volvo skids 14.3 m before colliding with a parked BMW. The cars become locked together and skid another 6.6 m before stopping. The coefficient of friction between the road and the tyres is 0.75. The mass of the Volvo (and driver) is 1470 kg and the mass of the BMW is 1260 kg. Was the Volvo speeding?

5. In a 60km/hr zone, a Volvo skids 14.3 m before colliding with a parked BMW. The cars become locked together and skid another 6.6 m before stopping. The coefficient of friction between the road and the tyres is 0.75. The mass of the Volvo (and driver) is 1470 kg and the mass of the BMW is 1260 kg. Use the 4-step way to solve this problem

6. DIAGRAM Volvo BMW Before collision

7. DIAGRAM Volvo Volvo BMW BMW Before collision After collision

8. DIAGRAM Volvo Volvo BMW BMW Before collision After collision Momentum before collision = Momentum after collision mVvV+mBvB = (mV+mB)uV+B KNOWNS sV= 14.3 m sV+B = 6.6 m μ = 0.75 mV = 1470 kg uB= 0 ms-1vV+B = 0 ms-1 mB= 1260 kg PLAN UNKNOWNS vV= ?

9. mVvV+mBvB = (mV+mB)uV+B

10. vV mV+mBvB = (mV+mB)uV+B What can we say about vB , the velocity of the BMW? It’s zero, so … mV vV+ mB0 = (mV+mB)uV+B mVvV = (mV+mB)uV+B Make vV the subject Divide both sides by mV …

11. Put in all the numbers for the knowns … What do we need to find now? uV+B, the velocity of the Volvo and BMW after the collision.

12. uV+B, is actually the initial velocity of the Volvo and BMW after the collision. To find this we use the skid marks. Let’s rewrite some knowns … KNOWNS sV+B = 6.6 m vV+B = 0 ms-1 UNKNOWNS uV+B = ? PLAN Which kinematics equation relates initial velocity, final velocity and displacement (but NOT time)? v2 – u2 = 2as

13. Since vV+B = 0, it’s better to write Make u the subject … Put in all the numbers for the knowns …

14. What do we need to find now? a, the acceleration. Apart from Kinematics, in what other relationship have we seen acceleration? F = ma, Newton’s Second Law. What clue in the question relates to Force? The coefficient of friction, . How? F=mg

15. F = maand F=mg So ma= mg How do we make athe subject? Divide both sides by m. a=g Put in all the numbers for the knowns … a =0.75 10 a =7.5 ms-2 Actually, a = - 7.5 ms-2 because it is decelerating.

16. Substitute a = -75 ms-2 into our earlier equation for the velocity of the two cars together. uV+B = 9.95 ms-1 Substitute this into our earlier equation for the final velocity of the Volvo (just before the collision). vV = 18.48 ms-1

17. Is this the velocity of the Volvo before it started skidding? No, this is the velocity of the Volvo just before the collision. To work out its initial velocity we need to take into consideration the skid mark distance of 14.3 m. Based on earlier work using skid marks, calculate the initial velocity of the Volvo and determine whether it was speeding. ms-1

18. ms-1 Convert this to km/hr. kmhr-1 kmhr-1

19. ms-1 Convert this to km/hr. kmhr-1 kmhr-1 Yes, the Volvo was speeding!