Question

Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.

Answer 1

Answer:

a)Car E = Car D  > (Car F = Car B = Car C) > Car A

b)Car E = Car D  > (Car F = Car B = Car C) > Car A

Explanation:

Car A: mass = 500 kg; speed = 10 m/s

Car B: mass = 2000 kg;speed = 5 m/s

Car C:mass = 500 kg; speed = 20 m/s

Car D: mass = 1000 kg; speed = 20 m/s

Car E:mass = 4000 kg; speed = 5 m/s

Car F: mass = 1000 kg; speed = 10 m/s

Part a) Now we know that momentum of each car is product of mass and velocity , so we will have

CarA:

$P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s$

Car B:

$P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s$

Car C:

$P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s$

Car D:

$P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s$

Car E:

$P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s$

Car F:

$P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s$

So the momentum is given as ,

Car E = Car D  > (Car F = Car B = Car C) > Car A

Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car

so the order of impulse from largest to least is given as

Car E = Car D  > (Car F = Car B = Car C) > Car A