Question

Two identical 1000-kg cars skid off the highway: car A hits a side rail and slows from 30 m/s to a complete stop in 0.1 seconds, while car B hits a protective shock absorber and slows from 30 m/s to a complete stop in 0.6 seconds. The following questions refer to the time during which the cars slow down from 30 m/s to 0 m/s. During this time: a) What is the magnitude of change in momentum (i) for A? Ans:__________ ; (ii) for B? Ans: __________b) What is the magnitude of impulse acting (i) on A? Ans.:__________ _ (ii) on B? Ans.:__________ c) What is the magnitude of the average net force acting (i) on A? Ans.: ___________ ;(ii) on B? Ans.:__________

Answer 1

Answer:

a) For both:

ΔP = -30000 kg*m/s

b) For both:

I = 30000 kg*m/s

c) i. F = 300000 N

ii. F = 50000 N

Explanation:

We know that the linear momentun is calculated by the next equation:

P = MV

Where M is the mass and V the velocity.

Also, the impulse is calculated like the change in the linear momemtun, so:

I = $P_f-P_i$

or:

I = Ft

where F is the force and t is the time.

a) i. We will calculate the initial linear momentum Pi and the final momentum Pf for A as:

$P_ i =MV_i$

$P_ i =(1000 kg)(30 m/s)$

$P_ i = 30000 kg*m/s$

$P_f = MV_f$

$P_f = (1000 kg)(0 m/s)$

$P_f = 0$

So, the change in the momentum for A is:

ΔP = Pf - Pi = 0 - 30000 = -30000 kg*m/s

ii. We will calculate the initial linear momentum Pi and the final momentum Pf for B as:

$P_ i =MV_i$

$P_ i =(1000 kg)(30 m/s)$

$P_ i = 30000 kg*m/s$

$P_f = MV_f$

$P_f = (1000 kg)(0 m/s)$

$P_f = 0$

so, the change in the momentum for B is:

ΔP = -30000 kg*m/s

b) We know that the impulse is equal to the change in the momentum, so:

I = ΔP

Then, I = 30000 kg*m/s for both cars

c) i. Now we will use the equation:

I = Ft

Replacing values and solving for F, we get:

30000 kg*m/s = F(0.1 s)

F = 300000 N

ii. Now we will use the equation:

I = Ft

Replacing values and solving for F, we get:

30000 kg*m/s = F(0.6 s)

F = 50000 N