A1400 kg car moving at 6.6 m/s is initially traveling north in the positive y direction.
after completing a 90° right-hand turn to the positive x direction in 4.9 s, the inattentive operator drives into a tree, which stops the car in 430 ms.

what is the magnitude of the impulse on the car :

(a) due to the turn and (b) due to the collision?

what is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision?

(e) what is the angle between the average force in (c) and the positive x direction?

a) J = 13067 kg*m/s

b) J = 9240 kg*m/s

c) F = 2666.73 N

d) F = 21488.37 N

e) 135°

Explanation:

We know that the impulse could be calculated by:

J = ΔP

where ΔP is the change in the linear momentum:

ΔP =

Also:

P = MV

Where M is the mass and V is the velocity.

so:

J=

where is the final velocity and is the inicial velocity

a)The impluse from turn is:

On the turn, and , the magnitude of the impulse on direction x and y are:

So, using pythagoras theorem the magnitude of the impulse is:

J =

J = 13067 kg*m/s

b) The impluse from the collision is:

J = 0 - (1400)(6.6)

J = 9240 kg*m/s

c) Using the next equation:

FΔt = J

where F is the force, Δt is the time and J is the impulse.

Replacing J by the impulse due to the turn, Δt by 4.9s and solving for F we have that:

F = J / Δt

F = 13067 / 4.9 s

F = 2666.73 N

d) At the same way, replacing J by the impulse during the collision, Δt by 0.43s and solving for F we have that:

F = J / Δt

F = 9240 / 0.43

F = 21488.37 N

e) The force have the same direction than the impulse due to the turn, Then, if the impulse have a direction of -45°, the force have -45° or 135°

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr