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## Homework Statement

A 750 kg car is traveling north at 54 km/h while a 1200 kg truck travels south at 83 km/h. The two vehicles are about to have a head-on collision.

1.1) Convert the speed's of both car's into m/s.

1.2) After impact, the car rebounds backwards with a velocity of 18 km/h. Calculate the velocity of the truck after the collision.

## Homework Equations

Pbefore = Pafter

P = impulse

MaUa + MbUB = MaVa + MbVb

U = initial velocity

V = final velocity

M = mass in kilograms

## The Attempt at a Solution

1.1) km/h to m/s (kilometers per hour to metres per second): You divide the value in km/h by 3.6

Therefore the speed of the car in m/s:

54 / 3.6 = 15 m/s

And the speed of the truck in m/s:

83 / 3.6 = 23.06 m/s

1.2) So if we take north to be positive then the car is moving in the positive direction and the truck is approaching in a negative direction.

So Ua = +15 m/s

Ub = -23.06 m/s

And if car A rebounds in the opposite direction than Va is -5 m/s (18/3.6 = 5) and Vb will be +.

So if I then plug in all my values into the formula:

Pbefore = Pafter

MaUa + MbUB = MaVa + MbVb

750 * 15 + 1200 * -23,06 = 750 * -5 + 1200 * Vb

Therefore Vb = -10,56 or 10,56 in the original direction of the car (north and we took north to be positive) (+10,56 m/s)

The answer in the memo is 35,56 m/s

They say that both Ua and Ub are positive so Ua = 15 and Ub = 23,06 which conceptually does not make sense as they are heading towards each other from opposite directions and velocity is a vector so it has both magnitude and direction.

Is my answer correct or am I making a mistake in my logic or calculations?

Any help is appreciated, Thanks!

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