Two cars collide at right angles and stick together as they skid to a stop. Car 1 has a mass of 1300 kg and a speed of 21 km/h to the east. Car 2 has a mass of 1000 kg and a speed of 36 km/h to the north. The average coefficient of friction between the tires and the road is 0.65. How far from the collision point, and in what direction, do the cars stop?

The cars stop 2.34 m from the collision point and move in the direction dictated by the velocity, 52.8°.

Explanation:

Mass of first car = 1300 kg

Velocity of first car = (21î) km/h = (5.833î) m/s

Mass of second car = 1000 kg

Velocity of second car = (36j) km/h = (10j) m/s

Let the velocity of both cars after collision be v

Using the law of conservation of Momentum.

Momentum before collision = momentum after collision

Momentum of first car before collision = (1300)(5.833î) = (7582.9î) kgm/s

Momentum of second car before collision = (1000)(10j) = (10000j) kgm/s

Momentum of both cars after collision = (1300+1000)v = 2300v

Momentum before collision = momentum after collision

(7582.9î) + (10000j) = 2300v

v = (3.30î + 4.35j) m/s

Magnitude of v = √(3.30² + 4.35²) = 5.46 m/s

Direction of v = tan⁻¹ (4.35/3.30)

Direction of v = 52.8°

Then using the work-energy theorem,

The kinetic energy of the two joint cars = Work done by frictional forces in stopping the joint cars.

Kinetic energy of the cars after collision = (1/2)(2300)(5.46²) = 34,283.34 J

Work done by frictional force = F × d

where F = frictional force = μmg = (0.65)(2300)(9.8) = 14,651 N

d = distance over which the frictional force acts before stopping the two cars = ?

34,283.34 = 14,651d

d = (34,283.34/14,651)

d = 2.34 m

Therefore, the cars stop 2.34 m from the collision point and move in the direction dictated by the velocity, 52.8°.

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