Answer:
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°.
Hope this Helps!!!
