5. A 1400 kg car sliding on frictionless ice at 20 m/s hits a stationary 2200 kg minivan. The two vehicles are locked together after impact on the ice. What is their speed after impact?
a) 20 m/s
b) 7.8 m/s
c) 13 m/s
d) 12 m/s
e) 0.0 m/s
To solve this problem, we can use the conservation of momentum principle. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity.
Before the collision, the car has a momentum of (1400 kg) * (20 m/s) = 28000 kg·m/s, and the minivan has a momentum of (2200 kg) * (0 m/s) = 0 kg·m/s.
After the collision, the two vehicles are locked together and will move with the same velocity. Let's assume this velocity is v.
Since momentum is conserved, the total momentum after the collision is equal to the sum of the individual momentum of each vehicle:
(1400 kg + 2200 kg) * v = 1400 kg * v + 2200 kg * v
Simplifying the equation:
3600 kg * v = 1400 kg * v + 2200 kg * v
Combining like terms:
3600 kg * v = 3600 kg * v
This equation shows that the mass of the vehicles does not affect the velocity after the collision. Therefore, the velocity after the collision will be the same as the velocity before the collision, which is 20 m/s.
Therefore, the speed of the two vehicles after the impact is 20 m/s.
The correct answer is a) 20 m/s.