Answer :
Answer: 3 m/s east
Explanation:
F=ma
The acceleration needed to move the 2500 kg car by 4.2 m/s from 3.0 m/s is 1.2 m/s.
This is F=2500*1.2=3000
Therefore, the car requires a force of 3000 N. Since every force has an equal and opposite reaction force by Newton's Third Law, there is also a 3000 N force acting on the 1500 kg car.
3000 N would decelerate the 1500 kg car by:
F=ma
3000=1500*a
a=3000/1500=2 m/s^2
Therefore, the 1500 kg is decelerated by 2 m/s^2 to the west or -2 m/s^2 if you consider east to be the positive direction.
The final answer is 5-2 m/s east or 3 m/s east.
Answer:
1.33 m/s east
Explanation:
The total momentum is conserved. The momentum of each vehicle is equal to its mass times its velocity.
p = p
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(1500 kg) (5 m/s) + (2500 kg) (2 m/s) = (1500 kg) v₁ + (2500 kg) (4.2 m/s)
v₁ = 1.33 m/s
The final velocity of the car is 1.33 m/s east.
