Am I correct?
Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its contents, of 5.0 kg, and the other has total mass of 3.0 kg. If the 3.0 kg cart is moving at 3.0 m/s after the spring is allowed to push the carts apart, what is the velocity of the 5.0 kg cart after release? The system is closed.
(Points : 1)
-1.8 m/s <-------
-5.0 m/s
-0.56 m/s
-0.0 m/s
A car traveling north at 10.0 m/s crashes into a car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. The two cars have the same mass. What is their combined speed after the collision? (Points : 1)
5.0 m/s
15 m/s
9 m/s <-------
25 m/s
Once again, it is true again and again:
Momentum before = momentum after if no EXTERNAL force on system, like here:
0 = 5 v + 3 * 3
v = -9/5 = -1.8
north momentum = 10 m
east momentum = 15 m
final mass = 2 m
final north v = 10 m/2 m = 5 m/s
final east v = 15 m/2 m = 7.5 m/s
|v| = sqrt(25+56.25) = 9.1 so yes
OK, thank you, I understand :)
Good, you are welcome :)
To find the velocity of the 5.0 kg cart after release, you can use the principle of conservation of momentum. According to this principle, the total momentum of a closed system remains constant before and after an interaction.
In the given scenario, the system consists of two carts with a compressed spring between them. Initially, both carts are at rest, so their total momentum is zero. After the spring is released and pushes the carts apart, the momentum is still conserved.
The momentum of an object is given by the product of its mass and velocity. Let's assume that the velocity of the 5.0 kg cart after release is v (in m/s). The momentum of the 3.0 kg cart after release would be (3.0 kg) * (3.0 m/s) = 9.0 kg m/s (since it is given that the 3.0 kg cart is moving at 3.0 m/s after release).
The total momentum is the sum of the momenta of the two carts after release. Therefore, the total momentum of the system after release is (5.0 kg) * v + (3.0 kg) * (3.0 m/s).
Since momentum is conserved, the total momentum before release is equal to the total momentum after release. Since both carts are initially at rest, the total momentum before release is zero.
So, we can set up the equation:
0 = (5.0 kg) * v + (3.0 kg) * (3.0 m/s)
Simplifying the equation:
0 = 5.0v + 9.0
Solving for v:
5.0v = -9.0
v = -9.0 / 5.0
v = -1.8 m/s
Therefore, the velocity of the 5.0 kg cart after release is -1.8 m/s.