Asked by Me and I
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
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
Answers
Answered by
Damon
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
Momentum before = momentum after if no EXTERNAL force on system, like here:
0 = 5 v + 3 * 3
v = -9/5 = -1.8
Answered by
Damon
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
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
Answered by
Me and I
OK, thank you, I understand :)
Answered by
Damon
Good, you are welcome :)
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