A single 1,000 kg train car moving at 5.0 m/s collides with the back of two 1,000 kg train cars linked together. It latches onto the car it strikes as the two parts of the coupled cars exert forces on each other. What is the speed of the cars immediately afterward?
Solution: When the two train cars are linked together, they will move as one unit, so their total mass is 2000kg. The momentum of the train cars before the collision is 20005 = 10000kg m/s. The momentum of the train cars after the collision is 3000v, where v is the final velocity of the train cars after the collision.
According to the law of conservation of momentum, the momentum of the train cars before the collision is equal to the momentum of the train cars after the collision. Therefore, we can write the following equation:
20005 = 3000v
Solving for v, we get:
v = (2000*5)/3000 = 5/3 m/s
So the speed of the cars immediately afterward is 5/3 m/s.
A firecracker stuck into a 150 g apple explodes and sends five apple fragments in different directions. The vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components px = 1.20 kg m/s, py = 0.80 kg m/s (with no other component). The mass of the fifth fragment is 0.050 kg. What is its velocity right after the explosion?
Solution: According to the principle of conservation of momentum, the total momentum of the system before the explosion is equal to the total momentum of the system after the explosion. Therefore, we can write the following equation:
0 = p1 + p2 + p3 + p4 + p5
where p1, p2, p3, p4, p5 are the momenta of the five fragments.
As the vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components px = 1.20 kg m/s, py = 0.80 kg m/s. Therefore we can write the following equation:
p1 + p2 + p3 + p4 = -p5 = (-1.20, -0.80) kg m/s
We can use these values to find the velocity of the fifth fragment using the relationship between momentum and velocity:
p = mv
v = p/m = (-1.20 kg m/s, -0.80 kg m/s) / 0.050 kg = (-24.0 m/s, -16.0 m/s)
So the velocity of the fifth fragment right after the explosion is (-24.0 m/s, -16.0 m/s).
A 1,600 kg car traveling north at 10.0 m/s crashes into a 1,400 kg car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. What is their speed after the crash?
Solution: The total momentum of the cars before the collision is:
m1v1 + m2v2 = 160010 + 140015 = 16000 + 21000 = 37000 kgm/s
The total momentum of the cars after the collision is:
(m1+m2)*vf = (1600+1400)vf = 3000vf