A 12100 kg railroad car travels alone on a level frictionless track with a constant speed of 20.5 m/s. A 5850 kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
An application of the law of conservation of linear momentum, in the horizontal direction, will tell you the answer.
Give it a try.
Adding a load by dropping from above will not change the horizontal momentum.
12100kg-5850kg=6250kg/100
1/2kv2
1/2(625)(20.5)^2=
131328.125m/s
umm, im sorry about that answer, but really? 131 thousand meters per second? might want to check your math there, that's 131 kilometers per second which is pretty fast...
To find the car's new speed after the load is dropped onto it, we need to apply the law of conservation of momentum. According to this law, the total momentum before and after the collision remains the same as long as no external forces are acting.
First, let's calculate the initial momentum of the railroad car before the load is dropped. We can find momentum by multiplying the mass of an object by its velocity. Therefore, the initial momentum of the railroad car is:
Momentum1 = mass1 × velocity1
= 12100 kg × 20.5 m/s
= 248,605 kg·m/s
Now, let's calculate the momentum of the load after it is dropped. Since the load was initially at rest, its initial momentum is zero.
Momentum2 = mass2 × velocity2
= 5850 kg × velocity2
According to the conservation of momentum, the total initial momentum equals the total final momentum. So, we can equate the two momentum values:
Momentum1 = Momentum2
248,605 kg·m/s = 5850 kg × velocity2
Now we can solve for velocity2, which will be the new speed of the car after the load is dropped.
velocity2 = (248,605 kg·m/s) / 5850 kg
= 42.5 m/s
Therefore, the car's new speed after the load is dropped will be 42.5 m/s.