A train carriage travelling at 40 km/h collides and links up with two identical carriages and the three?

Question

contunie moving together. At what velocity will they travel together?
Help my physic homework please guys!!!!!

Answer 1

Momentum is conserved. Figure out the total momentum of all three carrianges before the collision, taking into account the direction of the momentums.

After the collision, the momentum of the three linked carrianges must be equal to the momentum of the system before the collision.

If the other two carriages are not moving before the collision, then the velocity of the linked system will be 40/3.

Answer 2

This is a conservation of linear momentum problem. Since they are all identical they all have the same mass. When they link up we can think of this as a single train with three times the mass as a single carriage.

Since momentum is conserved we know that

Pf = Pi

where Pf is the final momentum and Pi is the initial momentum. We also know that

P = m v

where m is the mass and v is the velocity. So now we need to find out how much linear momentum the single moving carriage had before the collision. We know the velocity, but not the mass. That is not a problem, as we shall see. Normally, I would convert the speed to m/s but that is not necessary here, since the final answer will most likely also be in these units.

Pi = m (40km/h) + m (0) + m (0)

The extra two terms are the momentum contributed by the other two carriages. Since their velocity is zero, the do not contribute anything to the system.

Now we need to figure out how to represent Pf. Since we can think of the three carriage system as a single carriage with 3 times the mass, we can say that

Pf = (3m) (v)

Since we know that Pf = Pi we have an equation that we can solve.

(m) (40km/h) = (3m) (v)

dividing both sides by (3m) gives us

(m)/(3m) (40km/h) = v

v = (1/3) (40km/h)
= 13.3 km/h
= 13 km/h (with the proper amount of significant figures)

So, assuming there is no energy lost due to sound, friction or other forces, the three carriage system will be moving at about 13 km/h.

Answer 3

I'm assuming the two carriages that the moving one links up with where stationary before this event.

Using the conservation of momentum:
Since 2 of the carriages where motionless before the collision, they had 0 momentum. So the momentum of the system (3 carriages) where equal to the product of the velocity and the mass of the moving carriage. If m is the mass of a carriage we have, 40 * m as the momentum (ignoring units).

After the collision, all three carriages are moving and so the momentum is v * (3*m), where v is the velocity we want to calculate.
v * (3*m) = 40*m, which gives
v = 40/3 = 10.333 km/h

Answer 4

Momentum=mv. I think you know this formula.

Momentum before collision=Momentum after collision. This is based on the principle of conservation of momentum.

Momentum before collision, of train no.1=m*40
Momentum before collision, of train no.2&3=2m*0

Momentum after collision, of train No.1,2,&3=3mv

Therefore,

m*40+2m*0=3mv

(m cancels out, right ?)

40=3v
v=40/3
=13.33km/h

Note: Can you imagine this situation really happening in real life? At 40km/h train no. 1 will likely damage the 2 other trains that the 3 together might not be able to travel together at once. They'd need repairmen first. I wonder what the person who wrote this problem has to say about this.