Question

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

a) what is the change in momentum?

b.) What is the magnitude of the force which Superman exerts on the train?

Answer 1

Hello!

a) What is the change in momentum?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

`\boxed{a=(V-Vi)/(t) }`

Like you see, the final velocity will be zero, so:

`a = (0m/s-40m/s)/(4s)`

`a = (-40m/s)/(4s)`

`a = -10\ m/s^(2)`

Like you see, the aceleration applicate by superman is of -10 m/s^2.

If the aceleration is negattive, means the velocity will decrease.

b.) What is the magnitud of the force wich superman exerts on the train?

For calculate the force applicate for superman, lets applicate the second law of Newton:

`\boxed{F=ma}`

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of -80000 Newtons.

When the force is negative means the force applicated in the another direction of the object what is traveling.