Calculating Angular Momentum: Converting Units and Solving for Mass and Velocity

[Sneakatone]

I used kg*m/s*m

so 1500 metric tons=1500000 kg
85 km/h=23.61 m/s

( 1500000 )*23.61*53=1.877x10^9 kg*m^2/s

I feel like this is correct.

 

[gneill]

1.877 x 10^9 kg*m^2/s looks right. Perhaps there's an issue with your rounding of the answer.

 

[BruceW]

I agree with you. Maybe because you rounded down, they didn't give the mark? The one other thing that it might be, is that they don't actually say which direction the train is moving in. They just say where the point is in relation to the track. The natural assumption is to do as you did, and assume they mean that if the point is 'to the left' of the train, then that means someone who is facing forward on the train will see the point on their left. But there is an ambiguity here. Maybe they took it to be the other way around.

 

[Sneakatone]

I origionally put 1.8 instead of 1.877 but your right, thanks!

 

[gneill]

I agree with you. Maybe because you rounded down, they didn't give the mark? The one other thing that it might be, is that they don't actually say which direction the train is moving in. They just say where the point is in relation to the track. The natural assumption is to do as you did, and assume they mean that if the point is 'to the left' of the train, then that means someone who is facing forward on the train will see the point on their left. But there is an ambiguity here. Maybe they took it to be the other way around.

The direction "upwards" for the angular momentum vector was marked correct, so it looks like the assumption panned out.

 

[BruceW]

oh yeah. good, good.