Momentum conservation problem — Man standing on a trolley throwing sandbags off

[Alsfc]

Homework Statement:

A man stands on a trolley that is free to move on a horizontal surface and subject to no
resistance forces. The mass of the man is 70 kg and the mass of the trolley is 10 kg.
He has two sandbags, each of mass 20 kg, on the trolley with him. Initially, the trolley
and the man are at rest.

(a) The man throws one bag off the trolley, and after this the trolley moves at a constant
velocity. The bag is thrown so that it has a horizontal velocity of 2 m/s. Find the speed
of the trolley, man and the remaining bag.

(b) After the first bag has been thrown, the man then throws the second bag, in the same
direction as the first. Immediately after the bag is thrown, it has a horizontal velocity
of 2 m/s relative to the trolley. Find the speed of the trolley and the man after the
second bag has been thrown.

Relevant Equations:

m1v1 +m2v2 = m1v3 + m2v4

I am having trouble with part b). I found the answer to part a) is -0.4 and for part b) the mark scheme says that the velocity of the 2nd bag is 1.6m/s but how do I find that out? If someone could perhaps show step-by-step of part b)?
Thank you!

 

[phystro]

Hello, obviously you will need to use again conservation of momentum. Let MV be the total momentum before (with M= 100, and V=- 0.4) and after he throws the bag the total momentum will be 1.6(2) + 80v', hence you can calculate v' easily.

 

[jbriggs444]

Homework Statement:: A man stands on a trolley that is free to move on a horizontal surface and subject to no
resistance forces. The mass of the man is 70 kg and the mass of the trolley is 10 kg.
He has two sandbags, each of mass 20 kg, on the trolley with him. Initially, the trolley
and the man are at rest.

(a) The man throws one bag off the trolley, and after this the trolley moves at a constant
velocity. The bag is thrown so that it has a horizontal velocity of 2 m/s. Find the speed
of the trolley, man and the remaining bag.

(b) After the first bag has been thrown, the man then throws the second bag, in the same
direction as the first. Immediately after the bag is thrown, it has a horizontal velocity
of 2 m/s relative to the trolley. Find the speed of the trolley and the man after the
second bag has been thrown.
Relevant Equations:: m1v1 +m2v2 = m1v3 + m2v4

I am having trouble with part b). I found the answer to part a) is -0.4 and for part b) the mark scheme says that the velocity of the 2nd bag is 1.6m/s but how do I find that out? If someone could perhaps show step-by-step of part b)?
Thank you!

Without seeing the mark scheme myself, I cannot be sure that it is saying what you think it is.

The final forward velocity of the thrown bag in part b will not be 1.6 m/s forward relative to the initial rest frame of the trolley+man+2 bags in part a.

Consider part b by itself. Adopt a frame of reference in which the 70 kg trolley, the 10 kg man and the remaining 20 kg bag begin at rest.

We are told that after the bag is flung forward, its velocity relative to the trolley+man will be 2 m/s.

After it is flung, we do not know the bag's forward velocity relative to our chosen frame. It is unknown. So give this velocity a name and call it v.

If the bag moves forward at v, how fast must the bag+man move rearward?

 

[PeroK]

for part b) the mark scheme says that the velocity of the 2nd bag is 1.6m/s but how do I find that out?

If the trolley has velocity $-0.4m/s$ (units are important, by the way) relative to the ground and the bag has a velocity of $+2m/s$ relative to the trolley, then the trolley has a velocity of $-0.4m/s + 2m/s$ relative to the ground.

 

[jbriggs444]

If the trolley has velocity $-0.4m/s$ (units are important, by the way) relative to the ground and the bag has a velocity of $+2m/s$ relative to the trolley, then the trolley has a velocity of $-0.4m/s + 2m/s$ relative to the ground.

The bag's final velocity is given relative to the trolley's final velocity, not relative to the trolley's intermediate velocity.

[Which is a bit bizarre, since it means that the two throws have significantly different relative velocities]

 

[Alsfc]

If the trolley has velocity $-0.4m/s$ (units are important, by the way) relative to the ground and the bag has a velocity of $+2m/s$ relative to the trolley, then the trolley has a velocity of $-0.4m/s + 2m/s$ relative to the ground.

Thank you!