Momentum: Solving Qs on Man Throwing Rocks in Lake

[Arbitrary]

Homework Statement

A powerful, 100-kg man is kneeling in the middle of a frozen lake, holding two large rocks. One rock has a mass of 25 kg, and the other a mass of 15 kg. The man wants to quickly get to the south edge of the lake, and he knows that he needs to throw his rocks to the north to propel himself there. He always throws each rock away from himself at 20 m/s.

What will be his final speed if he
a) throws both rocks simultaneously?
b) throws the heavy rock and then the lighter rock?
c) throws the lighter rock and then the heavy rock?

Homework Equations

m$$\Delta$$v$$_{initial}$$ = m$$\Delta$$v$$_{final}$$

The Attempt at a Solution

(a)

m$$_{man}$$v = (m$$_{rock1}$$+m$$_{rock2}$$)v$$_{rock}$$

100v = (20+15)20
v = 8 m/s

(b)

mv = m$$_{rock1}$$v$$_{rock}$$
100v = 25*20
v = 5

mv$$_{final}$$-(m+m$$_{rock2}$$)v$$_{initial}$$ = m$$_{rock2}$$v
100v$$_{final}$$-(100+15)*5 = 15*20
v$$_{final}$$ = 8.75 m/s

(c) I get the same answer as (b), using the same methods, which is quite weird.

The correct answers are 5.71 m/s for (a), 6.18 m/s for (b), and 6.14 m/s for (c). I'm not sure what I'm doing wrong for each part...

 

[Vijay Bhatnagar]

Your solution for a) is correct. In solution for b) when the heavier rock is thrown the man is still holding the other rock. So his net weight will be 115 kg. The second rock is thrown with a speed of 20 m/s relative to the man, in the direction opposite to in which the man is moving after throwing the first rock. Its speed relative to the ground will be (20 - speed of the man after throwing the first rock) in the opposite direction. Solve on the above basis. Answers given appear to be incorrect.

 

[ogb p]

Hello, it seems like you are on the right track with your calculations. However, I believe the mistake lies in how you are applying the momentum equation. Remember that momentum is a vector quantity, so it has both magnitude and direction. In this case, the direction of the man's final velocity will depend on the direction in which he throws the rocks.

For part (a), when throwing both rocks simultaneously, the man's initial momentum is zero since he is at rest. However, the combined momentum of the rocks will be in the opposite direction of his final velocity. Therefore, the correct equation to use would be:

m_{man}v_{man,initial} = -(m_{rock1}+m_{rock2})v_{rock,final}

Plugging in the values, we get:

100(0) = -(25+15)(20)
0 = -800
v_{man,final} = 8 m/s

For parts (b) and (c), the man's initial momentum will be in the direction of his throw, and his final momentum will be in the opposite direction. Therefore, the correct equations to use would be:

m_{man}v_{man,initial} = m_{rock1}v_{rock1,final}
and
m_{man}v_{man,initial} = m_{rock2}v_{rock2,final}

Plugging in the values, we get:

m_{man}v_{man,initial} = 25(20)
v_{man,initial} = 5 m/s

And for part (c):

m_{man}v_{man,initial} = 15(20)
v_{man,initial} = 3 m/s

Now, using the momentum equation again, we can find the man's final velocity in each case. For part (b):

m_{man}v_{man,final} = -(m_{rock1}+m_{rock2})v_{rock1,initial} - m_{rock2}v_{rock2,initial}
100v_{man,final} = -(25+15)(20) - 15(5)
100v_{man,final} = -800
v_{man,final} = 8 m/s

And for part (c):

m_{man}v_{man,final} = -(m_{rock1}+m_{rock2})v_{