Find the Distance (Momentum)

[Tom2]

Homework Statement

You are 60kg and you are holding a 10kg weight as you stand at rest on a frictionless surface. you can throw the weight 5m on ground where there is friction to keep you from moving backwards. How far will the 10kg weight be from you after you throw it on the frictionless surface?

Homework Equations

M1V1 + M2V2 = 0

The Attempt at a Solution

Total momentum of the system (you + weight) after the throw must be the same as it was before so it will be 0. [/B]

M1 = 60kg
M2= 10kg
V1 = ?
V2 = ?

Please help! Thanks!

 

[haruspex]

Homework Statement

You are 60kg and you are holding a 10kg weight as you stand at rest on a frictionless surface. you can throw the weight 5m on ground where there is friction to keep you from moving backwards. How far will the 10kg weight be from you after you throw it on the frictionless surface?

There isn't really enough information. To answer this you will need to make some assumptions about how arms work.
Which of the following do you think will be approximately the same in the two cases:

• the force exerted by the muscles
  
• the distance over which the force is exerted (as far as the muscles are concerned)
• the power delivered by the muscles
• the time for which the power is delivered

?

 

[PhanthomJay]

Homework Statement

You are 60kg and you are holding a 10kg weight as you stand at rest on a frictionless surface. you can throw the weight 5m on ground where there is friction to keep you from moving backwards. How far will the 10kg weight be from you after you throw it on the frictionless surface?

Homework Equations

M1V1 + M2V2 = 0

The Attempt at a Solution

Total momentum of the system (you + weight) after the throw must be the same as it was before so it will be 0. [/B]

Yes.

M1 = 60kg
M2= 10kg
V1 = ?
V2 = ?

Please help! Thanks!

Try solving for V1 in terms of V2. Assume you the throw the weight horizontally.

 

[haruspex]

Hi Tom2,

Just had a private chat with PhanthomJay. His model is like firing a gun. That's rather different because the speed of the bullet would be the same with or without friction. For that to apply to a thrower, the thrower would have to absorb the recoil initially by flexing the body, so as to avoid any horizontal force on the feet until the mass has been released (with and without friction).
I regard that as the least realistic of three models (the other two being constant force and constant power), but if the question is not intended to be much advanced then it may be what the questioner has in mind.
The constant force model can also be analysed without calculus, but I regard constant power as the most realistic.

 

[HalfLostAndSearching]

To find the distance that the weight will be from you after you throw it on the frictionless surface, we need to use the equation for momentum, which is:

p = mv

Where p is the momentum, m is the mass, and v is the velocity.

We know that the total momentum of the system (you + weight) must be 0 after the throw, so we can set up the following equation:

(60kg)(0m/s) + (10kg)(v2) = 0

Solving for v2, we get:

v2 = 0m/s

This means that the weight will have a velocity of 0m/s after the throw on the frictionless surface. Since the weight is not moving, it will remain 5m away from you, the same distance as when you threw it.