How Does Conservation of Momentum Apply in a Frictionless Sled Problem?

[PatsObi]

I'm stuck on this problem.

A sledge of mass 120 kg is at rest on a horizontal icy surface. A man of mass 70 kg stands at one end of the sledge so the initially the distance from the man to the shore is 20 m. The man now walks 3 m relative to the sledge, towards the shore. The he stops. Assuming the moves w/out friction, how far from the shore is the man when he stops.

THis is pretty much all i know:

70(Rm)+120(Rs)=0

where Rm and Rs are the distances from the man to the cm and the cm of the sledge to the cm respectfully.

 

[Shooting Star]

THis is pretty much all i know:

70(Rm)+120(Rs)=0

where Rm and Rs are the distances from the man to the cm and the cm of the sledge to the cm respectfully.

That is actually all you need, but with some understanding. Can you tell me what this eqn signifies? And how to find the CM of the system wrt the shore? After that, we can form the necessary eqns.

Also, the word you were looking for is "respectively", not respectfully.

 

[Moetasim]

Hello, I understand that you are stuck on the frictionless sled problem. It seems like you have already set up the equation using the distances from the man to the center of mass (cm) and the center of mass of the sledge to the cm. However, in order to solve this problem, we also need to consider the conservation of momentum and the conservation of energy.

First, we can start by setting up the conservation of momentum equation:

m1v1 + m2v2 = (m1 + m2)v'

Where m1 and v1 are the mass and velocity of the man, m2 and v2 are the mass and velocity of the sled, and v' is the final velocity of the combined system.

Since the problem states that the man initially stands at one end of the sledge, we can assume that the initial velocity of the sled is zero. This means that the equation becomes:

m1v1 = (m1 + m2)v'

Substituting in the values given in the problem, we get:

(70 kg)(3 m/s) = (70 kg + 120 kg)v'

Solving for v', we get:

v' = 1.82 m/s

Next, we can use the conservation of energy to find the distance from the shore where the man stops. We can set up the equation:

mgh = 1/2(m1 + m2)v'^2

Where m is the total mass (70 kg + 120 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the distance from the shore where the man stops, and v' is the final velocity we calculated earlier.

Substituting in the values, we get:

(190 kg)(9.8 m/s^2)(h) = 1/2(190 kg)(1.82 m/s)^2

Solving for h, we get:

h = 0.17 m

Therefore, the man stops 0.17 m from the shore when he walks 3 m relative to the sled. I hope this helps you solve the problem. Remember to always consider the conservation of momentum and energy when solving physics problems involving motion. Good luck!