Two kids sledding down frictionless hill- conservation of momentum/energy

[grantaere]

Homework Statement

Gayle runs at a speed of 4.20m/s and dives on a sled, which is initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 4.77m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 17.4m? Gayle's mass is 50.8kg, the sled has a mass of 5.50kg and her brother has a mass of 33.7kg.

Homework Equations

conservation of momentum: m1v1+ m2v2 = m1+m2(v3)
conservation of energy: KE1 + PE1 = KE2 + PE2

The Attempt at a Solution

I tried using conservation of momentum to find the velocity of Gayle+sled, then used that to find Gayle+sled's PE and KE at a vertical distance of 12.63m --> total energy of the system.
That's when things become fuzzy... are we supposed to include the potential energy of Gayle's unmoving brother within the system? (In which case to find the final velocity at the bottom of the hill, PE of both her and her brother = zero, and KE of Gayle+sled+brother = total energy of the system?)
Any help would be incredibly appreciated!

 

[gneill]

Hi grantaere, Welcome to Physics Forums.

It's usually best to solve these sort of problems in stages. The first stage involves an inelastic collision between Gayle and the sled. There will be a resulting velocity of the pair, and thus a KE that goes into the next stage.

The next stage involves a vertical drop of 4.77 m, so you can work out the change KE via conservation of energy: you have a change in vertical height, so a change in gravitational PE which converts to a change in KE. The KE at the end of the stage gives you a velocity that Gayle and the sled take into an inelastic collision with the brother (the next stage).

The inelastic collision takes place with some resulting new velocity for them all (Gayle, the sled and her brother are now one combined object).

The instant after the collision, Gayle, the sled, and her brother continue at some new velocity, and so with some new KE, proceeding to fall another 12.63 m in vertical height, gaining KE from the new change in gravitation PE associated with that change in elevation.

Note that you don't have to pin your zero reference for gravitational PE to the bottom of the slope for all the stages. It is enough that you can determine the change in PE involved for a given stage, and that is tied to the change in vertical height that occurs during that stage.

 

[_N3WTON_]

First you would use conservation of momentum to find the velocity of Gayle plus the sled(at the instant she jumps on). Then, use the final velocity you found above in combination with conservation of energy to find her speed at the instant bro jumps on. Next, use the velocity found above and conservation of momentum to find the sled's velocity after the collision (her brother jumping on the sled). Finally, use the conservation of energy to find the speed at the bottom of the hill. You are basically breaking this problem up into four parts, and using the velocity you find in each part to help you find a new velocity in the next step of the problem. I hope that makes sense...

 

[Head_Unit]

Um. At the top you have Gayle's kinetic energy = 0.5*50.8*4.20^2 = 448.05 by my calculations.

And a potential energy = m*g*h = 50.8*9.81*17.4 = 8671.25. BUT WAIT! I forgot the sled! PE= (50.8+5.50)*9.81*17.4 =9610.07

As for the brother, he has a potential energy=33.7*9.81 *(17.4-4.77)=4175.44
I think we must assume when he "hops" he does not donate kinetic energy.

Then yeah, all those energies turn into KE at the bottom, and you end up taking a square root to get the velocity

 

[haruspex]

Um. At the top you have Gayle's kinetic energy = 0.5*50.8*4.20^2 = 448.05 by my calculations.

And a potential energy = m*g*h = 50.8*9.81*17.4 = 8671.25. BUT WAIT! I forgot the sled! PE= (50.8+5.50)*9.81*17.4 =9610.07

As for the brother, he has a potential energy=33.7*9.81 *(17.4-4.77)=4175.44
I think we must assume when he "hops" he does not donate kinetic energy.

Then yeah, all those energies turn into KE at the bottom, and you end up taking a square root to get the velocity

No, you need to take into account those inelastic collisions, as mentioned in posts #2 and #3. Work is not conserved during those.

 

[haruspex]

The first stage involves an inelastic collision between Gayle and the sled. There will be a resulting velocity of the pair, and thus a KE that goes into the next stage.

That's surely the way to handle it, but there is an awkwardness here. That velocity will be horizontal. If the slope starts suddenly, the sled will necessarily become airborne for a while. Another inelastic collision occurs when it contacts the slope. To solve it, it is necessary to assume that the slope initially curves down gently, so that the vertically acceleration is less than g.

 

[gneill]

That's surely the way to handle it, but there is an awkwardness here. That velocity will be horizontal. If the slope starts suddenly, the sled will necessarily become airborne for a while. Another inelastic collision occurs when it contacts the slope. To solve it, it is necessary to assume that the slope initially curves down gently, so that the vertically acceleration is less than g.

The problem statement doesn't specify the collision angle, but given the question level I assumed no added complications; Gayle would be running along the slope and jumping onto the sled parallel to the slope's surface. Why the sled would be at rest on a frictionless slope until the collision occurred is left to the the imagination of the reader :)

 

[haruspex]

Gayle would be running along the slope and jumping onto the sled parallel to the slope's surface. Why the sled would be at rest on a frictionless slope until the collision occurred is left to the the imagination of the reader :)

I read it a little differently. I took the sled to be at rest on the horizontal just before the slope.

 

[gneill]

I read it a little differently. I took the sled to be at rest on the horizontal just before the slope.

Yes, I agree that is a valid interpretation of the problem statement. Of course the horizontal section could be of any length before the slope begins since there are no time constraints and the surface is frictionless

 

[grantaere]

I've figured it out now, thank you all so much for the help! :)