Conservation of momentum and energy

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SUMMARY

The discussion focuses on the application of conservation of momentum and energy in a physics problem involving Gayle, her brother, and a sled on a frictionless hill. The initial speed of the sled was calculated to be 3.89 m/s using momentum conservation. After Gayle descends 5.33 m, the speed of the sled should be approximately 10.94 m/s before her brother hops on. The final velocity after the brother joins was determined to be 12.85 m/s, but the correct approach involves recalculating the kinetic energy and potential energy after the brother's mass is accounted for.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Knowledge of kinetic energy (K.E = 1/2mv^2) and potential energy (P.E. = mgy)
  • Ability to perform calculations involving mass and velocity
  • Familiarity with gravitational acceleration (g = 9.81 m/s²)
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  • Calculate the velocity of the sled at various heights using conservation of energy
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  • Learn about advanced applications of conservation laws in complex systems
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Homework Statement


Hi, new to the forums and I was wondering if someone could help me figure out what I am doing wrong. Thanx in advance.
Gayle runs at a speed of 4.38 m/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 5.33 m, 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 12.6 m? Gayle's mass is 46.5 kg, the sled has a mass of 6.00 kg and her brother has a mass of 32.4 kg.

Homework Equations



K.E=1/2mv^2
P.E.=mgy
p=mv
Mgyle= 46.5 kg
M(gyle+sled)= 52.5 kg
M(gyle+sled+bro) = 84.9 KG

The Attempt at a Solution



Ok so what i did was i found out the initial speed of the sled using conservation of momentum.
p=p!(prime)
Mgyle*Ugyle = M(gyle+sled)V
V=3.89 m/s
then i used conservation of energy to find out the velocity of the sled at 5.35m.
Ebefore= Eafter
mgYo + 1/2mV1^2 = mgY1 + 1/2mV2^2
( 2(52.5)(9.81)(12.6) + (52.5*3.89^2) - (2*52.5*9.81*5.35) ) / 52.5 = v2^2
v2= 12.55 m/s
Now when i calculated this velocity. Should i include the brothers mass or do i find it after using conservation of momentum using the new velocity (12.55 m/s).
AND if i use conservation of momentum i get the new velocity to be 7.76 m/s (when brothers jumps on) at 5.35m.
After that, I use conservation of energy again to find final velocity.
Ebefore = Eafter
2MgY1 + MV2^2 = MV3^2
( (2*84.9*9.81*5.35) + (84.9*7.76^2) ) / 84.9 = v3^2 (USED NEW VELOCITY AFTER COLLISION WITH BROTHER)
v3 = 12.85 m/s but this isn't right. Can any1 tell me where i went wrong?
 
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then i used conservation of energy to find out the velocity of the sled at 5.35m.
Ebefore= Eafter
mgYo + 1/2mV1^2 = mgY1 + 1/2mV2^2
( 2(52.5)(9.81)(12.6) + (52.5*3.89^2) - (2*52.5*9.81*5.35) ) / 52.5 = v2^2

After she has descended a vertical distance of 5.33 m,

I believe I have found your problem. When the sled descends a vertical distance of 5.33 m, the sled is a vertical distance of (12.6 - 5.33) m above the ground. You should come up with about 10.94 m/sec for the speed at that point before the brother hops on.

Now when i calculated this velocity. Should i include the brothers mass or do i find it after using conservation of momentum using the new velocity (12.55 m/s).
Find the velocity of the sled without the brother's mass at (12.6 - 5.33) m above the ground. Then use conservation of momentum with the new velocity to get the velocity after the brother gets on. Then use the result to calculate the total energy. The reasoning being that the sled is already traveling at a certain velocity before the brother gets on. When the brother jumps on, he not only changes the configuration of the potential and kinetic energy, but he adds to the total energy too. In order to figure out the new total energy, you need to find both the new potential energy that he adds and the new kinetic energy of the sled after he jumps on, but in order to calculate the new kinetic energy, you need to use the conservation of momentum.
 
Last edited:
thank you, worked perfectly.
 

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