Conservation of momentum and energy

  • Thread starter Zubz
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  • #1
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Homework Statement


Hi, new to the forums and I was wondering if someone could help me figure out what im 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???
 

Answers and Replies

  • #2
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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:
  • #3
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thank you, worked perfectly.
 

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