- #1
Zubz
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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?