-/1 points SerCP10 6.P 031 0/4 Submissions Used My Notes OAsk Your Te Gayle runs

Question

-/1 points SerCP10 6.P 031 0/4 Submissions Used My Notes OAsk Your Te Gayle runs at a speed of 3.60 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hil. After she has descended a vertical distance of 5.00 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 15.0 m? Gayle's mass is 46.0 kg, the sled has a mass of 5.50 kg and her brother has a mass of 30.0 kg m/s Need Help? RN Submit Answer Save Progress

Explanation / Answer

Initial Data:

The velocity of Gayle = 3.60 m/s

The velocity of sled = 0

The velocity of here brother = 0

After first collision (Collision of Gayle with sled):

The momentum will be conserved as there is no external impulse.

46* 3.60 = 46* v1 + 5.5*v2 => 8.36v1 + v2= 30.11 -------(i)

The velocity of aproaching will be equal to the velocity of separation.

3.60 = v2-v1 ----(ii)

v1 = 2.83 m/s and v2 = 6.43 m/s

before she collides with her brother,

speed of Gayle = sqrt(v1^2 + 2*9.81*5) = 10.30 m/s

after collision the common speed:

v = (46/(46+40))*10.30 = 5.51 m/s

The velocity at the bottom will be sqrt(5.51^2 + 2*9.81*10) = 15.05 m/s

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