Question 4, chap 109, sect 6 part l of 1 10 points Gayle runs at 4.64 m/s and di

Question

Question 4, chap 109, sect 6 part l of 1 10 points Gayle runs at 4.64 m/s and dives on a sled, initially at rest onto the top of a frictionless snow covered After she and the sled have descended a vertical distance of 3.27 m her brother who is initially at rest hops on her back and together they continue down the The acceleration of gravity is 9.8 m/s. What is their speed at the bottom if the total vertical drop is 14.9 m? Gayle's mass is 42.3 kg, the sled's is 2.7 kg and her brother's is 39.1 kg. Answer in units of m/s.

Explanation / Answer

Here, first of all, starting at the top of the hill, find the speed of the sled plus Gayle at the instant she lands on it. Apply conservation of momentum -

So,
p(i) = p(f)
mv(i) + mv(i) = (m + m)v(f)
v(f) = [mv(i) + mv(i)] / (m + m)
= [42.3kg)(4.64m/s) + 0] / (42.3kg + 2.7kg)
= 4.36 m/s

Now, her brother jumps on when Gayle and the sled have descended 3.27 m. The velocity of Gayle plus sled at the instant her brother jumps on is found from the law of conservation of energy:

E(i) = E(f)
KE(i) + PE(i) = KE(f) + PE(f)
0.5mv²(i) + mgh(i) = 0.5mv²(f) + mgh(f)
v(f) = [v²(i) + 2g{h(i) - h(f)}]

Initial velocity here, is final velocity from the first stage. Therefore:

v(f) = [(4.36 m/s)² + 2(9.80m/s²)(3.27 m - 0)]
= 9.12 m/s

At this instant. her brother jumps on, so we have another perfectly inelastic collision as in the first stage. Initial velocity here is 9.12 m/s, mis 45.0kg, and mof course is 39.1 kg :

v(f) = [mv(i) + mv(i)] / (m + m)
= [(45.0kg)(9.12 m/s) + 0] / (45.0kg + 39.1 kg)
= 4.88 m/s

Again, apply conservation of energy again gets you the final velocity at the bottom of the hill -

v(f) = [v²(i) + 2g{h(i) - h(f)}]
= [(4.88 m/s)² + 2(9.80m/s²)(11.63 m - 0)]
= 15.87 m/s
Therefore, the final speed of Gayle and her brother along with sledge at the bottom is 15.87 m/s.

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