Question

Gayle runs at a speed of 4.20 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. 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 52.0 kg, the sled has a mass of 5.35 kg and her brother has a mass of 30.0 kg.

What is ______ m/s

Answer 1

Given,

The mass of Gayle, m_G = 52 kg

The mass of sled, m_s = 5.35 kg

The mass of a brother, m_b = 30 kg

The initial speed of Gayle, u = 4.20 m/s

Let the final velocity of Gayle's and sled just after Gayle dives on it be ' v₁'

By using the moment of conservation,

m_G u₁ =(m_G + m_s) v₁

52 * 4.20 = (52 + 5.35) v₁

57.35 v₁ = 218.4

v₁ = 3.81 m/s

Let the velocity of after she descended a vertical distance of 5 m be v₂

By the conservation of energy,

1/2 * (m_G + m_s) v₁² + (m_G + m_s) g h = 1/2 * (m_G + m_s) v₂²

(v₁² )/ 2 + gh =  (v₂²) / 2

3.81² / 2 + 9.8 * 5 = (v₂²) / 2

0.5 (v₂²) = 56.26

v₂² = 112.52

v₂ = 10.61 m/s

Let the velocity just after her brother jumps be ' v₃'

By using the conservation of momentum

(m_G + m_s) v₂ = ( m_G + m_s + m_b ) * v₃

(52 + 5.35) * 10.61 = (52 + 5.35 + 30) * v₃

87.35 * v₃ = 608.484

v₃ = 6.97 m/s