Question

A toboggan having a mass of 10 kg starts from rest at A and carries a girl and boy having a mass of 40 kg and 45 kg, respectively. When the toboggan reaches the bottom of the slope at B , the boy is pushed off from the back with a horizontal velocity of vb/tv_{b / t}=2 m/s , measured relative to the toboggan. Determine the velocity of the toboggan afterwards. Neglect friction in the calculation.

Solution

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Answered 2 months ago
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Conservation of energy

T1+V1=T2+V2T_1+V_1=T_2+V_2

12(mt+mg+mb)v12+(mt+mg+mb)gh=12(mt+mg+mb)v22+0\dfrac{1}{2}(m_t+m_g+m_b)v_1^2+(m_t+m_g+m_b)gh=\dfrac{1}{2}(m_t+m_g+m_b)v_2^2+0

0+(10+40+45)(9.81)(3)=12(10+40+45)v22+00+(10+40+45)(9.81)(3)=\dfrac{1}{2}(10+40+45)v_2^2+0

v2=7.672  m/s\therefore v_2=7.672\;\text{m/s}

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