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 $v_{b / t}$=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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Conservation of energy

$T_1+V_1=T_2+V_2$

$\dfrac{1}{2}(m_t+m_g+m_b)v_1^2+(m_t+m_g+m_b)gh=\dfrac{1}{2}(m_t+m_g+m_b)v_2^2+0$

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

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

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