Question

A toboggan having a mass of 12.5 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/t = 2 m/s, measured relative to the toboggan. Suppose that h = 3.2 m . Neglect friction in the calculation.Part ADetermine the velocity of the toboggan afterwards.vt =

Answer 1

about 9.63 m/s

Step-by-step explanation:

You want the speed of a toboggan at the bottom of a frictionless slope of height 3.2 m if it has a mass of 12.5 kg and carries a girl and boy of mass 40 kg and 45 kg respectively. The speed of interest is the speed after the boy is ejected from the back at a speed of 2 m/s relative to the toboggan.

Speed

The speed at the bottom of the slope will be the result of converting the initial potential energy to kinetic energy. It will be ...

v = √(2gh) = √(2·9.8·3.2) m/s ≈ 7.920 m/s

Momentum

The momentum of the toboggan and riders will be the same before and after the boy is ejected. This relation can be solved for the final velocity of the toboggan.

`(m_t+m_g+m_b)v_0=(m_t+m_g)v_1+m_b(v_0-2)\\\\(m_t+m_g)v_0+2m_b=(m_t+m_g)v_1\qquad\text{add $2m_b$}\\\\v_1=v_0+(2m_b)/(m_t+m_g)\qquad\text{divide by $m_t+m_g$}\\\\v_1=√(62.72)+(2\cdot45)/(12.5+40)\approx9.634`

The final speed of the toboggan and girl is about 9.63 m/s.