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>t = 2 m>s, measured relative to the toboggan. Determine the velocity of the toboggan afterwards. Neglect friction in the calculation.

Answer 1

Answer:

The velocity of the toboggan is 7.51 m/s.

Explanation:

Given that,

Mass of toboggan = 10 kg

Mass of girl =40 kg

Mass of boy = 45 kg

Velocity = 2 m/s

Suppose that h = 3.2 m

We need to calculate the velocity

Using conservation of energy

$K.E_{1}+P.E_{1}=K.E_{2}+P.E_{2}$

$\dfrac{1}{2}(m_{t}+m_{g}+m_{b})v^2+(m_{t}+m_{g}+m_{b})gh=\dfrac{1}{2}(m_{t}+m_{g}+m_{b})v^2+0$

Put the value into the formula

$0+(10+40+45)\times9.8\times3.2=\dfrac{1}{2}(10+45+40)v^2$

$v^2=\dfrac{95\times9.8\times3.2\times2}{95}$

$v=\sqrt{62.72}\ m/s$

$v=7.91\ m/s$

The velocity of boy is

$v_{b/t}=v_{t}-v_{b}$

$v_{b}=v_{t}-2$

We need to calculate the velocity of the toboggan

Using principle of momentum

$(m_{t}+m_{g}+m_{b})v=(m_{t}+m_{g})v_{t}+m_{b}v_{b}$

Put the value into the formula

$(10+40+45)\times7.91=(10+45)\times v_{t}+45\times(v_{t}-2)$

$751.45=55v_{t}+45v_{t}-90$

$751.45+90=130v_{t}$

$v_{t}=\dfrac{751.45}{100}$

$v_{t}=7.51\ m/s$

Hence, The velocity of the toboggan is 7.51 m/s.