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Chapter 9: Problem 20

A toboggan of mass \(8.6 \mathrm{kg}\) is moving horizontally at \(23 \mathrm{km} / \mathrm{h}\). As it passes under a tree, \(15 \mathrm{kg}\) of snow drop onto it. Find its subsequent speed.

Short Answer

Expert verified
The subsequent speed of the toboggan after the snow drops onto it is 2.33 m/s.

Step by step solution

01

Calculate the Initial Momentum of the Toboggan

Start by finding the initial velocity in m/s by converting from km/h to m/s. This is done by dividing the initial velocity in km/h by 3.6. So, the initial velocity is \(23km/h ÷ 3.6 = 6.39 m/s\). Then we calculate the initial momentum of the toboggan before the snow falls which is the product of the mass and the initial velocity. So, \(p_initial= m_toboggan × v_initial = 8.6kg × 6.39m/s = 54.95 kg*m/s\)
02

Calculate the Total Momentum of the System After the Snow Falls

After the snow falls on the toboggan, the snow is stationary hence it does not contribute any momentum. The total momentum of the system after the snow fall (p_final) is then the sum of the initial momentum of the toboggan (p_initial) and the zero momentum of the snow. So, \(p_final = p_initial + m_snow × 0 = 54.95 kg*m/s\)
03

Calculate Final Speed

Now, we will calculate the final speed of the toboggan. As per the conservation of momentum principle, the momentum before the snow falls is equal to the momentum after the snow falls. Therefore,\(p_final = p_initial\), from where we get, \(m_total × v_final = p_initial\). The total mass (m_total) is the mass of the toboggan plus the mass of the snow, \(m_total = m_toboggan + m_snow = 8.6 kg + 15 kg = 23.6 kg\). Therefore, the final velocity is: \(v_final = p_initial ÷ m_total = 54.95 kg*m/s ÷ 23.6 kg = 2.33 m/s\)

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