Question

A (trolley + child) of total mass $200\,{\rm{kg}}$ is moving with a uniform speed of $36\,{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ on a frictionless track. The child of mass $20\,{\rm{kg}}$ starts running on the trolley from one end to the other ($10\,{\rm{m}}$ away) with a speed of $10\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$ relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
A. Final speed= $9\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$
B. Final speed= $4\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$
C. Trolley moves $9\,{\rm{m}}$
D. Trolley moves $4\,{\rm{m}}$

Answer 1

The law of conservation of momentum shows that the initial momentum is equal to final momentum of the system. Find out the initial momentum of the system and final momentum of the system and equate them to find the final velocity of the trolley. Distance covered by trolley is the product of velocity and time taken.

Complete step by step answer:
The mass of the system is $200\,{\rm{kg}}$; the initial speed of system is $36\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$. The mass of child is $20\,{\rm{kg}}$, the distance covered by the child is $10\,{\rm{m}}$ and the speed of child relative to trolley is$10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$.
The formula to calculate the initial momentum of the system is
${p_i} = m{v_i}$
Where, ${p_i}$ is the initial momentum, $m$ is the total mass of the system and ${v_i}$ is the initial velocity of the system.
Substitute $200\,{\rm{kg}}$ for $m$ and $36\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${v_i}$ in the formula to calculate the initial momentum of the system.
{p_i} = \left( {200\,{\rm{kg}}} \right)\left( {36\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}} \right)\left( {\dfrac{{\dfrac{5}{{18}}\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{1\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}}}} \right)\\\
\implies {p_i} = \left( {200\,{\rm{kg}}} \right)\left( {10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)\\\
$\implies {p_i} = 2000\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$
The total mass of the system is $200\,{\rm{kg}}$ and the mass of the child is $20\,{\rm{kg}}$ so the formula to calculate the mass of the trolley is
${m_t} = m - {m_c}$
Where, ${m_t}$ is the mass of the trolley and ${m_c}$ is the mass of the child.
Substitute $200\,{\rm{kg}}$ for $m$ and $20\,{\rm{kg}}$ for ${m_c}$ in the formula to calculate the mass of trolley.

{m_t} = 200\,{\rm{kg}} - 20\,{\rm{kg}}\\\ {\rm{ = 180}}\,{\rm{kg}} \end{array}$$ The speed of child relative to trolley is $$10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$ and the speed of the trolley relative to ground is $$10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$, therefore the speed of child relative to ground is $$10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} + 10\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} = 20\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$. The formula to calculate the final momentum of the system is $${p_f} = {m_t}{v_t} + {m_c}{v_c}$$ Where, $${p_f}$$is the final momentum, $${v_t}$$ is the final speed of the trolley and $${v_c}$$ is the final speed of the child. Substitute $$180\,{\rm{kg}}$$ for$${m_t}$$, $$20\,{\rm{kg}}$$ for $${m_c}$$ and $$20\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$for $${v_c}$$ in the formula to obtain an equation for final momentum of the system. $$\begin{array}{c} {p_f} = \left( {180\,{\rm{kg}}} \right)\left( {{v_t}} \right) + \left( {20\,{\rm{kg}}} \right)\left( {20\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)\\\ = \left( {180\,{\rm{kg}}} \right)\left( {{v_t}} \right) + 400\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{array}$$ According to the law of conservation of linear momentum, the initial momentum is equal to the final momentum of the system. The formula of conservation of the linear momentum is $${p_i} = {p_f}$$. Substitute $$2000\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$ for $${p_i}$$ and $$\left( {180\,{\rm{kg}}} \right)\left( {{v_i}} \right) + 400\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$ for $${p_f}$$ in the formula and solve for the final velocity of the trolley. $ 2000\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} = \left( {180\,{\rm{kg}}} \right)\left( {{v_t}} \right) + 400\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}\\\ \left( {180\,{\rm{kg}}} \right)\left( {{v_t}} \right) = 2000\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} - 400\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}\\\ {v_t} = \dfrac{{1600\,{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{180\,{\rm{kg}}}}\\\ \approx 9\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} $ Thus, the final speed of the trolley is $$9\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$ which means option (a) is correct. Therefore, the formula to calculate the distance covered by the trolley is $$d = {v_t}t$$ Where, $$d$$ is the distance covered and $$t$$ is the time taken. Substitute $$9\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$ for $$d$$ and $$1\,{\rm{s}}$$ for $$t$$ in the formula to calculate the distance covered. $ d = \left( {9\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)\left( {1\,{\rm{s}}} \right)\\\$ $ d= 9\,{\rm{m}}$ **So, the correct answer is “Option C”.** **Note:** The initial momentum of the system includes both trolley and child and the final momentum of the system includes different velocity of both trolley and child. The child covers $$10\,{\rm{m}}$$ distance in $$1\,{\rm{s}}$$ therefore the time should be $$1\,{\rm{s}}$$.