When the bomb Q explodes,the trolley of mass | vedclass

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(C) The kinetic energy $K$ imparted to a trolley of mass $m$ by the explosion is $K = \frac{P^2}{2m}$,where $P$ is the momentum.Since the explosion pr

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$16$

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(C) The kinetic energy $K$ imparted to a trolley of mass $m$ by the explosion is $K = \frac{P^2}{2m}$,where $P$ is the momentum.Since the explosion provides the same impulse to both trolleys,the momentum $P$ is the same for both.The work done by friction to stop the trolley is $W = f \cdot s = \mu mg \cdot s$.By the work-energy theorem,$K = W$,so $\frac{P^2}{2m} = \mu mg \cdot s$.This gives $s = \frac{P^2}{2 \mu g m^2}$,which implies $s \propto \frac{1}{m^2}$.Therefore,$\frac{s_2}{s_1} = \left( \frac{m_1}{m_2} ight)^2$.Given $m_1 = 200 \, kg$,$s_1 = 36 \, m$,and $m_2 = 300 \, kg$:$\frac{s_2}{36} = \left( \frac{200}{300} ight)^2 = \left( \frac{2}{3} ight)^2 = \frac{4}{9}$.$s_2 = 36 	imes \frac{4}{9} = 16 \, m$.

When the bomb $Q$ explodes,the trolley of mass $200 \, kg$ travels a distance of $36 \, m$ before coming to rest. What distance will the trolley of mass $300 \, kg$ travel before coming to rest (in $, m$)?

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