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There are two bodies of masses 10 kg and 5 kg respectively. Find the ratio of the force acting on them if their velocity-time graphs are as shown below: [Hint: \[slope=\tan \theta \] ]

Answer:

\[{{m}_{1}}=10\,kg\] \[{{m}_{2}}=5\,kg\] \[{{F}_{1}}:{{F}_{2}}=?\] We know that,\[F=ma\] How to get acceleration (a) ? Slope of velocity-time graph gives acceleration. \[\Rightarrow \]Slope\[=a=\tan \theta \] .........(A) \[\therefore F=ma=m(\tan \theta )\] .........(1) [\[\therefore \] from (A)] Applying (1) to both the cases, we get \[{{F}_{1}}={{m}_{1}}(tan{{\theta }_{1}})\] \[{{F}_{2}}={{m}_{2}}(tan{{\theta }_{2}})\] \[\Rightarrow \,\,{{F}_{1}}=m(tan30{}^\circ )\] \[\Rightarrow \,\,{{F}_{2}}=\frac{m}{2}(\tan 60{}^\circ )\] \[\Rightarrow {{F}_{1}}=\times \frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}\] ??(2) \[\Rightarrow {{F}_{2}}=5\times \sqrt{3}=\frac{5\sqrt{3}}{\sqrt{3}}\] ??(3) Dividing (2) by (3), we have \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{}^{10}/{}_{\sqrt{3}}}{{}^{5\sqrt{3}}/{}_{2}}=\frac{10}{3}\times \frac{2}{5\sqrt{3}}=\frac{4}{3}\Rightarrow {{F}_{1}}:{{F}_{2}}=4:3\]Therefore, the ratio of the forces acting on the two bodies is 4 : 3.