question_answer

A lorry and a car are moving with the same velocity. When the same force is applied to stop them, (i) Which one takes more time to come to a halt? (ii) Which one travels more distance before coming to rest?

Answer:

The mass of the lorry is greater than the mass of the car. Given, velocity of lorry \[({{v}_{lorry}})=\] velocity of car \[({{v}_{car}})\] Force applied to stop lorry \[({{F}_{lorry}})=\] Force applied to stop car \[({{F}_{car}})\] (i) Which one takes more time = ? Let us define a relation connecting force and time. We know, \[F=ma\Rightarrow F=m\left( \frac{v-u}{T} \right)\] \[\left[ \therefore a=\left( \frac{v-u}{t} \right) \right]\] As the bodies are coming to rest, their final velocities are zero. \[\Rightarrow F\times t=m\times v-u\] (or) \[\frac{F}{v-u}=\frac{m}{t}\] \[\Rightarrow \frac{m}{t}=\]constant         (\[\because \]F, v, u are same for both car and lorry) \[\Rightarrow m=t\times cons\tan t\] (or) \[m\propto t\] From the above equation it is clear that time taken to stop the body with greater mass is more. Therefore, the lorry takes more time to come to rest. (ii) Using the above, we can simply say that the lorry travels greater distance before coming to rest. This is because F, u, v are same for both lorry and car. Another explanation We know, \[F=ma\] \[\Rightarrow F=m\left( \frac{{{v}^{2}}-{{u}^{2}}}{2s} \right)\] \[\left( \because {{v}^{2}}-{{u}^{2}}=2as\,\,(or)\,\,a=\frac{{{v}^{2}}-{{u}^{2}}}{2s} \right)\] \[\Rightarrow 2FS=m({{v}^{2}}-{{u}^{2}})\] \[\Rightarrow \frac{2F}{{{v}^{2}}-{{u}^{2}}}=\frac{m}{s}\](or) \[\frac{m}{s}=\]constant (\[\because \]\[F,v,u\]are same for both lorry and car) \[\Rightarrow m=s\times \]constant (or) \[m\propto s\] From the above equation, it is clear that body with greater mass travels greater distance before coming to rest.