A) 8
B) 6
C) 4
D) 2
Correct Answer: C
Solution :
Given: \[{{m}_{1}}=m\]and \[{{m}_{2}}=\frac{m}{4}\] Initial velocity of heavier body \[={{u}_{1}}\] Final velocity of heavier body \[={{u}_{1}}+4\] Initial velocity of lighter body \[={{u}_{2}}\] Final velocity \[={{v}_{2}}\] Now, initial kinetic energy of heavier body is given by \[{{E}_{1}}=\frac{1}{2}mu_{1}^{2}\] ?(i) Initial kinetic energy of lighter body \[{{E}_{2}}=\frac{1}{2}{{m}_{2}}u_{2}^{2}\] \[{{E}_{2}}=\frac{1}{2}\frac{m}{4}u_{2}^{2}\] ?(ii) \[\left( But\,given:\,{{E}_{1}}=\frac{1}{4}\,{{E}_{2}} \right)\] Hence, putting the values \[{{E}_{1}},\] as \[{{E}_{2}}\] in Eq. (iii) \[\frac{1}{2}mu_{1}^{2}=\frac{1}{4}\times \frac{1}{8}mu_{2}^{2}\] \[u_{1}^{2}=\frac{u_{2}^{2}}{16}\]or \[{{u}_{1}}=\frac{{{u}_{2}}}{4}\] ?(iv) Final KE of the heavier body \[=\frac{1}{2}m{{({{u}_{1}}+4)}^{2}}\] So, \[\frac{1}{2}m{{({{u}_{1}}+4)}^{2}}=\frac{1}{2}\frac{m}{4}u_{2}^{2}\] or \[{{({{u}_{1}}+4)}^{2}}=\frac{u_{2}^{2}}{4}\] \[{{u}_{1}}+4=\frac{{{u}_{2}}}{2}\] \[{{u}_{1}}+4=\frac{4{{u}_{1}}}{2}\] \[\left( \begin{align} & \because \,{{u}_{1}}=\frac{{{u}_{2}}}{4} \\ & {{u}_{2}}=4{{u}_{1}} \\ \end{align} \right)\] \[{{u}_{1}}=4\] \[{{u}_{1}}=4\,m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
