question_answer

A body of mass\[30\times {{10}^{-3}}kg\], when acted upon by a force for 5s, attains a velocity of 100 m/s. If the same force is applied for 2 minutes on a body of mass 10 kg at rest, what will be its velocity?

Answer:

\[{{m}_{1}}=30\times {{10}^{-3}}kg\,\,\,{{m}_{2}}=10kg\] \[=3\times {{10}^{-2}}kg\,\,{{t}_{2}}=2\,\,\min =120s\] \[{{t}_{1}}=5s\] \[{{v}_{2}}=?\] \[{{v}_{1}}=100\,m/s\] \[{{u}_{2}}=0\] \[{{u}_{1}}=0\] impulse on first body \[=F{{t}_{1}}={{m}_{1}}{{v}_{1}}-{{m}_{1}}{{u}_{1}}\] \[={{m}_{1}}{{v}_{1}}-0\] \[\Rightarrow F\times 5=3\times {{10}^{-2}}\times 100\] \[\Rightarrow F\times 5=3\] ____ (1) Impulse on 2nd body \[=F{{t}_{2}}={{m}_{2}}{{v}_{2}}-{{m}_{2}}{{u}_{2}}\] \[={{m}_{2}}{{v}_{2}}-0\] \[\Rightarrow F\times 120=10{{v}_{2}}\] ____ (2) Dividing (1) by (2) we have \[\frac{F\times 5}{F\times 120}=\frac{3}{10{{V}_{2}}}\] \[\Rightarrow \frac{1}{24}=\frac{3}{10{{v}_{2}}}\Rightarrow 10{{v}_{2}}=72\] \[\Rightarrow {{v}_{2}}=\frac{72}{10}=7.2\,m/s\]