A) 0, 1
B) 1, 1
C) 1, 0.5
D) 0, 2
Correct Answer: A
Solution :
If two bodies collide head on with coefficient of restitution \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\] ?(i) From the law of conservation of linear momentum \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[\Rightarrow \]\[{{v}_{1}}\left[ \frac{{{m}_{1}}-e{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right]{{u}_{1}}+\left[ \frac{(1+e){{m}_{2}}}{{{m}_{1}}{{m}_{2}}} \right]{{u}_{2}}\] Substituting \[{{u}_{1}}=2m{{s}^{-1}},{{u}_{2}}=0,{{m}_{1}}=m\]and \[{{m}_{2}}=2\,m,\]\[e=0.5\] We get \[{{v}_{1}}=\frac{m-m}{m+2m}\times 2\] \[\Rightarrow \]\[{{v}_{1}}=0\] Similarly, \[{{v}_{2}}=\left[ \frac{(1+e){{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right]{{u}_{1}}+\left[ \frac{{{m}_{2}}-e{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right]{{u}_{2}}\] \[=\left[ \frac{1.5\times m}{3m} \right]\times 2=1\,m{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
