| This question has statement I and Statement II. Of the four choice given after the statements, choose the one that best describes the two statements. |
| Statement - I: A Point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as |
| \[f\left( \frac{1}{2}m{{v}^{2}} \right)\]then\[f=\left( \frac{m}{M+m} \right)\] |
| Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision. [JEE MAIN 2013] |
A) Statement - I is true, Statement - II is true, statement - II is a correct explanation of Statement I
B) Statement - I is true, Statement - II is true, statement - II is not a correct explanation of Statement I
C) Statement - I is true, Statement - II is false
D) Statement - I is false, Statement - II is true
Correct Answer: D
Solution :
| [d] Maximum energy loss when inelastic collision takes place |
| \[mv=(m+M)v\prime \] |
| \[v\prime =\frac{m}{m+M}v\] |
| \[{{k}_{i}}=\frac{1}{2}m{{v}^{2}}\] |
| \[{{k}_{f}}=\frac{1}{2}(m+M)v{{'}^{2}}=\frac{1}{2}(m+M)\frac{{{m}^{2}}{{v}^{2}}}{{{(m+M)}^{2}}}\] |
| \[=\frac{1}{2}m{{v}^{2}}\left( \frac{m}{M+m} \right)\] |
| Loss of energy\[={{k}_{i}}-{{k}_{f}}=\frac{1}{2}m{{v}^{2}}\left[ 1-\frac{m}{M+m} \right]\] |
| \[=\frac{M}{M+m}\times \frac{1}{2}m{{v}^{2}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
