A) \[+\text{ }4\text{ }m/s\]for both
B) \[-\,3\,\,m/s\] and \[+\text{ }5\text{ }m/s\]
C) \[-\text{ }4\text{ }m/s\] and \[+4\text{ }m/s~\]
D) \[-\text{ }5\text{ }m/s\] and \[+\text{ }3\text{ }m/s\]
Correct Answer: D
Solution :
| Key Idea: In an elastic collision, linear momentum remains conserved. |
| Given: \[{{u}_{1}}=3\,m/s,\,{{u}_{2}}=-\,5m/s,\,{{m}_{1}}={{m}_{2}}=m\] |
| According to principle of conservation of linear momentum |
| \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] |
| \[m\times 3-m\times 5=m{{v}_{1}}+m{{v}_{2}}\] |
| or \[{{v}_{1}}+{{v}_{2}}=-2\] ...(i) |
| In an elastic collision, |
| \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\] |
| \[\Rightarrow \] \[{{v}_{2}}-{{v}_{1}}=e\,({{u}_{1}}-{{u}_{2}})\] |
| \[\Rightarrow \] \[{{v}_{2}}-{{v}_{1}}=(1)\,(3+5)\] \[(\because \,\,e=1)\] |
| \[\Rightarrow \] \[{{v}_{1}}-{{v}_{2}}=-8\] ...(ii) |
| Adding Eqs. (i) and (ii), we obtain |
| \[2{{v}_{1}}=-\,10\,\] |
| \[\Rightarrow \] \[{{v}_{1}}=-5\,m/s\] |
| From Eq. (i), |
| \[{{v}_{2}}=-2-{{v}_{1}}=-2+5=3\,m/s\] |
| Thus, \[{{v}_{1}}=-5m/s\,,\,{{v}_{2}}=+3\,m/s\] |
| Alternative: If two bodies collide elastically, then their velocities are interchanged. Since, it is an elastic collision hence, velocities after collision will be 5 m/s and 3 m/s. |
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