A) \[12\,\,m/s,\,\,6\,\,m/s\]
B) \[12\,\,m/s,\,\,25\,\,m/s\]
C) \[6\,\,m/s,\,\,12\,\,m/s\]
D) \[8\,\,m/s,\,\,20\,\,m/s\]
Correct Answer: C
Solution :
Since, collision is elastic momentum remains conserved, hence we have Momentum before collision = Momentum after collision Initially, \[p={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=10\times 10+5\times 4=120\] ... (i) Final\[,\] \[p'={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=10{{v}_{1}}+5{{v}_{2}}\] ... (ii) Equating Eqs. (i) and (ii), we get \[10{{v}_{1}}+5{{v}_{2}}=120\] \[\Rightarrow \] \[2{{v}_{1}}+2{{v}_{2}}=24\] ... (iii) Since collision is elastic relative velocity remains unchanged in magnitude but is reversed in direction, \[e=1=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\] \[\Rightarrow \] \[10-4={{v}_{2}}-{{v}_{1}}\] ? (iv) Solving Eqs. (iii) and (iv), we get \[{{v}_{1}}=6\,\,m/s,\,\,{{v}_{2}}=12\,\,m/s\]
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