| In the figure \[{{m}_{A}}={{m}_{B}}=1kg\]. Block A is neutral while\[{{q}_{B}}=-1C\].Sizes of A and B are negligible. B is released from rest at a distance \[1.8\,\,m\] from A. Initially spring is neither compressed nor elongated. |
 |
| Smooth x=0 x=1.8m x-axis |
| If collision between A and B is perfectly inelastic, what is velocity of combined mass just after collision? |
A) 6 m/s
B) 3 m/s
C) 9 m/s
D) 12 m/s
Correct Answer: B
Solution :
| Electrostatics force an A is zero, while on B is \[\left| {{F}_{B}} \right|=\left| {{q}_{B}} \right|E=(1)(10)=10N\](along negative x-direction) |
| \[{{a}_{B}}=\frac{{{F}_{B}}}{{{m}_{B}}}=10m/{{s}^{2}}\] |
| Just before collision, |
| \[{{V}_{B}}=\sqrt{2{{a}_{B}}s}=\sqrt{2\times 10\times 1.8}=6m/s\] |
| Now, from conservation of linear momentum, velocity of combined mass will be, |
| \[({{m}_{A}}+{{m}_{B}})V={{m}_{B}}{{v}_{B}}\] or \[v=3m/s\] |
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