A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. |
When hit by a bullet of mass 50 g moving with speed r, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. |
If a freely falling object were to acquire speed \[\frac{\upsilon }{10}\] after being dropped from height H, then neglecting energy losses and taking \[g=10m{{s}^{-2}},\]the value of H is close to: [JEE ONLINE 10-04-2015] |
A) 0.2 km
B) 0.4 km
C) 0.5 km
D) 0.3 km
Correct Answer: B
Solution :
If block to come to rest after 2m. |
So, \[d=\frac{{{u}^{2}}}{2a}\] |
\[a=\mu g=0.05\times 10=0.5m/{{\sec }^{2}}\] |
\[{{u}^{2}}=2\times 2\times 0.5=2\] |
\[u=\sqrt{2}m/\sec \] |
by momentum conservation |
\[\sqrt{2}\left( 10+50\times {{10}^{-3}} \right)=\left( 50\times {{10}^{-3}} \right)\times V\] |
approximately \[\frac{\sqrt{2}\times 10}{50\times {{10}^{-3}}}=V\] |
\[V=200\sqrt{2}m/\sec \]so \[\frac{V}{10}=20\sqrt{2}\] |
to get \[V=20\sqrt{2}m/\sec \] \[{{V}^{2}}=2gh\] |
\[h=\frac{800}{20}=40m\] |
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