question_answer

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 \[V=20\sqrt{2}m/\sec \]            \[{{V}^{2}}=2gh\]

\[h=\frac{800}{20}=40m\]