Laws Of Motion Question 220

Question: The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates into the wood for a distance of 12 cm is [CBSE PMT 2000; DPMT 2003]

Options:

A) $ 2.2\times 10^{3}N $

B) $ 3.2\times 10^{3}N $

C) $ 4.2\times 10^{3}N $

D) $ 5.2\times 10^{3}N $

Show Answer

Answer:

Correct Answer: D

Solution:

$ u=250m/s $ , $ v=0 $ , $ s=0.12metre $

$ F=ma=m( \frac{u^{2}-v^{2}}{2s} )=\frac{20\times {{10}^{-3}}\times {{(250)}^{2}}}{2\times 0.12} $

$ \therefore $ $ F=5.2\times 10^{3}N $



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