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 $
BETA
