Work Energy And Power Question 288
Question: A bullet of mass 20g and moving with 600 m/s collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision?
Options:
A) 200mA
B) 150 m/s
C) 400 m/s
D) 300 m/s
Show Answer
Answer:
Correct Answer: A
Solution:
Initial, $ K.E.=\frac{1}{2}mv^{2}=\frac{1}{2}\times \frac{20}{1000}\times 60^{2} $ Change in $ K.E.=P.E. $
$ \frac{1}{2}m( v^{2}-v{{’}^{2}} )=mgh $
$ \Rightarrow 3600-\frac{1}{2}\times \frac{20}{1000}\times v_1^{2}=4\times 10\times 80 $
$ \Rightarrow {v_1}=200,m/s $
BETA
