Work Energy And Power Question 160
Question: A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be [MP PET 1994]
Options:
A) 100%
B) 95%
C) 5%
D) 50%
Show Answer
Answer:
Correct Answer: B
Solution:
Initial K.E. of system = K.E. of the bullet = $ \frac{1}{2}m _{B}v_B^{2} $
By the law of conservation of linear momentum $ m _{B}v _{B}+0={m _{sys\text{.}}}\times {v _{sys\text{.}}} $
therefore $ {v _{sys\text{.}}}=\frac{m _{B}v _{B}}{{m _{sys\text{.}}}}=\frac{50\times 10}{50+950}=0.5\ m/s $
Fractional loss in K.E. = $ \frac{\frac{1}{2}m _{B}v_B^{2}-\frac{1}{2}{m _{sys\text{.}}}v _{sys\text{.}}^{2}}{\frac{1}{2}m _{B}v_B^{2}} $
By substituting $ m _{B}=50\times {10^{-3}}kg,\ v _{B}=10\ m/s $
$ {m _{sys\text{.}}}=1kg,\ v _{s}=0.5\ m/s $
we get Fractional loss = $ \frac{95}{100} $
Percentage loss = 95%
BETA
