Work Energy And Power Question 289
Question: A block of mass 0.50 kg is moving with a speed of $ 2.00m{s^{-1}} $ on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Options:
A) 0.16 J
B) 1.00 J
C) 0.67 J
D) 0.34 J
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Initial kinetic energy of the system
$ K.E _{i} =\frac{1}{2}mu^{2}+\frac{1}{2}M{{(0)}^{2}}=\frac{1}{2}\times 0.5\times 2\times 2+0=1J $
For collision, applying conservation of linear momentum $ m\times u=( m+M )\times v $
$ \therefore 0.5\times 2= (0.5+1) \times v\Rightarrow v=\frac{2}{3}m/s $
Final kinetic energy of the system is $ K.{E _{f}} =\frac{1}{2}( m+M )v^{2}=\frac{1}{2}( 0.5+1 )\times \frac{2}{3}\times \frac{2}{3}=\frac{1}{3}J $
$ \therefore $ Energy loss during collision $ =( 1-\frac{1}{3} )J=0.67J $
BETA
