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 $



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