Two identical 5kg blocks are moving with same speed of $2 m s^{- 1}$ towards each other along a frictionless horizontal surface. The two blocks collide, stick together, and come to rest. Consider the two blocks as a system. The work done by external and internal forces are respectively,

0, 0

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0, 20J

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0, -20J

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20J, -20J

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Solution

The correct option is C 0, -20J
Initial kinetic energy $K E_{1} = \frac{1}{2} \times 5 \times 2^{2} + \frac{1}{2} \times 5 \times 2^{2}$
$= 20 J$
Final kinetic energy $K E_{2} = 0$
Change in kinetic energy $△ K E = K E_{2} - K E_{1}$
$= 0 - 20$
$= - 20 J$
Work done by external and internal forces are the same as they are action-reaction pairs according to newton's third law of motion.
Hence, the answer is $0, - 20 J .$

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