Two identical $4$ kg blocks are moving with same velocity $1$ m $/$ s towards each other along a smooth horizontal surface, when they collide they stick together and comes to rest. The work done by external and internal forces will be respectively.

0, 4 J

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4 J, 9 J

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4 J, 0

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0, 9 J

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Solution

The correct option is A $0, 4 J$
For these two blocks $F_{e x t} = 0$
So, $W_{e x t} = F_{e x t} \cdot s = 0$
Aslo from work-energy theorem
$W = K_{f} - K_{f}$
$\Rightarrow W_{e x t} + W_{i n t} + 0 - 2 (\frac{1}{2} m u^{2})$
$[∵ u_{1} = u_{2} = u, v_{1} = v_{2} = 0, m = 4]$
$0 + W_{i n t} = - 4 \times 1^{2}$
$\Rightarrow W_{i n t} = - 4$ J
Here, negative sign shows that internal forces on each block are opposite to its motion.

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Two identical 4kg blocks are moving with same velocity 1m/s towards each other along a smooth horizontal surface, when they collide they stick together and comes to rest. The work done by external and internal forces will be respectively.

The correct option is A 0,4JFor these two blocks Fext=0So, Wext=Fext⋅s=0Aslo from work-energy theoremW=Kf−Kf⇒Wext+Wint+0−2(12mu2)[∵u1=u2=u,v1=v2=0,m=4]0+Wint=−4×12⇒Wint=−4JHere, negative sign shows that internal forces on each block are opposite to its motion.

Two identical $4$ kg blocks are moving with same velocity $1$ m $/$ s towards each other along a smooth horizontal surface, when they collide they stick together and comes to rest. The work done by external and internal forces will be respectively.