Laws of Motion

JEE Main level, single correct, medium

29 of 210

Question 29 of 210

A block of metal weighing 2kg2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1kgs−11kg{s}^{-1} and at a speed of 10ms−110{ms}^{-1}. Then, the initial acceleration of the block, in ms−2m{s}^{-2}, will be

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Answer: (C) 55

Let velocity of water jet be vv.
So, force applied is F=ΔpΔt=(∆mv)ΔtF=\dfrac{\Delta p}{\Delta t}=\dfrac{\left(∆mv\right)}{\Delta t}
⇒ma=(∆mv)Δt\Rightarrow ma=\dfrac{\left(∆mv\right)}{\Delta t}
Acceleration of block, a=1×102a=\dfrac{1\times 10}{2}
a=5ms−2a=5m{s}^{-2}

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