A block of metal 2 kg is in rest on a smooth plane. It is striked by a jet releasing water of $1…

A block of metal 2 kg is in rest on a smooth plane. It is striked by a jet releasing water of $1 \mathrm{~kg} \mathrm{~s}^{-1}$ at a speed of $5 \mathrm{~ms}^{-1}$, then the acceleration of the block is
  1. $2 \mathrm{~ms}^{-2}$
  2. $2.5 \mathrm{~ms}^{-2}$
  3. $0.25 \mathrm{~ms}^{-2}$
  4. $50 \mathrm{~ms}^{-2}$

Solution


$\begin{aligned} & \frac{\Delta \mathrm{m}}{\Delta \mathrm{t}}=1 \mathrm{~kg} / \mathrm{s}, \mathrm{M}=2 \mathrm{~kg} \\ & \mathrm{v}=5 \mathrm{~ms}^{-1} \end{aligned}$
Force on the block due to water jet $\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{\Delta(\mathrm{mv})}{\Delta \mathrm{t}}=\mathrm{v} \cdot \frac{\Delta \mathrm{~m}}{\Delta \mathrm{t}}=5 \times 1=5 \mathrm{~N}$ $\therefore \quad$ Acceleration of the block is $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}}=\frac{5}{2}=2.5 \mathrm{~m} / \mathrm{s}^2$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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