Question

A block of mass 5 kg is kept on a smooth horizontal surface. A horizontal stream of water coming out of a pipe of area of cross-section \( 5 \, \text{cm}^2 \) hits the block with a velocity of \( 5 \, \text{ms}^{-1} \) and rebounds back with the same velocity. The initial acceleration of the block is: (Density of water is \( 1 \, \text{g/cc} \)) \vspace{0.5cm}

• A. \( 10 \, \text{ms}^{-2} \)
• B. \( 2.5 \, \text{ms}^{-2} \)
• C. \( 12.5 \, \text{ms}^{-2} \)
• D. \( 5 \, \text{ms}^{-2} \) \vspace{0.5cm}

Hint:

For problems involving fluid impact on surfaces, use mass flow rate \( \dot{m} = \rho A v \) and force equation \( F = \dot{m} \Delta v \) to determine acceleration.

Answer 1

The Correct Option is D

Step 1: Find the Mass Flow Rate of Water The density of water: \[ \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \] Cross-sectional area: \[ A = 5 \times 10^{-4} \, \text{m}^2 \] Velocity of water: \[ v = 5 \, \text{ms}^{-1} \] Mass flow rate is given by: \[ \dot{m} = \rho A v \] \[ = (1000) (5 \times 10^{-4}) (5) \] \[ = 2.5 \, \text{kg/s} \] \vspace{0.5cm} Step 2: Find the Force on the Block The force exerted by the water on the block is: \[ F = \text{Rate of change of momentum} \] \[ = \dot{m} (v_{\text{final}} - v_{\text{initial}}) \] Since the water rebounds with the same velocity: \[ F = (2.5) (5 - (-5)) \] \[ = 2.5 \times 10 = 25 \, \text{N} \] \vspace{0.5cm} Step 3: Compute the Acceleration of the Block Using Newton’s second law: \[ F = ma \] \[ 25 = 5a \] \[ a = \frac{25}{5} = 5 \, \text{ms}^{-2} \] Thus, the initial acceleration of the block is: \[ \mathbf{5 \, \text{ms}^{-2}} \] \vspace{0.5cm}