Question

A block of metal 2 kg is in rest on a smooth plane. It is struck by a jet releasing water of 1 kg s\(^{-1}\) at a speed of 5 m s\(^{-1}\), then the acceleration of the block is

• A. \(2 \, {ms}^{-2}\)
• B. \(2.5 \, {ms}^{-2}\)
• C. \(0.25 \, {ms}^{-2}\)
• D. \(50 \, {ms}^{-2}\)

Hint:

When calculating the force due to a fluid striking an object, always consider the rate of momentum transfer, which depends on the mass flow rate and the velocity of the fluid.

Answer 1

The Correct Option is B

The force exerted by the water jet can be calculated using the formula: \[ F = \Delta p / \Delta t \] where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval. Since the water is striking the block, the change in momentum per second (\(\Delta p\)) is given by the mass flow rate multiplied by the velocity of the water: \[ \Delta p = (1 \, {kg/s}) \times (5 \, {m/s}) = 5 \, {kg m/s}^2 \] Now, using Newton's second law, \( F = ma \), where \( m \) is the mass of the block and \( a \) is its acceleration: \[ 5 \, {N} = 2 \, {kg} \times a \] \[ a = \frac{5 \, {N}}{2 \, {kg}} = 2.5 \, {ms}^{-2} \]