Question

A block of metal 4 kg is in rest on a frictionless surface. It was targeted by a jet releasing water of 2 kg/s at a speed of 10 ms\(^{-1}\). The acceleration of the block is:

• A. 10 ms\(^{-2}\)
• B. 15 ms\(^{-2}\)
• C. 20 ms\(^{-2}\)
• D. 5 ms\(^{-2}\)

Hint:

To find the acceleration when a jet of fluid strikes a block, use the principle of conservation of momentum. The rate of momentum transfer equals the force applied to the object.

Answer 1

The Correct Option is D

Step 1: Applying the principle of conservation of momentum The force on the block due to the jet of water can be found using the principle of conservation of momentum. The rate of momentum transfer from the jet to the block is given by: \[ F = \dot{m} v \] Where: - \( \dot{m} \) is the mass flow rate of the water, and - \( v \) is the velocity of the water. Substituting the given values: \[ F = (2 \, \text{kg/s}) \times (10 \, \text{ms}^{-1}) = 20 \, \text{N} \] Step 2: Calculate the acceleration of the block The acceleration \( a \) of the block is given by Newton's second law: \[ F = ma \] Substituting the known values: \[ 20 = 4 \times a \quad \Rightarrow \quad a = \frac{20}{4} = 5 \, \text{ms}^{-2} \] Thus, the acceleration of the block is \( 5 \, \text{ms}^{-2} \).

Answer 2

Given:
Mass of block, \( m = 4 \, \text{kg} \)
Mass flow rate of water, \( \dot{m} = 2 \, \text{kg/s} \)
Speed of water jet, \( v = 10 \, \text{ms}^{-1} \)
Initial velocity of block = 0 (at rest)

Step 1: Calculate the force exerted on the block by the water jet (rate of change of momentum):
\[ F = \dot{m} \times v = 2 \times 10 = 20 \, \text{N} \]

Step 2: Calculate acceleration of the block using Newton’s second law:
\[ a = \frac{F}{m} = \frac{20}{4} = 5 \, \text{ms}^{-2} \]

Therefore, the acceleration of the block is:
\[ \boxed{5 \, \text{ms}^{-2}} \]