Examkin

Question

Difficulty: HardNewton's Laws of Motion and Linear Momentum

A horizontal jet of water issuing from a nozzle with a cross-sectional area of 2.0×10−3 m22.0 \times 10^{-3}\text{ m}^2 at a speed of 20 m s−120\text{ m s}^{-1} strikes a vertical wall perpendicularly. If the water rebounds horizontally in the opposite direction at a speed of 5.0 m s−15.0\text{ m s}^{-1}, what is the magnitude of the force exerted by the water stream on the wall? (Density of water = 1000 kg m−31000\text{ kg m}^{-3})

  1. A
    600 N600\text{ N}
  2. B
    800 N800\text{ N}
  3. 1000 N1000\text{ N}Answer
  4. D
    50 N50\text{ N}

Answer

The magnitude of the force exerted by the water jet on the wall is 1000 N1000\text{ N}.
According to Newton's second law, force is the rate of change of linear momentum. The mass of water hitting the wall each second is ΔmΔt=ρAv1=1000×2.0×10−3×20=40 kg s−1\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000 \times 2.0 \times 10^{-3} \times 20 = 40\text{ kg s}^{-1}. Taking the initial direction as positive (+20 m s−1+20\text{ m s}^{-1}), the rebounding velocity is opposite in direction (−5.0 m s−1-5.0\text{ m s}^{-1}). The magnitude of the change in velocity is ∣−5.0−20∣=25 m s−1|-5.0 - 20| = 25\text{ m s}^{-1}. Multiplying mass flow rate by the change in velocity gives a force magnitude of 40×25=1000 N40 \times 25 = 1000\text{ N}.

Step-by-Step Solution

1
Calculate the mass of water striking the wall per second (mass flow rate, ΔmΔt\frac{\Delta m}{\Delta t}).
ΔmΔt=ρAv1=1000 kg m−3×(2.0×10−3 m2)×20 m s−1=40 kg s−1\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000\text{ kg m}^{-3} \times (2.0 \times 10^{-3}\text{ m}^2) \times 20\text{ m s}^{-1} = 40\text{ kg s}^{-1}.
The volume of water reaching the wall per second is given by the cross-sectional area multiplied by its initial speed.
2
Determine the change in velocity vector per unit mass of water (Δv)(\Delta v).
Taking the direction towards the wall as positive, v1=+20 m s−1v_1 = +20\text{ m s}^{-1} and v2=−5.0 m s−1v_2 = -5.0\text{ m s}^{-1}. Thus, Δv=v2−v1=−5.0−20=−25.0 m s−1\Delta v = v_2 - v_1 = -5.0 - 20 = -25.0\text{ m s}^{-1}.
Velocity is a vector quantity; rebounding in the opposite direction requires assigning opposite signs to the initial and final velocities.
3
Apply Newton's Second Law (F=ΔpΔtF = \frac{\Delta p}{\Delta t}) to find the magnitude of force on the wall.
F=ΔmΔt∣Δv∣=40 kg s−1×25.0 m s−1=1000 NF = \frac{\Delta m}{\Delta t} |\Delta v| = 40\text{ kg s}^{-1} \times 25.0\text{ m s}^{-1} = 1000\text{ N}.
By Newton's third law, the magnitude of force exerted on the wall equals the rate of change of momentum of the water stream.

Key Concept

Newton's Second Law and Linear Momentum Rate of Change
Rate this question

Topics

Newton's Laws of Motion and Linear Momentum

More questions from this topic