Question

A jet of water hits a flat stationary plate perpendicular to its motion. The jet ejects 500g of water per second with a speed of 1m//s. Assuming that after striking, the water flows parallel to the plate, then the force exerted on the plate is

Answer 1

Answer:

F = 0.5N

Explanation:

The given question statement relates to impulsive force example, where a sharp and sudden force is applied as in case of water jet.

The formula for impulsive force is

F = Δp / Δt

F = m (vf − vi) / Δt

Here Vi is the initial velocity, Vf is the final velocity

As per given information

vi=1m/s

vf=0

m=0.5kg

Putting the values in formula, we get

F = 0.5N

Answer 2

As given that water is perpendicular to the stationary plate.

• As the given that amount of water ejected is 500 g let convert it into Kg that is 1/2 Kg
• The speed of water is given that is 1 m/sec.
• Now use the relation of force and momentum as it is changing with time
• F=ΔP/Δt=mΔv/Δt
• As the velocity is changing that is initial v is 0 and final v=1 m/sec
• Put in above equation
• F=m(1-0)
• F=1/2
• F=0.5 N
• That is the force exerted when the plate is parallel.