Question

Question 5.28 A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectional area 10–2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Chapter Laws Of Motion Page 112

Answer 1

Hi

Speed of the water stream, v = 15 m/s

Cross-sectional area of the tube,
A = 10^-2 m^2

Volume of water coming out from the pipe per second,
V = Av = 15 × 10^-2 m^3 /s

Density of water= 10^3 kg/m^3

Mass of water flowing out through the pipe per second = density × V = 150 kg/s

we knowthat
according to Newton’s second law of motion as:

F = Rate of change of momentum =
∆P / ∆t
= mv / t
= 150 × 15 = 2250 N
hope it help u

Answer 2

speed of stream water = 15 m/sec
cross section area of tube = 10^-2 m^2
we know,
volume = cross section area * length of tube
hence,
volume of water coming out per second from the tube = cross section area *speed of stream water
= 10^-2 * 15 = 15* 10^-2 m^3/s

density of water = 10^3 kg/m^3
hence, mass of water coming out per second ( m) = volume per second * density
= 15*10^-2*10^3 kg/s
= 150 kg/s

now,
force exerted on wall by the impact of water = change in linear momentum per second = m* speed of stream water
= 150 * 15 N
= 2250 N