The diagram shows a man standing on a platform that is attached to a flexible pipe.

Question 8

The diagram shows a man standing on a platform that is attached to a flexible pipe. Water is pumped through the pipe so that the man and platform remain at a constant height.

The resultant vertical force on the platform is zero. The combined mass of the man and platform is 96 kg. The mass of water that is discharged vertically downwards from the platform each second is 40 kg.

What is the speed of the water leaving the platform?

A 2.4 m s^-1                        B 6.9 m s^-1                        C 24 m s^-1                         D 47 m s^-1

Reference: Past Exam Paper – June 2016 Paper 11 Q11

Solution:

Answer: C.

Mass flow rate (= mass / time = Δm / Δt) = 40 kg s^-1

The weight of the man and the platform acts downwards.

Weight = mg = (96 × 9.81) N

The resultant vertical force on the platform is zero. So, there should be an upward force on the platform that is equal in magnitude to the weight.

This upward force occurs as the momentum of the water changes (since it is flowing).

Upward force = Δp / Δt = Δ(mv) / Δt

Assuming the water flows at a constant speed,

Upward force = v (Δm/Δt)

The flow rate Δm/Δt is 40 kg s^-1.

Upward force = 40 v

The resultant vertical force on the platform is zero.

Upward force = Weight

40 v = 96 × 9.81

Speed v = (96 × 9.81) / 40 = 23.5 = 24 m s^-1