Question

A water pipe in a building delivers 1000 liters (with mass 1000 kg) of water per second. The water is moving through the pipe at 2.6 m/s. The pipe has a 90° bend, and the pipe will require a supporting structure, called a thrust block, at the bend, as in (Figure 1). We can use the ideas of momentum and impulse to understand why. Each second, 1000 kg of water moving at vz = 2.6 m/s changes direction to move at vy = 2.6 m/s = N. what is the magnitude of the change in momentum of the 1000 kg of water?​

Answer 1

The magnitude of the change in momentum is 3676.95 kgm/s

Since the water initially moves in the z- direction and vz = 2.6 m/s.

So, its initial velocity is V = (2.6 m/s)k

Also, it changes direction to vy = 2.6 m/s.

So, its final velocity is V' = (2.6 m/s)j

So, its change in velocity ΔV = V' - V = (2.6 m/s)j - (2.6 m/s)k

So, the magnitude of the change in velocity is |ΔV| = √(vx² + vy² + vz²)

= √[(0 m/s)² + (2.6 m/s)² + (-2.6 m/s)²]

= √[(0 m/s)² + (2.6 m/s)² + (2.6 m/s)²]

= √[(2(2.6 m/s)²

= 2.6√2 m/s

So, the magnitude of the change in momentum is |Δp| = m|ΔV| where m = mass of water = 1000 kg

So, |Δp| = m|ΔV|

|Δp| = 1000 kg × 2.6√2 m/s

|Δp| = 2600√2 kgm/s

|Δp| = 3676.95 kgm/s

The magnitude of the change in momentum is 3676.95 kgm/s

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