Question

An empty 50-g bowl rests on a scale that measures force (in newtons). Water is then poured into the scale from a height of 0.50 m and rate of 10 mL/s. Assuming that the water does not splash as it comes to rest, calculate the reading on the scale at t = 5.0 s

Answer 1

Answer:

1 Newton

Explanation:

Mass of bowl (Mb) = 50 g = 0.05 kg

Initial Velocity (Vo) = 10 mL/s = 10g/s

time (t) = 5.0s

Mass of water (Mw) = 10 g/s * 5s = 50 g = 0.05 kg

*Vob = initial velocity of bowl

*Vow = initial velocity of water

*Vf = final velocity

*g = gravity = 9.8

Conservation of momentum [Inelastic Collision]:

• (m1)(Vo1) + (m2)(Vo2) = (M1+M2)Vf
• (Mw)(Vow) +  (Mb)(Vob) = (Mw+Mb)Vf
• (50g)(10g/s) + (50g)(0) = (50g + 50g)(Vf)
• Vf = 500/100 = 5g/s = 0.005kg/s

Impulse-Momentum Theorem:

• Δp = mΔv

Scale Reading:

• Δp + Mw(g) + Mb(g)
• = (0.05kg + 0.05kg)*(0.05kg/s) + (0.05kg*9.8) +(0.05kg*9.8)
• = 0.9805 = 1 Newton