Power produced by Niagara Falls, the Falls pours an average of 5200 m^3 of water

Question

Power produced by Niagara Falls, the Falls pours an average of 5200 m^3 of water over a drop of 62.0 m every second If all the potential energy of that water could be convened to electric energy, how much electrical power could the Falls generated? A block of mass m = 7.8 kg, moving on a frictionless surface with a speed v_1 = 7.8 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 7.8 kg block recoils with a speed of V_f = -2.8 m/s I am asking you to calculate speed of V and mass M a) V = _____ b) M = _____

Explanation / Answer

3) The electric power produced by the falling of water is

P = Electric Energy generated (E) / time (t)

here Electric energy generated by 5200 m3 volume of water is

E = Potential energy of 5200 m3 volume of water at height h =62.0 m

E = ( V d ) g h   

here V =volume of the water =5200 m3

g = Acceleration due to gravity = 9.81 m/s2

h=Height from which the water falls = 62.0 m

d = density of water = 1000 kg /m3  

on substituting all values we get

E = (5200 m3 ) (1000 kg /m3 ) (9.81 m/s2 )(62.0 m)

E =3.163 x 109 J

The electric power produced is

P = E / t = (3.163 x 109 J ) / (1 s)

P = 3.163 x 109 W or 3.163 GW

4) From the conservation of energy of the system in the elastic collision we can write as

Momentum of the system before the collision =Momentum of the system after the collision.

mvi + M (0 m/s) = m vf  + M v

here m =7.8 kg , vi =7.8 m/s , vf = -2.8 m/s

(7.8 kg) (7.8 m/s) = (7.8 kg) (-2.8 m/s) + M v

M v = 82.68 kg m/s ------------(1)

We know that in the elastic collision the kinetic energy of the system is also conservative so

Intial kinetic energy of the system = Final kinetic energy of the system

(1/2)mvi2 + (1/2) M (0 m/s)2 =(1/2) m vf2  + (1/2) M v2

mvi2  = m vf2  + M v2

(7.8 kg) (7.8 m/s)2 = (7.8 kg) (-2.8 m/s)2 + M v2

M v2 = 413.4 J -----------(2)

By doing (2) / (1) we get

v = ( 413.4 J ) / (82.68 kg m/s)

v = 5.0 m/s

and from (1)

M v = 82.68 kg m/s

M( 5.0 m/s) = 82.68 kg m/s

M = 16.536 kg

Therefore the mass of the resting object is  16.536 kg and it's final velocity after the collision is 5.0 m/s (towards right)

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