Water flows out from A pipe at 5 kg /sas and its velocity change from 4 m per second to zero on striking the wall find the force exerted by the water on the wall
To find the force exerted by the water on the wall, we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (dp/dt).
Momentum is defined as the product of an object's mass (m) and its velocity (v). Therefore, the change in momentum (dp) can be calculated by subtracting the initial momentum (m1v1) from the final momentum (m2v2):
dp = m2v2 - m1v1
Given:
Initial mass, m1 = 5 kg/s
Initial velocity, v1 = 4 m/s
Final velocity, v2 = 0 m/s
Since the water comes out of the pipe continuously and its mass flow rate is given as 5 kg/s, the change in mass (dm) over a short time interval (dt) can be given by:
dm = m2 - m1 = 5 kg/s * dt
Since the final velocity is zero, we can consider this situation as a perfectly inelastic collision. In this case, the final mass (m2) of the water will be the sum of the initial mass (m1) and the change in mass (dm):
m2 = m1 + dm = 5 kg/s * dt + 5 kg/s
Now let's substitute the values into the equation for the change in momentum:
dp = m2v2 - m1v1 = [(5 kg/s * dt + 5 kg/s) * 0 m/s] - (5 kg/s * 4 m/s)
Simplifying the equation:
dp = -20 kgm/s
Finally, we divide the change in momentum by the time interval (dt) to find the force exerted by the water on the wall:
F = dp/dt = -20 kgm/s / dt
The force exerted by the water on the wall is equal to -20 kgm/s divided by the time interval (dt).