A wooden spherical ball with SG of 0.42 and diameter of 30 cm is dropped from height 4.2m above surface of water of pool. If the ball barely touches bottom of pool, what is water level in pool at that point ?
The answer is 3.19m..
Pls.. help me on how to show the solution ... tnx :)
The answer is 3.19m..
Pls.. help me on how to show the solution ... tnx :)
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I (or anyone else who wants to help) need to know what assumptions are being made. After working through and getting a close answer I'm going to guess that the buoyant force is ignored until the ball is totally submerged (this will make is drop a bit further)? Also no air or fluid resistance? I'll just show you what I got. (Also ignore the reference to kinetic energy, I didn't end up using that.)
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Going to assume air and fluid resistance are ignored here. Also going to make the assumption that the buoyant force is ignored until the center of mass passes the surface and even then I'm going to simplify the buoyant force by assuming the ball is submerged totally at that point.
Specific gravity = density wood / density water = 0.42
The density of water is 1000kg/m^3 so we have:
density wood = 0.42 (1000kg/m^3)
density wood = 420kg/m^3
diameter ball = 0.30m
radius ball = 0.15m
volume ball = (4/3)pi*(0.15)^3 meters= V meters^3
mass ball = 420*V kg
Potential energy at 4.2m above pool = Pi = 420V*g*(4.2) = kinetic energy at the surface of the pool
420V*g*(4.2) = (1/2)m*v^2
g*(4.2) = (1/2)*v^2
g*8.4 = v^2
v = sqrt(g*8.4)
v is the initial downwards velocity of the ball at the surface of the pool.
The net force (in the direction of v) is the force of gravity minus the buoyant force:
buoyant force = mass of water displaced * g = volume sphere * density water * g,
buoyant force = V*(1000)*g = 9800V Newtons = B
Net force = mg - B = 420V*g - 9800V = -5684V Newtons
To find how far down the sphere drops, note that it is gaining potential energy as it sinks (if you consider the surface of the water as the equilibrium point). We want the magnitude of this potential energy to equal the initial potential energy of the ball above the water:
420V*g*(4.2) = 5684V*d, where d is the depth to which the ball sinks
420*g*4.2 = 5684*d
d = 3.04m
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Going to assume air and fluid resistance are ignored here. Also going to make the assumption that the buoyant force is ignored until the center of mass passes the surface and even then I'm going to simplify the buoyant force by assuming the ball is submerged totally at that point.
Specific gravity = density wood / density water = 0.42
The density of water is 1000kg/m^3 so we have:
density wood = 0.42 (1000kg/m^3)
density wood = 420kg/m^3
diameter ball = 0.30m
radius ball = 0.15m
volume ball = (4/3)pi*(0.15)^3 meters= V meters^3
mass ball = 420*V kg
Potential energy at 4.2m above pool = Pi = 420V*g*(4.2) = kinetic energy at the surface of the pool
420V*g*(4.2) = (1/2)m*v^2
g*(4.2) = (1/2)*v^2
g*8.4 = v^2
v = sqrt(g*8.4)
v is the initial downwards velocity of the ball at the surface of the pool.
The net force (in the direction of v) is the force of gravity minus the buoyant force:
buoyant force = mass of water displaced * g = volume sphere * density water * g,
buoyant force = V*(1000)*g = 9800V Newtons = B
Net force = mg - B = 420V*g - 9800V = -5684V Newtons
To find how far down the sphere drops, note that it is gaining potential energy as it sinks (if you consider the surface of the water as the equilibrium point). We want the magnitude of this potential energy to equal the initial potential energy of the ball above the water:
420V*g*(4.2) = 5684V*d, where d is the depth to which the ball sinks
420*g*4.2 = 5684*d
d = 3.04m
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