1) A body freely falling from rest has a velocity v after it falls through a height h . The distance it has to fall further for its velocity to become double is
a)3h b)6h c)8h d)10h
2) A body starts falling from height h and travels distance h/2 during the last second of motion. Then f ind the travel(in sec) is
a)under root 2.1 b)2+under root 2 c)under root 2+under root 3 d)under root 3+2
3) A cylinder of wood floats vertically in water with one-fourth of its length out of water.the density of wood is
a)0.5g/cm3 b) 0.5g/cm3 c)0.75g/cm3 d)1g/cm3
4) Two solids X and Y float on water,X floats with half of its volume submerged while Y floats with one-third of its volume out of water. The density of X and Y are in the ratio of
a)4:3 b)3:4 c)2:3 d)1:3
1) The potential energy it looses in falling a distance h is given by
GPE = m*g*h
After falling that distance, the kinetic energy will have increased from 0 to a value equal to m*g*h. Therefore we can say that
m*g*h = (1/2)*m*v1^2 --------------------------------- 1
where v1 is the velocity after falling 1*h.
The condition when the velocity is 2*v1 is
m*g*n*h = (1/2)*m*(2*v1)^2
where n*h is the distance it has fallen -- total.
Notice that (1/2)*m*(2*v1)^2 = 4*(1/2)*m*v1^2
And 4*(1/2)*m*v1^2 is just 4 times the kinetic energy expression in equation 1. So
m*g*n*h = 4*m*g*h
So n = 4. The question asks for the distance "further", beyond a distance h. So the further distance is 3h.
2) Sorry, I don't know the meaning of "under root" and searches only turned up things in computing, dentistry, and other contexts that do not apply to your question.
3) For simplicity, assume the volume of the cylinder is 1 cm^3. The volume of water displaced is 0.75cm^3. So the mass of the water displaced is
0.75cm^3*1g/cm^3 = 0.75 g
Therefore the mass of the cylinder is 0.75g and its density is
mass/volume = 0.75g/cm^3
4) From #3, we can conclude that X has density 1/2 g/cm^3 and Y has density 2/3 g/cm^3.
So we're comparing 1/2 and 2/3
(1/2) : (2/3)
Multiply thru by 6
3 : 4
I hope this helps,