physics olympiad Q (urgent)

2) Taking the surface area for which the wind is blowing on the man = 0.6 m2, and density of air = 1.3 kg/m3

Mass of wind blowing to the man per sec = 1.3 x 0.6 x 100/3.6 = 21.7 kg/s

So average force on the man is given by 21.7 x 100/3.6 = 463 N


8) When the inclination angle changes from α to β, vertical height fallen by the sphere = L(sin α - sin β)

So GPE loss of the sphere = MgL(sin α - sin β)

which is also the rotational k.e. of the rod + translational k.e. of the block by conservation of energy.

Suppose that ω is the angular speed of the rod at that moment, then:

v = -ωb csc2 β where v is the speed of the block.

ω = -(v sin2 β)/b

So the total k.e. is:

(ML2ω2/2) + mv2/2

= (1/2) [(Mv2L2 sin4 β)/b2 + mv2]

= (v2/2) [(ML2 sin4 β)/b2 + m]

Then

MgL(sin α - sin β) = (v2/2) [(ML2 sin4 β)/b2 + m]

v2 = [2MgL(sin α - sin β)]/[(ML2 sin4 β)/b2 + m]

= [2MgLb2(sin α - sin β)]/[(ML2 sin4 β) + mb2]

v = b√{[2MgL(sin α - sin β)]/(ML2 sin4 β + mb2)}

14) Suppose that A has fallen by a distance of x, then B has risen a distance of h - x.

Speed of A at that moment = √(2gx)

So speed of B = √(gx/2)

Hence

[√(gx/2)]2 - u2 = -2g(h - x) where u is the initial speed of B.

u2 = g(2h - 3x/2)

Also time elapsed = [√(2gx)]/g = √(2x/g)

So:

√[g(2h - 3x/2)] - g√(2x/g) = √(gx/2)

√(2h - 3x/2) - √(2x) = √(x/2)

2h - 3x/2 = 9x/2

x = h/3

So collision occurs at the height of h - x = 3h/2

16) The gravitaitional force on the station (including the bubble and water drop) is just equal to the centripetal force on it and hence the apparent g value in the station is zero. Thus the bubble will not move relative to the water drop.

17) By grav force = centripetal force:

GMm/a2.1 = mu2/a

u2 = GMm/a1.1

So when the distance changes to b, let v be its speed:

Loss in ke = m(u2 - v2)/2

which is equal to the work done against the grav force:

∫(x = a → b) (GMm/x2.1) dx = (GMm/a1.1 - GMm/b1.1)/1.1 = GMm(1/a1.1 - 1/b1.1)/1.1

Thus

m(u2 - v2)/2 = GMm(1/a1.1 - 1/b1.1)/1.1

u2 - v2 = (2/1.1)GM(1/b1.1 - 1/a1.1)

u2 = v2 + (2/1.1)GM(1/b1.1 - 1/a1.1)

u = √[v2 + (2/1.1)GM(1/b1.1 - 1/a1.1)]

2012-01-26 11:02:45 補充：
1) The spring also exerts an upward force on the block = 80N, so the acc. of the block is (100 - 80)/10 = 2 m/s2 downward, which is also the acc. of the parachute.

Suppose that the air resistance on the parachute is R, then:

(10 + 80 - R)/1 = 2

R = 88 N

2012-01-26 11:02:51 補充：
7) Recoil force = ρAv2 = 1000 x 10-3 x 588 = 588 N