From F = G M m / r^2, we get G = F r^2 / (M m). Dimensions: [G] = ([F][r^2]) / ([M][m]) = (M L T^-2 * L^2) / (M * M) = M^-1 L^3 T^-2.

G = F r^2 / (M m)

At maximum height the velocity is zero. (a) h = u^2 / (2g) = (20)^2 / (2*10) = 400/20 = 20 m. (b) Time of flight T = 2u/g = (2*20)/10 = 4 s.

h = u^2/2g ; T = 2u/g

Force of friction f = mu * m g = 0.2 * 2 * 10 = 4 N. Net force = applied - friction = 10 - 4 = 6 N. Acceleration a = net force / mass = 6 / 2 = 3 m/s^2.

f = mu m g ; a = (F - f)/m

By conservation of energy, KE at the ground = loss in potential energy = m g h = 5 * 10 * 20 = 1000 J. (Check: v = sqrt(2gh) = 20 m/s, KE = (1/2)(5)(20^2) = 1000 J.)

KE = m g h

Parallel axes theorem: I = I_cm + M d^2, where I_cm is the moment of inertia about a parallel axis through the centre of mass and d is the distance between the axes. For a rod, I_cm = M L^2 / 12 and d = L/2, so I_end = M L^2/12 + M (L/2)^2 = M L^2/12 + M L^2/4 = M L^2/3.

I = I_cm + M d^2

Escape velocity v_e = sqrt(2 g R) = sqrt(2 * 9.8 * 6.4 x 10^6) = sqrt(1.2544 x 10^8) = 1.12 x 10^4 m/s = 11.2 km/s.

v_e = sqrt(2 g R)

Isothermal: temperature stays constant, so delta U = 0; obeys P V = constant; occurs slowly with good thermal contact; heat can flow in or out. Adiabatic: no heat is exchanged (Q = 0); obeys P V^gamma = constant; occurs quickly or with perfect insulation; any work done changes the internal energy (delta U = -delta W).

Isothermal: PV = const ; Adiabatic: P V^gamma = const

v_rms = sqrt(3 R T / M) = sqrt(3 * 8.31 * 300 / 0.032) = sqrt(7479 / 0.032) = sqrt(233719) = 483 m/s (approximately).

v_rms = sqrt(3 R T / M)

For a bob displaced by a small angle, the restoring force is F = -m g sin(theta). For small theta, sin(theta) ~ theta = x/L, so F = -(m g / L) x. This is of the form F = -k x with k = m g / L, hence the motion is SHM. Then omega^2 = k/m = g/L, so the time period T = 2 pi / omega = 2 pi sqrt(L/g).

T = 2 pi sqrt(L / g)

For a product/power, percentage errors add with the powers as weights: (dP/P) = (dA/A) + 2 (dB/B) = 2% + 2(3%) = 2% + 6% = 8%.

dP/P = dA/A + 2 dB/B

On a v-t graph for constant acceleration, the velocity rises linearly from u to v = u + a t. The displacement equals the area under the graph, which is a rectangle (height u, width t) plus a triangle (base t, height a t): s = (u * t) + (1/2)(t)(a t) = u t + (1/2) a t^2.

s = u t + (1/2) a t^2

Using v^2 = u^2 - 2 a s with v = 0: 0 = (20)^2 - 2(5) s, so 10 s = 400, giving s = 40 m.

v^2 = u^2 - 2 a s

Parallelogram law: if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, their resultant is represented by the diagonal from that point. Magnitude: R = sqrt(P^2 + Q^2 + 2 P Q cos(theta)), and the resultant makes an angle alpha with P where tan(alpha) = (Q sin(theta)) / (P + Q cos(theta)).

R = sqrt(P^2 + Q^2 + 2 P Q cos(theta))

R = u^2 sin(2 theta) / g = (30)^2 * sin(60 deg) / 10 = 900 * 0.866 / 10 = 77.9 m (about 78 m).

R = u^2 sin(2 theta) / g

By conservation of linear momentum (initial total momentum = 0): m_bullet * v_bullet = m_gun * v_gun. So 0.020 * 400 = 4 * v_gun, giving v_gun = 8 / 4 = 2 m/s (opposite to the bullet).

m1 v1 = m2 v2

On a smooth incline the acceleration along the plane is a = g sin(theta) = 10 * sin(30 deg) = 10 * 0.5 = 5 m/s^2, directed down the incline.

a = g sin(theta)

Work done W = m g h = 100 * 10 * 10 = 10000 J. Power P = W / t = 10000 / 5 = 2000 W = 2 kW.

P = W / t = m g h / t

When the net external torque on a system is zero, its total angular momentum L = I omega remains constant. A spinning skater experiences almost no external torque, so I omega is constant. On pulling the arms in, the moment of inertia I decreases, so the angular speed omega increases to keep L constant - the skater spins faster.

I1 omega1 = I2 omega2

Perpendicular axes theorem (for a planar body): the moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes in the plane meeting at the same point: Iz = Ix + Iy. For a ring, Iz (about central axis) = M R^2, and by symmetry Ix = Iy = I(diameter). So M R^2 = 2 I(diameter), giving I(diameter) = M R^2 / 2.

Iz = Ix + Iy

1) Law of orbits: every planet moves in an ellipse with the Sun at one focus. 2) Law of areas: the line joining a planet to the Sun sweeps out equal areas in equal intervals of time (so a planet moves faster when nearer the Sun); this follows from conservation of angular momentum. 3) Law of periods: the square of the orbital period is proportional to the cube of the semi-major axis, T^2 is proportional to a^3.

T^2 = k a^3