Shortcut Methods and Tricks to Solve Numerical Exercises

Galilean Laws

1. Uniformly Accelerated Motion

Concept

In uniformly accelerated motion, the object covers equal distance in equal intervals of time.

Tricks

• The object covers 1 unit of distance in the first second, 3 units in the next second, 5 units in the third second and so on.

• Total displacement covered in (n) seconds is given by (s = n(n+1)/2 ) units.

• The relation between initial velocity (u), final velocity (v) and displacement (s) is given by ( v2 - u2 = 2as).

Example

A train travels 10 m in the first second, and gains velocity of 5 m/s every second. How far will it travel in the 10th second?

Sol 1 (Simple Method) Distance traveled in the first 9 seconds = 10 + (3 * 5) = 25 m Distance traveled in the 10th second = (10 + 45)/2 =27.5 m Total distance = 25 m + 27.5 m = 52.5 m

Sol 2 (Using trick) Distance covered in 10 th second = (10(10 + 1)/2 = 55) m

Sol 3 (Using formula) ( v = u + at = (0 + 5 * 9) = 45 m/s), (s = ut +(1/2) at2 => 1/2 * 45 * 10 = 225m)

Answer: 52.5 m, 55m, 225m

Newton’s Law of Gravitation

Concept

Force of attraction between two point masses (M) and (m), separated by a distance (r), is given by: ( F = GmMm2 / r2)

Tricks

• If the masses are in kilograms and distance in meters, then (G=6.67\times10−11 )

• The direction of gravitational force is along imaginary line joining the centers of the two masses.

• Gravitational force is an action-reaction pair.

• Gravitational force between two bodies does not depend on presence or absence of any other object.

• Acceleration due to gravity is the gravitational force per unit mass and for the earth, (g = 9.8 m/s2)

Example

Calculate the force of gravitational attraction between the earth and a body of mass 100 kg. The mass of the earth is 6 × 1024 kg and its radius is 6.4 × 106 m.

Sol 1 (Simple method) Using (F = GmMr^2), plugging the values we get F = 6.67 * 10 - 11 x (6 * 1024) x 100/ (6.4 * 106)2 F = 667 * 10/ 40.96 *1012 = 16.3 N

Sol 2 (Using tricks) ( F = mg ), where (g=6.67\times10−11\times6\times1024/6.42×1062 ) = 9.8 m/s2 (F = 100 * 9.8 = 980 N)

Sol 3 (Using G) ( F = GmMr2 = 6.67 * 10 - 11 x (6 * 1024) x 6.42 × 106^2 = 980 N)

**Answer: **16.3 N, 980 N, 980N

Kepler’s Laws

Concept

Kepler’s Three Laws of Planetary Motion:

Law 1 (Law of Ellipses): Each planet’s orbit around the sun is an ellipse with the sun at one focus. Law 2 (Law of Equal Areas): A line from the sun to a planet sweeps out equal areas in equal intervals of time. Law 3 (Law of Harmonies): The ratio of the squares of the orbital period of two planets is equal to the ratio of the cubes of their semi-major axes. (T2/T1)2/(R2/R1)3

Tricks

• The closest point in an elliptical orbit is called perihelion, and the furthest point from the sun is called aphelion.

• The major axis of an elliptical orbit is the longest diameter of the ellipse, while the minor axis is the shortest.

• The semi-major axis is half of the major axis, while the semi-minor axis is half of the minor axis.

• The eccentricity of an orbit is a measure of its shape, and is determined by the following formula: (e = (c)/a , (c = \sqrt{ a^2 - b^2})).

• The period of an elliptical orbit is the time it takes for a body to complete one full orbit.

Example

In the solar system, the planet Mercury has an orbital period of 0.24 years and has a semi-major axis of 0.39AU (1 AU=1.5*108km). What is the orbital period of Venus if its semi-major axis is 0.72AU?

Sol 1 (Simple Method) Orbital periods are related by using (T^2=KR^3), where K is a constant: Orbital periods are related by using T2 = KR3, where K is a constant: T2/T1 = (R2/R1)3 = (0.72/0.39)3 = 7.462 So, T2 = T1 * (7.462)1/2 = 0.24 * 2.73 = 0.65 years

Sol 2 (Using Trick) (T2/T1=(R2/R1)3/2 , hence T2=(T1*(R2/R1)3/2) ), given (R2=0.72,R1=0.39 ), and (T1=0.24) ( T2=(0.24 * (0.72/0.39)3/2 ) = (0.24 * 1.8 * 1.259) = 0.659 years

**Answer: **0.65 years , 0.654 years.