Problem 2 : A force (2i + 3j) Newton acts on a particle. If force and displacement both are are in same plane, then find work done by the force in moving particle from (1 m,2 m) to (3 m,4 m) along a straight line.

Solution : We can infer from the expression of force, which involves two unit vectors, that it acts in xy - plane. Since force and displacement both are in same plane, the displacement of the particle is also in xy-plane.

Figure 3

Force and displacement

The figure above shows force vector (2i + 3j) and displacement end points A (1,2) and B(3,4). Note that we have used two coordinate systems - one for displacement and one for force.

Now, the work in two dimensions is given by :

W = F x x + F y y W = F x x + F y y

Here,

F x = 2 N ; F y = 3 N ; x = 3 − 1 = 2 m ; y = 4 − 2 = 2 m F x = 2 N ; F y = 3 N ; x = 3 − 1 = 2 m ; y = 4 − 2 = 2 m

W = 2 x 2 + 3 x 2 = 10 J W = 2 x 2 + 3 x 2 = 10 J

Problem 3 : A particle is acted upon by a two dimensional force (2i+3j). The particle moves along a straight line in xy - plane, defined by the equation, ax + 3y = 5, where "a" is a constant. If work done by the force is zero, then find "a".

Solution : As both force and displacement are non-zero, the work can be zero, only if force and displacement are perpendicular to each other. In order to compare the directions of force and displacement, we write equation of a straight line, which is parallel to line of action of force. From the expression of the force, it is clear that its slope, " m 1 m 1 ", is given by :

m 1 = y x = 3 2 m 1 = y x = 3 2

The equation of a straight line parallel to the line of action of force is :

y = m 1 x + c = 3 2 x + c y = m 1 x + c = 3 2 x + c

Now, rearranging equation of trajectory as given in the question, we have :

y = - a 3 + 5 3 y = - a 3 + 5 3

Let " m 2 m 2 " be the slope of the line representing trajectory. Then,

m 2 = - a 3 m 2 = - a 3

For two lines to be perpendicular, the product of slopes is m 1 X m 2 = - 1 m 1 X m 2 = - 1 ,

( 3 2 ) ( - a 3 ) = - 1 ⇒ a = 2 ( 3 2 ) ( - a 3 ) = - 1 ⇒ a = 2

Problem 5 : A body, having a mass of 10 kg, moves in a straight line with initial speed of 2 m/s and an acceleration of 1 m / s 2 m / s 2 . Find the initial time rate of change of kinetic energy.

Solution : Kinetic energy of the particle is given by :

K = 1 2 m v 2 K = 1 2 m v 2

We need to obtain a relation for "dK/dt" to obtain the required rate. Now,

ⅆ K ⅆ t = 1 2 m x 2 v ⅆ v ⅆ t ⇒ ⅆ K ⅆ t = m a v ⅆ K ⅆ t = 1 2 m x 2 v ⅆ v ⅆ t ⇒ ⅆ K ⅆ t = m a v

Putting values, we have :

ⅆ K ⅆ t = 10 X 1 X 2 = 20 J ⅆ K ⅆ t = 10 X 1 X 2 = 20 J

Problem 6 : If the ratio of the magnitude of linear momentum and kinetic energy of a particle at any given instant is inversely proportional to time, then find the nature of motion.

Solution : The ratio of linear momentum and kinetic energy of a particle at any given instant is :

p K = m v 1 2 m v 2 = 2 v p K = m v 1 2 m v 2 = 2 v

However, it is given that the ratio of linear momentum and kinetic energy is inversely proportional to time.

p K ∝ 1 t p K ∝ 1 t

Comparing two equations,

2 v ∝ 1 t 2 v ∝ 1 t

⇒ v ∝ t ⇒ v ∝ t

We know that "v = u + at" for motion with constant acceleration. In this case, v ∝ t, which is same as the relation obtained above. Hence, motion considered in the question is characterized by constant acceleration.