Kerala Medical Entrance (KEAM) 2008 Questions from Newton’s Laws of Motion

The following questions which appeared in Kerala Medical Entrance 2008 question paper are of the type popular among question setters:

(1) Two blocks of masses 7 kg and 5 kg are placed in contact with each other on a smooth surface. If a force of 6 N is applied on the heavier mass, the force on the lighter mass is

(a) 3.5 N

(b) 2.5 N

(c) 7 N

(d) 5 N

(e) 6 N

The common acceleration a of the system of masses is given by

a = Driving force/Total mass moved = 6 N/(7+5) kg = 0.5 ms^–2

The force on the lighter mass = Ma = 5×0.5 = 2.5 N.

(2) A body of mass 60 kg suspended by means of three strings P, Q and R as shown in the figure is in equilibrium. The tension in the string P is

(a)130.9g N

(b) 60g N

(c) 50g N

(d) 103.9g N

(e) 109g N

Let us denote the tensions in the three strings by P, Q and R. The concurrent forces P, Q and R keep the common meeting point of the strings in equilibrium so that they can be represented by the three sides of a triangle taken in order as shown in the adjoining figure. From this right angled triangle,

Q/P = tan 30º where Q = weight of the mass M = Mg =60g newton.

[This equation is often written as W/_H = tan θ where W is the weight and H is the horizontal force and is known as tangent law].

Substituting for Q, we have 60g/P = 1/√3, from which P = 60g×√3 = 103.9g N.