Kerala Engineering Entrance 2008 Questions on One Dimensional Motion

The following questions (on one dimensional kinematics) numbered 1, 2 and 3 appeared in KEAM (Engineering) 2008 question paper:

(1) A particle starts from rest at t = 0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t = 3 s is

(a) 2 ms^–1

(b) 4 ms^–1

(c) 6 ms^–1

(d) 8 ms^–1

(e) 1 ms^–1

According to the acceleration – time graph shown, the particle has an acceleration of 4 ms^–2 during the first two seconds. Therefore, its velocity (v₂) at the end of 2 seconds is given by

v₂ = v₀ + at = 0 + 4×2 = 8 ms^–1

From 2 second to 3 second the particle has a retardation of 4 ms^–2. Hence its velocity (v₃) at the end of 3 seconds is given by

v₃ = v₂ – at = 8 – 4×1 = 4 ms^–1 [Option (b)].

(2) Two cars A and B are moving with same speed of 45 km/hr along same direction. If a third car C coming from the opposite direction with a speed of 36 km/hr meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)

(a) 6.75

(b) 7.25

(c) 5.55

(d) 8.35

(e) 4.75

The relative velocity of car A with respect to car B (and that of car B with respect to A) is zero since they have the same speed in the same direction. The relative velocity of car C with respect to A and B is 45 + 36 = 81 km/hr. Since the car C takes a time of 5 minutes to cover the distance between A and B, the separation between A and B is 81×(5/60) km = 6.75 km.

(3) An object is dropped from rest. Its v - t graph is

The correct option is (a) since the velocity ‘v’ is directly proportional to the time t in accordance with the equation v = v₀ + at where v₀ is the initial velocity which is zero and a is the acceleration which is the constant acceleration due to gravity. The graph should evidently pass through the origin.

Suppose you were asked to draw the v – t graph in the case of a ball projected vertically upwards with a velocity u. If you are asked to draw the graph from the instant the ball leaves your hand to the instant you catch it while returning, you can do it as shown, ignoring the air resistance and the time taken for the velocity to reduce to zero on hitting your hand.

Questions involving Viscosity and Buoyancy

The following simple question involving viscous force and force of buoyancy is meant for checking whether you have a good understanding of basic points:

A wooden ball of relative density 0.5 is released from the bottom of a still water reservoir. Its acceleration while moving up will

(a) go on decreasing initially

(b) go on increasing initially

(c) remain constant at g/2 (in magnitude)

(d) remain constant at g (in magnitude)

(e) be zero throughout

The net force acting on the ball at the moment of releasing is its apparent weight which is upwards. [The apparent weight = Real weight – up thrust = Vρg – Vσg. Since the density of wood (ρ) is less than that of water (σ), the apparent weight is negative which means it is directed upwards]

The sphere therefore moves up with an acceleration. But, when the velocity (v) increases from zero, the opposing viscous force (6πrηv) also increases thereby reducing the net upward force. The upward acceleration therefore goes on decreasing. So, the correct option is (a).

The following question which appeared in Kerala Engineering Entrance 2008 question paper also involves viscous force and force of buoyancy; but the latter is negligible:

Eight drops of a liquid of density ρ and each of radius ‘a’ are falling through air with a constant velocity of 3.75 cm s^–1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop will be

(a) 1.5×10^–2 ms^–1

(b) 2.4×10^–2 ms^–1

(c) 0.75×10^–2 ms^–1

(d) 25×10^–2 ms^–1

(e) 15×10^–2 ms^–1

The viscous force acting on a sphere of radius ‘a’ is 6πahv where h is the coefficient of viscosity of the fluid and ‘v’ is the velocity of the sphere. Terminal velocity is attained when the opposing viscous force is equal in magnitude to the apparent weight of the sphere. Therefore,

6πahv = (4/3)πa³(ρ – σ)g where ρ is the density of the sphere, σ is the density of the fluid and ‘g’ is the acceleration due to gravity.

The above equation shows that the terminal velocity ‘v’ is directly proportional to the square of the radius of the sphere.

In the above problem, eight identical drops coalesce to form a single drop. The new drop thus formed has eight times the volume of each small drop. The radius ‘R’ of the new drop is given by

(4/3)πR³ = 8×(4/3)πa³

Therefore, R =2a

Since the radius of the new drop is twice that of each small drop, the terminal velocity of the new drop must become four times. The correct option therefore is 15×10^–2 ms^–1.