MCQ on Newton’s Laws of Motion- KEAM (Engineering) 2008 Questions

The following multiple choice questions appeared in Kerala Engineering Entrance 2008 question paper:

(1) A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. The force time graph is shown below

The velocity of the particle after 10 s is

(a) 20 ms^–1

(b) 10 ms^–1

(c) 75 ms^–1

(d) 26 ms^–1

(e) 50 ms^–1

The area under the force-time graph gives the impulse received which is equal to the change in momentum of the particle. Since the initial momentum of the particle is zero, the area under the graph gives the momentum of the particle after 10 s.

The area under the curve is made of rectangles and triangles and is equal to100 newton second. [Note that Ns is the same as kg ms^–1].

The velocity of the particle after 10 s = (100 kg ms^–1)/2 kg = 50 ms^–1

(2) An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with with an acceleration of 0.25 ms^–2 is (g = 10 ms^–2)

(a) 51.25 N

(b) 48.75 N

(c) 52.75 N

(d) 47.25 N

(e) 55 N

The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign is to be taken as positive. Since the lift is moving upwards the sign of its acceleration is negative so that the weight of the object as indicated by the spring balance is m[g-(-a)] = m(g+ a) = 5×10.25 = 51.25 N.

(3) A bullet of mass 0.05 kg moving with a speed of 80 ms^–1 enters a wooden block and is stopped after a distance of 0.4 m. The average resistance force exerted by the block on the bullet is

(a) 300 N

(b) 20 N

(c) 400 N

(d) 40 N

(e) 200 N

The bullet is retarded within the wooden block. The retardation is given by the usual equation of uniformly accelerated motion, v² = u² + 2as. Here v = 0, u = 80 ms^–1 and s = 0.4 m.

Therefore, 0 = 80² + 2a×0.4 from which a = – 8000 ms^–2

The retardation is 8000 ms^–2 and the resistance force exerted by the wooden block is 0.05×8000 = 400 N.

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.

Multiple Choice Questions (MCQ) on Newton’s Laws of Motion

The following questions appeared in Kerala Engineering Entrance 2007 question paper:

A shell at rest at the origin explodes into three fragments of masses 1 kg, 2 kg and ‘m’ kg. The 1 kg and 2 kg pieces fly off with speeds of 5 ms^–1 along the X- axis and 6 ms^–1 along the Y-axis respectively. If the m kg piece flies off with speed 6.5 ms^–1, the total mass of the shell must be

(a) 4 kg (b) 5 kg (c) 3.5 kg (d) 4.5 kg (e) 5.5 kg

The initial momentum of the shell is zero. Explosion of the shell involves internal forces only. Since there are no external forces, the total momentum of all the fragments of the shell taken together must be zero.

The momentum of the 1 kg fragment is 5 kg ms^–1 and that of the 2 kg fragment is 12 kg ms^–1. Since these are at right angles, the net momentum of these two fragments taken together is √(5² + 12²) = 13 kg ms^–1. The momentum of the m kg fragment must be equal and opposite to this (to make the net momentum of the three fragments zero).

Therefore, m×6.5 = 13, from which m = 2 kg.

The mass of the shell is therefore 1 + 2 + 2 = 5 kg.

The following question is simple, but many among you will have certain doubts about it:

A uniform rope of length ‘L’ is resting on a smooth horizontal surface. If it is pulled at one end by applying a horizontal force ‘F’ parallel to its length, what will be the tension in the rope at a distance ‘d’ from the other end?
(a) F (b) F/d (c) FL/d (d) Fd (e) Fd/L

When you work out questions on bodies connected by strings, you will assume that the string is light and therefore the tension in a segment of the string connecting two bodies will be the same everywhere. The acceleration of the system in such cases will be determined by the masses of the connected bodies only.

In the above problem, there are no bodies connected to the rope and hence the acceleration ‘a’ of the rope is determined by the mass of the rope only.

Therefore, a = F/ mL where ‘m’ is the mass per unit length of the rope.

The tension ‘T’ in the rope at the distance ‘d’ from the other end is responsible for pulling the portion of length ‘d’ of the rope. Therefore we have

T = mass × acceleration = md × F/mL = Fd/L

The following question on connected bodies is a popular one and you are expected to know how to solve it:

Three blocks of masses m₁, m₂ and m₃ are connected by light inextensible strings and carried by a light frictionless pulley as shown. If (m₁+m₂) > m₃, the tension in the string connecting m₁ and m₂ is

(a) 2m₁m₃g/ (m₁+m₂+m₃)

(b) 2(m₁+m₂)m₃g/(m₁+m₂+m₃)

(c) m₁g

(d) (m₁+m₂)

(e) 2m₁m₂m₃g/ (m₁+m₂+m₃)

Note that the tension in the string connecting m₁ and m₂ is different from the tension in the string connecting m₂ and m₃.

The common acceleration ‘a’ of the system of masses is the ratio of the net driving force to the total mass moved.

Therefore, a = (m₁+m₂ – m₃)g/(m₁+m₂+m₃)

The tension (T) in the string connecting m₁ and m₂ is produced by the apparent weight ( as in the case of motion in a lift) of the mass m₁ which moves downwards with acceleration ‘a’.

Therefore, T = m₁(g – a) = m₁[g – (m₁+m₂ – m₃)g/(m₁+m₂+m₃)]

= m₁g[1– (m₁+m₂ – m₃)/(m₁+m₂+m₃)]

= 2m₁m₃g/ (m₁+m₂+m₃)

You will find more questions (with solution) in this section by clicking on the label ‘Newton’s laws’ below this post or on the side of this page.

Similar multiple choice questions with solution can be found at physicsplus: Multiple Choice Questions on Newton’s Laws of Motion

Questions involving motion in a lift

You may often encounter questions involving motion in a lift. Generally such questions are simple, but you should have no confusion in this context. So keep the following points in mind:
The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign is taken as positive. The following cases can arise in this connection:
(1) Lift moving down with acceleration of magnitude ‘a’:
In this case ‘a’ also is positive and the weight is m(g-a) which is less than the real weight of the body (when it is at rest).
(2) Lift moving up with acceleration:
In this case ‘a’ is negative and the weight is m[g-(-a)] = m(g+a).
(3) Lift moving down with retardation (going to stop while moving down):
In this case also ‘a’ is negative and the weight is m[g-(-a)] = m(g+a) which is greater than the actual weight.
(4) Lift moving up with retardation (going to stop while moving up):
In this case ‘a’ is positive and the weight is m(g-a)
(5) Lift moving up or down with uniform velocity:
In this case ‘a’ is zero and the weight is mg.
(6) Lift moving down with acceleration of magnitude ‘g’ (falling freely under gravity as is the case when the rope carrying the lift breaks):
In this case ‘a’ is positive and the weight is m(g-g) which is zero.
Let us now consider the following question:

A light, frictionless pulley is suspended from the ceiling using a light string. Another light string is passed over the pulley. If unequal masses M₁ and M₂ (M₁ >M₂) are attached to the ends of this string and the system is released, what will be tension ‘T’ in the string that carries the pulley?
(a) (M₁+M₂)g (b) (M₁- M₂)g (c) M₁M₂g/(M₁+M₂) (d) 2M₁M₂g/(M₁+M₂) (e) 4M₁M₂g/(M₁+M₂)

The tension in the string that caries the pulley is due to the weights of the masses m1 and m2. But these weights are not M1g and M2g but M₁(g-a) and M₂(g+a) because M₁ is moving down with acceleration ‘a’ and M2 is moving up with acceleration ‘a’. The magnitude of ‘a’ is given by
a = Net driving force/ Total mass moved = (M1- M2)g/(M1+M2)
Therefore, tension, T = M₁(g-a) + M₂(g+a) = 4M₁M₂g/(M₁+M₂) on substituting for ‘a’. The correct option is (e)
Now consider the following simple question:
A man of mass 70 kg records his weight on a weighing scale placed inside a lift. The ratio of the weight of the man recorded when the lift is ascending with a uniform speed of 1m/s to the weight recorded when the lift is descending with a uniform speed of 2m/s will be
(a) 2 (b) 0.5 (c) 3 (d) 1.5 (e) 1
You should note that his weight is the same as his real weight (when he is at rest). There is no distinction between state of rest and uniform motion. To put this in a different manner, the acceleration of the lift is zero and hence his weight is unchanged. The ratio of the weights is therefore equal to unity [Option (e)].
Now consider the following MCQ:
What is the minimum acceleration with which a fire man can slide down a rope whose breaking strength is 80% of his weight?
(a) 0.2g (b) 0.1g (c) ) 0.9g (d) g (e) 1.1g
This is a situation where the fire man has a rope of insufficient strength. He wants to come down from an unsafe height. He can use the rope to slide down with acceleration so that his weight is reduced (as in a lift moving down with acceleration). His minimum acceleration ‘a’ is given by
m(g-a) = 0.8mg
We have equated his reduced weight to the breaking strength of the rope. The value of ‘a’ is 0.2g [Option (a)]. Note that the rope will break if the acceleration of the fire man is less than this and also that he can use a weaker rope if his acceleration is greater than this.

Two Questions from Newton’s Laws of Motion

(1)The breaking strength of a rope is one and a half times the weight of a stone. The minimum time in which this stone can be raised through 10m using this rope is nearly
(a) 0.5s (b)1s (c) 2s (d) 2.5s (e) 3s
The minimum time will be obtained when the acceleration with which the stone is pulled up is maximum. But the maximum possible acceleration ‘a’ is that which will make the apparent weight of the stone(as in the case of a body in a lift) one and a half times its real weight. Therefore we have, m(g+a) = 1.5mg from which a=0.5g.
Substituting this value of ‘a’ in the equation, s = ut + ½ at^2, we have
10 = ½ ×0.5g t^2 since u=0. Taking ‘g’ to be nearly 10m/s^2 we obtain t = 2s.
(2) A rocket with a lift off mass 3.5×10^4 kg is blasted upwards with an initial acceleration of 10m/s^2. Then the initial thrust of the blast is
(a) 1.75×10^5N (b) 3.5×10^5N (c) 7×10^5N (d) 14×10^5N
The initial thrust of the blast is equal in magnitude to the apparent weight of the rocket while moving up with the given acceleration(as in the case of a body in a lift).Therefore, thrust = m(g+a) = 3.5×10^4(10+10) = 7×10^5N [Option (c)]This question appeared in the A.I.E.E.Exmination question paper of 2003.