Two Questions (MCQ) on Momentum Conservation

Often you will find questions based on the law of conservation of momentum. Here is a question which may be interesting to you:

Two skaters of masses 40 kg and 50 kg respectively are 4 m apart and are facing each other, holding the ends of a light rope. They advance by pulling on the rope, thereby reducing the separation between them. The approximate distances covered by them by the time their separation has reduced to zero are respectively

(a) 2.22 m and 1.78 m (b) 2 m and 2 m (c) 1.62 m and 2.38 m

(d) 4 m and zero (e) zero and 4 m

The initial momentum of the system containing the two skaters is zero. The pulling forces they exert are internal forces (within the system) and hence the total momentum of the system must be zero throughout their motion. If v₁ and v₂ are the average velocities of the skaters respectively, we have 40 v₁ + 50 v₂ = 0.

Therefore, 40 v₁ = – 50 v₂.

Since v₁ = s₁/t and v₂ = s₂/t where s₁ and s₂ are the displacements, we have

40 s₁ = –50 s₂

Considering the magnitudes of displacements, 40 s₁ = 50 s_2.

Since s₁+ s₂ (which is the total distance covered by the two scaters) is 4 m, s₂ = 4–s₁ so that

40 s₁ = 50(4–s₁) from which s₁ = 2.22 m and s₂ = 1.78 m [Option (a)]

The following MCQ is a simple one. But be careful not to be distracted to arrive at a wrong answer!

Two spheres of the same material have radii R and 2R. They are released in free space with initial separation between their centres equal to 15R. If the only force between the spheres is the gravitational pull between them, the distance covered by the smaller sphere before collision is approximately

(a) 6.67R (b) 10.67R (c) 13.33R (d) ) 2.33R (e) 4.33R

As in the previous question, the forces acting are internal forces and hence the total momentum of the system remains unchanged. The initial momentum of the system containing the two spheres is zero and hence the total momentum of the system must be zero throughout their motion.

Proceeding as before, we have m₁s₁ = m₂s₂ where m₁ and m₂ are the masses of the spheres and s₁ and s₂ are the distance covered by them before collision. But you have to be careful to note that when the spheres collide, the distance between their centres is (R+ 2R) = 3R so that the total distance covered by the spheres is 15R – 3R = 12R.

Therefore, s₁+s₂ = 12R so that m₁s₁ = m₂(12R–s₁).

Since the spheres are of the same material, their masses are directly proportional to their volumes which are proportional to the cubes of their radii. Therefore we have

R³s₁ = (2R)³(12R–s₁) from which s₁ = 10.67 R.

Kerala Engineering Entrance 2007 Questions on Heating Effect of Electric Current

Two questions on heating effect of electric current appeared in KEAM (Engineering) 2007 question paper:

(1) The resistance of a wire at room temperature 30º C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [Temperature coefficient of resistance of the material of the wire is 0.002 per º C]

(a) 36º C (b) ) 83º C (c) ) 63º C (d) ) 33º C (e) ) 66º C

The resistance (R_t) at tº C can be expressed in terms of the resistance (R₀) at 0º C and the temperature coefficient (α) of resistance as

R_t = R₀(1 + αt).

Therefore, we have

R₃₀ = 10 = R₀(1 + 0.002×30) and

R_t = 11 = R₀(1 + 0.002 t)

[The resistance R_t at the unknown temperature ‘t’ is greater by 10% and is therefore equal to (10 + 1) Ω = 11 Ω]

From the above equations we have

10/11 = (1 + 0.06)/ (1 + 0.002t) from which t = 83º C.

(2) If R₁ and R₂ be the resistances of the filaments of 200 W and 100 W electric bulbs operating at 220 V, then R₁/R₂ is

(a) 1 (b) 2 (c) 0.5 (d) 4 (e) 0.25

Since the power is V²/R where V is the operating voltage and R is the resistance, we have

220²/R₁ = 200 and

220²/R₂ = 100.

Dividing the second equation by the first, we obtain

R₁/R₂ = 0.5

Kerala Engineering Entrance 2007 Questions on Rotational Motion

The following questions appeared in KEAM (Engineering) 2007 question paper:

(1) A sphere of mass ‘m’ and radius ‘r’ rolls on a horizontal plane without slipping with speed ‘u’. Now if it rolls up vertically, the maximum height it would attain would be

(a) 3u²/4g (b) 5u²/2g (c) 7u²/10g (d) u²/2g (e) 11u²/9g

The initial kinetic energy (E) of the sphere while rolling on the horizontal surface is given by

E = ½ mu² + ½ Iω² where I is the moment of inertia of the sphere about its diameter[(which is equal to (2/5)mr²] and ‘ω’ is the angular velocity (which is u/r).

[Note that the first term is the translational kinetic energy and the second term is the rotational kinetic energy].

Therefore, E = ½ mu² + ½ ×(2/5)mr²u²/r² = 7mu²/10.

Since this energy is converted into gravitational potential energy (mgh) to attain the maximum height h, we have

7mu²/10 = mgh, from which h = 7u²/10g.

(2) If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of the full day would be

(a) 3 hours (b) 1.5 hours (c) 6 hours (d) 4 hours (e) 2 hours

If the radius of the earth becomes ‘n’ times he present value, without change in the mass, the duration of the day will become 24n² hours so that the answer to the above question is 24×(1/4)² = 1.5 hours.

Solution to Multiple Choice Questions on Centre of Mass

Two multiple choice questions on centre of mass were given to you for practice yesterday. Here is the solution along with the questions:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

You should have worked this out within seconds. The centre of mass will be unaffected with respect to external fixed points if there are no external forces acting on the system. So, the correct option is (e).

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L (b) 4L/3 (c) 3L/2 (d) 2L/3

The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.

The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the mid points of the horizontal and vertical portions of the T.

Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by

r = (m₁r₁ + m₂r₂)/(m₁ + m₂) = (m×2L + 2m×L)/(m + 2m) = 4L/3

[ Note that we have used the equation, r = (m₁r₁ + m₂r₂)/(m₁ + m₂) for the position vector r of the centre of mass in terms of the position vectors r₁ and r₂ of the point masses m₁ and m₂. We could use the simple equation involving the distances from C since the points are collinear].

Two Multiple Choice Questions on Centre of Mass (For practice)

Here are two questions on centre of mass. These are meant for checking whether you have a clear idea of the concept of centre of mass:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L
(b) 4L/3
(c) 3L/2
(d) 2L/3

This MCQ appeared in AIEEE 2005 question paper.

Try to find the answer to the above questions. If you have clear understanding of the centre of mass, you will be able to find the answers in a couple of minutes. I’ll be back with the solution shortly.

Multiple Choice Questions on Work and Energy

Here is a simple question which is meant for gauging your understanding of the work-energy principle. This MCQ appeared in AIEEE 2005 question paper:

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

(a) ML²/K (b) zero (c) KL²/2M (d) √(MK).L

If the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p²/2M. This must be equal to the maximum potential energy of the spring so that we have

p²/2M = ½ KL², from which p = L√(MK)., given in option (d).

Now consider the following MCQ which appeared in Kerala Medical Entrance 2006 question paper:

The work done by a force F = –6x³ î newton, in displacing a particle from x = 4m to x = –2m is

(a) 360 J (b) 240 J (c) – 240 J (d) – 360 J (e) 408 J

This is a case of variable force (in the X-direction), the point of application of which is moved in the X-direction. The work done is therefore given by

W = ∫F.dx = ₄ ∫^–2 (–6x³)dx = –6[x⁴/4] with x between limits 4 and –2.

Therefore, W = –(6/4)(16 – 256) = 360 joule, given in option (a).

The following MCQ appeared in Kerala Engineering entrance 2006 question paper:

A running man has the same kinetic energy as that of a boy of half the mass. The man speeds up by 2 ms^–1 and the boy changes his speed by ‘x’ ms^–1 so that the kinetic energies of the boy and the man are again equal. Then ‘x’ in ms^–1 is

(a) – 2√2 (b) + 2√2 (c) √2 (d) 2 (e) 1/√2

If the mass of the man is ‘m’, the mass of the boy is m/2. If v₁ and v₂ are the initial velocities of the man and boy respectively, we have

½ mv₁² = ½ (m/2)v₂²

Therefore, v₂ = v₁√2.

On changing the speeds, we have

½ m(v₁+2)² = ½ (m/2)(v₂+x)²

On substituting for v₂ (=v₁√2), the above equation simplifies to

(v₁+2)² = ½ (v₁√2+x)² from which x = 2√2 ms^–1, given in option (b).