AP Physics C- Multiple Choice Practice Questions on Centre of Mass

Let us consider some typical multiple choice questions on centre of mass. You know the concept of centre of gravity of a body which is the point through which the resultant gravitational force on the body (weight) acts. The concept of centre of mass is relevant even in situations of weightlessness even though it is true that the centre of mass and the centre of gravity coincide in a uniform gravitational field.

The centre of mass (R) of a system of particles of masses m₁, m₂, m₃, m₄, …etc. is defined by

R = (m₁r₁+ m₂ r₂ + m₃ r₃ + m₄ r₄ + …etc.)/ (m₁+ m₂+ m₃+ m₄+ …etc.) where r₁, r₂, r₃, r₄ etc. are the position vectors of the particles of masses m₁, m₂, m₃, m₄, …etc. respectively.

Note that R is the position vector of the centre of mass.

The above expression for the position vector of the centre of mass can be written in a compact form as

R = Σm_ir_i /M where M = Σm_i which is the total mass of the system of particles. The value of ‘i’_.should run from 1 to ‘n’ if there are ‘n’ particles in the system.

In the adjoining figure a system containing three point masses with position vectors r₁, r₂ and r₃ is shown. The position vector of the centre of mass also is shown in the figure. If the masses of the particles are respectively m₁, m₂ and m₃ the centre of mass has position vector R given by

R = (m₁r₁+ m₂ r₂ + m₃ r₃)/ (m₁+ m₂+ m₃)

The x, y and z coordinates of the position of the centre of mass are evidently given respectively by

x = (m₁x₁+ m₂ x₂ + m₃ x₃)/ (m₁+ m₂+ m₃),

y = (m₁y₁+ m₂ y₂ + m₃ y₃)/ (m₁+ m₂+ m₃) and

z = (m₁z₁+ m₂ z₂ + m₃ z₃)/ (m₁+ m₂+ m₃)

Even though we have considered a general case involving all the three components (x, y and z) for the position vectors of the particles and their centre of mass you will often find simple cases in which the particles are collinear or coplanar or having spherical symmetry in three dimensional arrangement. The following multiple choice questions will make you more confident:

(1) Two particles of masses 2 mg and 6 mg are separated by a distance of 6 cm. the distance of their centre of mass from the heavier particle is

(a) 1.5 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

(e) 4.5 cm

You may imagine the particles to be positioned on the X-axis, with the heavier one (of mass 6 mg) at the origin as shown.

6 mg ●----x-------------● 2 mg

If the centre of mass is at x we have R = (m₁r₁+ m₂ r₂)/M

Or, x = (6 mg ×0 + 2 mg×6 cm)/ 8 mg = 1.5 cm

You can easily arrive at this answer if you remember that in the case of a two particle system the moments of the masses about the centre of mass are equal. Therefore

6 × x = 2 ×(6 –x) from which x = 1.5 cm.

(2) A straight rod AB of length L has non-uniform linear density, λ = Kx/L where K is a constant and x is the distance from end A. The distance of the centre of mass of the rod from the end A is

(a) ^L/₃

(b) ^L/₂

(c) ^3L/₅

(d) ^2L/₃

(e) ^3L/₄

Consider an element of very small length dx of the rod at distance x from the end A. The mass of this element is dm = λdx = (Kx/L) dx

We have R = Σm_ir_i / Σm_i

If X is the distance of the centre of mass from the end A we can rewrite the above equation as

X = [₀∫^L dm x]/ [₀∫^L dm]

= [₀∫^L Kx²dx /L]/ [₀∫^L Kxdx /L] = (L³/3)/ (L²/2) = ^2L/₃

(3) A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an α-particle of mass m. If the α-particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the α-particle) will

(a) remain at rest

(b) move along the positive x-direction with speed less than v

(c) move along the positive x-direction with speed greater than v

(d) move in a direction inclined to the positive x-direction

(e) move along the positive x-direction with speed equal to v

The state of rest or of uniform motion of the centre of mass of a system of particles can be changed by external forces only. Since the α-emission is produced by internal forces, the centre of mass is unperturbed and it will continue to move along the positive x-direction with speed equal to v [Option (e)].

It will be interesting to note that the centre of mass of a shell which explodes in mid air will continue to move along the parabolic path originally followed by the unexploded shell.

(4) Three homogeneous solid spheres of masses 1 kg, 2 kg and 4 kg are arranged with their centres at (2i + j + k), (3i – 2 j + 2k) and (4i – j – 2k) respectively where i, j, k are unit vectors in the x, y and z directions. All distances are in metre. The y-coordinate of the centre of mass of the system of spheres is

(a) 3.3 m

(b) 2.6 m

(c) 0.33 m

(d) – 1 m

(e) – 2 m

By symmetry the centre of mass of a homogeneous sphere is its geometric centre. The three spheres can therefore be treated as three poin masses (for finding the centre of mass).

The position vector of the centre of mass is given by

R = (m₁r₁+ m₂ r₂ + m₃ r₃)/ (m₁+ m₂+ m₃)

Substituting proper values,

R = [1(2i + j + k) + 2(3i – 2 j + 2k) + 3(4i – j – 2k)] /(1+2+3)

= (20 i – 6 j – k)/ 6

The y-component of the position vector of the centre of mass is –1 metre. Therefore the correct option is (d).

(5) The figure shows a T-shaped portion cut from a plane sheet of uniform thickness (lamina) having areal density of 1 kgm^–2. Imagine the T to be made of the horizontal rectangular portion of sides 3 m and 1 m and the vertical rectangular portion of sides 2 m and 1 m. Neglect the thickness of the sheet. The centre of mass of the entire T with respect to the coordinate system shown in the figure is at the point

(a) (0.5, – 0.5, 0)

(b) (0.5, – 0.1, 0)

(c) (0.5, – 0.4, 0)

(d) (0.4, – 0.2, 0)

(e) (0.5, 0, 0)

The horizontal rectangle has its centre of mass at its centre A and the vertical rectangle has its centre of mass at its centre B. The mass of the horizontal rectangular portion is 3 kg since its area is 3 m². The mass of the vertical rectangular portion is 2 kg since its area is 2 m². The entire T can be imagined to be reduced to two particles of masses 3 kg and 2 kg located at A and B respectively. Evidently the coordinates of A and B are (0.5, 0.5, 0) and (0.5, –1, 0). The Z coordinate is zero since the T is placed in the XY plane.

The centre of mass has x coordinate given by

x = (m₁x₁+ m₂ x₂)/ (m₁+ m₂) = (3×0.5 + 2×0.5)/(3+2)

Or, x = 0.5 m

Similarly the centre of mass has y coordinate given by

y = (m₁y₁+ m₂ y₂)/ (m₁+ m₂ ) = [3×0.5 + 2×(–1)]/(3+2)

Or, y = – 0.1 m. The centre of mass is therefore at the point (0.5, – 0.1, 0) given in option (b).

You will find additional questions (with solution) in this section at physicsplus

Answer to Free Response Practice Question Involving Newtonian Mechanics and Electromagnetism for AP Physics B

Joy in looking and comprehending is nature’s most beautiful gift.

– Albert Einstein

In the post dated 16^th February 2009, I had given a free response practice question involving Newtonian mechanics and electromagnetism for AP Physics B aspirants. As promised, I give below the answer along with the question:

A cart carries a block of mass m₁ which is connected to a soft iron cylinder of mass m₂ using an inextensible string of negligible mass passing over a light frictionless pulley as shown in the adjoining figure. The soft iron cylinder is suspended vertically with its lower end inside a current carrying coil fixed to the cart. The cylinder and the coil are coaxial and their vertical sides touch each other without any friction between them. The current carrying coil exerts a magnetic force of constant magnitude f₂ on the soft iron cylinder and the entire system consisting of the cart of mass M (with the current carrying coil), and the connected masses m₁ and m₂ are moving with a constant acceleration ‘a’ in the positive x-direction under the action of a constant force F (fig.) in the same direction. There is no friction between the cart and the block m₁ and the block and the cylinder are at rest relative to the cart during the accelerated motion of the cart. Assume that g = 10 ms^–2.

Now answer the following:

(a) Write an expression for the acceleration ‘a’ of the cart in terms of the net force on the cart and the masses involved

(b) If m₁ = 11 kg, m₂ = 1 kg and a = g/10 calculate the magnitude of the magnetic force f₂ on the cylinder.

(c) If the current in the coil were in the opposite direction, will the direction of the magnetic force change? Justify your answer.

(d) If a cylindrical permanent magnet were used instead of the soft iron cylinder will your answer for part (c) be different? Justify your answer.

(a) The net force on the cart is F itself and the total mass moved is M+ m₁+ m₂. Therefore the acceleration a of the cart is given by

a = F/( M+ m₁+ m₂)

(b) The magnetic force on the soft iron cylinder is attractive since it will become a temporary magnet in the magnetic field produced by the coil, with the nearer end of the cylinder acquiring opposite polarity (compared to the polarity of the nearer end of the coil). The total downward force on the soft iron cylinder is therefore m₂g + f₂. Therefore we have

m₂g + f₂ = T where T is the tension in the string.

The inertial force on m₁ is m₁a so that we have

m₁a = T

The above equations give m₂g + f₂ = m₁a from which

f₂ = m₁a – m₂g = 11×(g/10) – 1×g = 0.1 g = 0.1× 10 =1 N.

(c) The magnetic force on the soft iron cylinder is always attractive (irrespective of the direction of the current) since it will become a temporary magnet in the magnetic field produced by the coil, with the nearer end of the cylinder acquiring opposite polarity (compared to the polarity of the nearer end of the coil). So the direction of the magnetic force is unchanged when the direction of the current is reversed.

(d) If a permanent magnet is used instead of the soft iron cylinder, the direction of the magnetic force will be reversed on reversing the direction of the current in the coil. This is due to the fact that the magnetic polarity of the nearer end of the cylinder (permanent magnet) is fixed where as the magnetic polarity of the nearer end of the current carrying coil changes on reversing the current.

Part (c) of the above question can be made a little more indirect by the following modification:

(c) If the current in the coil were in the opposite direction, will the force F applied on the cart be the same or different for keeping the masses m₁ and m₂ at rest relative to the cart?

Justify your answer.

Since the magnetic force on the soft iron cylinder is unchanged, the force F applied on the cart will be unchanged.

Obviously the force F will change if a permanent magnet is used instead of the soft iron cylinder.

A Free Response Practice Question Involving Newtonian Mechanics and Electromagnetism for AP Physics B

I give below a free response practice question involving fundamental principles in Newtonian mechanics and electromagnetism. Even though this question is meant for AP Physics B aspirants, I would advise AP Physics C aspirants also to try it. Here is the question:

A cart carries a block of mass m₁ which is connected to a soft iron cylinder of mass m₂ using an inxtensible string of negligible mass passing over a light frictionless pulley as shown in the adjoining figure. The soft iron cylinder is suspended vertically with its lower end inside a current carrying coil fixed to the cart. The cylinder and the coil are coaxial and their vertical sides touch each other without any friction between them. The current carrying coil exerts a magnetic force of constant magnitude f₂ on the soft iron cylinder and the entire system consisting of the cart of mass M (with the current carrying coil), and the connected masses m₁ and m₂ are moving with a constant acceleration ‘a’ in the positive x-direction under the action of a constant force F (fig.) in the same direction. There is no friction between the cart and the block m₁ and the block and the cylinder are at rest relative to the cart during the accelerated motion of the cart. Assume that g = 10 ms^–2.

Now answer the following:

(a) Write an expression for the acceleration ‘a’ of the cart in terms of the net force on the cart and the masses involved

(b) If m₁ = 11 kg, m₂ = 1 kg and a = g/10 calculate the magnitude of the magnetic force f₂ on the cylinder.

(c) If the current in the coil were in the opposite direction, will the direction of the magnetic force change? Justify your answer.

(d) If a cylindrical permanent magnet were used instead of the soft iron cylinder will your answer for part (c) be different? Justify your answer.

Try to answer this question which carries 15 points. You may take 15 minutes for answering this. I’ll be back soon with a model answer for you.

AP Physics B Multiple Choice Practice Questions on Temperature and Heat

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives”

– Albert Einstein

As promised in the last post dated 6^th February 2009, I give below a few multiple choice practice questions on temperature and heat relevant to AP Physics B:

(1) A soft iron rod has its length increased by 0.3% on increasing its temperature by ∆T. The percentage increase in the area of a circular hole in a sheet of the same material (soft iron) on increasing its temperature by ∆T will be

(a) 0.3%

(b) – 0.3%

(c) 0%

(d) 0.6%

(e) –0.6%

The radius of the hole will increase by 0.3% and hence the area of the hole will increase by 0.6%.

[The area A = πR² where R is the radius. Therefore, ∆A/A =2 ∆R/R. The fractional change in area is equal to twice the fractional change in radius. Therefore, the percentage change in area is equal to twice the percentage change in radius].

(2) The temperature of oil in a vessel is measured using a Celsius scale thermometer as well as a Fahrenheit scale thermometer. The readings are respectively tº C and 2tº F. The temperature of the oil is

(a) 32º C

(b) 100º C

(c) 120º C

(d) 142º C

(e) 160º C

We have t_C/100 = (t_F – 32) /180.

[This can be remembered as t_C = (5/9)(t_F – 32)]

Here t_C = t and t_F = 2t so that t/100 = (2t– 32) /180.

Or, 9t = 10t– 160 from which t = 160.

Therefore, the temperature of oil is 160º C.

(3) A copper rod and a steel rod are to have lengths Lc and L_s such that the difference between their lengths is the same at all ambient temperatures. If the coefficients of linear expansion of copper and steel are respectively α_c and α_s the lengths are related to the coefficients of linear expansion as

(a) Lc/L_s = α_c/α_s

(b) Lc – L_s = α_c – α_s

(c) Lc/L_s = α_s/α_c

(d) Lc/L_s = (α_c/α_s)^1/2

(e) Lc/L_s = (α_c/α_s)^–1/2

The changes in length must be equal on heating (or cooling) so that we have

Lc α_c ∆T = Ls α_s ∆T

This gives Lc/L_s = α_s/α_c

(4) Four identical brass rods AB, BC, CD and DA are joined to form a square (fig). If the diagonally opposite corners A and C are kept in melting ice and boiling water respectively what is the temperature difference between the corners B and D?

(a) 50º C

(b) 100º C

(c) 25º C

(d) 75º C

(e) 0º C

The corners B and D will be at the same temperature so that the difference will be zero.

(5) Thermal conductivity of metal A is three times the thermal conductivity of metal B. Two rods of the same dimensions made of metals A and B are joined end to end (fig.). The free end of A is maintained at 0º C and that of of B is maintained at 100º C. The temperature of the junction between A and B in the steady state will be

(a) 25º C

(b) 33.3º C

(c) 66.6º C

(d) 75º C

(e) 90º C

In the steady state the quantity of heat flowing through B per second will be equal to that flowing through A per second. If the temperature of the junction between A and B is tº C we have

kA(100 – t)/L = 3kA(t – 0)/L where A is the area of cross section, L is the length, k is the thermal conductivity of B and therefore 3k is the thermal conductivity of A.

Therefore 100 – t = 3t so that t = 25º C

[Since the rods are of identical dimensions and the thermal conductivity of A is three times that of B, you can easily argue that the temperature drop in A is only one-third of the temperature drop in B and arrive at the answer].

(6) A cup of coffee cools from 65º C to 63º C in one minute in a room of temperature 24º C. To cool from 58º C to 50º C the same cup of coffee will take nearly

(a) 2 min.

(b) 2.4 min.

(c) 3.8 min.

(d) 4.5 min

(e) 5.3 min

Since the rate of cooling is directly proportional to the excess of temperature (in accordance with Newton’s law of cooling), we have

S(65 – 63)/1 α (64 – 24) and

S(58 – 50)/t α (54 – 24)

where S is the heat capacity (thermal capacity) of the cup of coffee (which is the heat required to raise the temperature through 1K) and t is the time required to cool from 58º C to 50º C. Note that we have used the average temperatures (64º C and 54º C) of the cup of coffee to find the mean excess temperatures in the two cases.

On dividing, 2t/8 = 40/30 from which t = 5.3 min (nearly)

Temperature and Heat for AP Physics B- Equations to be Remembered

Here are the essential things you need to remember for solving multiple choice questions involving heat and temperature:

(1) Temperature t_C in Celsius scale can be converted to temperature t_F in Fahrenheit scale using the relation

t_C/100 = (t_F – 32) /180

Remember that the ice point and steam point of water are 32º F and 212º F respectively in the Fahrenheit scale. These temperatures in the Celsius scale are 0º C and 100º C respectively so that a temperature difference (∆t_F) of 180º in the Fahrenheit scale is equal to a temperature difference (∆t_C) of 100º in the Celsius scale. The significance of the numbers 100, 32 and 180 will be clear to you now.

(2) Temperature t_C in Celsius scale can be converted to temperature t_K in Kelvin scale (absolute scale) using the relation

t_C/100 = (t_K – 273.15) /100

so that t_C = t_K – 273.15 or, t_K = t_C + 273.15.

Often the above relation is written as t_K = t_C + 273 very nearly.

The above equation follows from the fact that the ice point and steam point of water are 273.15 K and 373.15 K respectively in the Kelvin scale.

A temperature difference of 1º in the Kelvin scale is equal to a temperature difference of 1º in the Celsius scale.

(3) The increase in length (∆ℓ) of a solid on raising its temperature by ∆T is given by

∆ℓ = α ℓ_o ∆T where α is the coefficient of linear expansion (linear expansivity) and ℓ_o is the original length.

(4) The increase in area (∆A) of a solid on raising its temperature by ∆T is given by

∆A = β A_o ∆T where β is the coefficient of area expansion (area expansivity) and A_o is the original area.

In the case of isotropic homogeneous solids β = 2α

(5) The increase in volume (∆V) of a solid on raising its temperature by ∆T is given by

∆V = γV_o ∆T where γ is the coefficient of volume expansion (volume expansivity) and V_o is the original volume.

In the case of isotropic homogeneous solids γ = 3α

(6) Specific heat capacity (specific heat) C of a substance is the quantity of heat absorbed or rejected by 1 kg of the substance to change the temperature by 1 K.

Therefore, the heat involved in changing the temperature of m kg of a substance through ∆T K (or ∆T º C) is mC ∆T where C is the specific heat of the substance.

(7) Molar specific heat of a substance is the quantity of heat absorbed or rejected by 1 mole of the substance to change the temperature by 1 K.

You know that in the case of gases there are two specific heats viz., specific heat at constant volume and specific heat at constant pressure. Similarly there are two molar specific heats viz., molar specific heat at constant volume and molar specific heat at constant pressure. In this context you will find a useful post here.

(8) Heat transfer takes place by three processes viz., conduction, convection and Radiation.

The quantity of heat Q conducted in a time t through a uniform rod of length L and area of cross section A when the ends of the rod are maintained at a temperature difference ∆T is given by

Q = KA (∆T/L) t where K is the thermal conductivity of the material of the rod

[The temperature gradient ∆T/L is often written as ∆T/∆x, replacing L by ∆x, which will be more appropriate in the case of heat conduction through slabs of thickness ∆x].

The rate of transfer of heat by conduction (time rate), H is given by

H = Q/t = KA ∆T/L

According to Stefan’s law the energy (E) radiated per second per unit surface area of a perfectly black body is directly proportional to the 4^th power of the absolute temperature of the body:

E α T⁴

Or, E = σT⁴ where σ is Stefan’s constant.

So if the temperature is doubled, the energy radiated from the body will become 16 times.

Newton’s law of cooling says that the rate of cooling (rate of loss of heat) of a body is directly proportional to the excess of temperature of the body over the surroundings:

dQ/dt α (T₂ – T₁) where dQ is the heat lost in a time dt when the temperatures of the body and the surroundings are respectively T₂ and T₁.

Note that the loss of heat mentioned in Newton’s law of cooling is due to all the three mechanisms viz., conduction, convection and radiation. Further, the excess temperature (T₂ – T₁) should be small.

In the next post we will discuss questions in this section. Meanwhile find a useful post here.