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

AP Physics B & C- Work, Energy & Power- Equations to be Remembered

You must remember the following points to make you strong in answering multiple choice questions involving work, energy and power:

(1) Work is done by a force if the point of application of the force is displaced. If the force vector F is constant and makes an angle θ with the displacement vector r, the work done (W) is given by

W = Fr cos θ

In vector notation this is given by

W = F.r

Thus work is a scalar quantity given by the scalar product of force and displacement.

If the force is not constant and is a function of the displacement (as for instance, in compressing or elongating a spring), the work done is given by W = ∫F.dr where dr is a small displacement for which the force F can be taken to be constant. The integration is to be carried out over the entire displacement. The work done by you in producing an elongation ‘x₁’ in a spring (which is not deformed initially) of force constant ‘k’ is given by

W = ₀∫^x1 F.dx = ₀∫^x1 kxdx = ½ kx₁²

[Note that the force with which you have to pull the spring to produce an extension ‘x’ is kx]

The work done in producing an extension ‘x’ in a spring is ½ kx².

If the elongation of a spring is to be increased from x₁ to x₂ the work required is ½ k(x₂² – x₁²).

If the force and the displacement are in the same direction and the magnitude of the force varies with displacement, the work done is given by the area under the force-displacement graph obtained by plotting the displacement on the x-axis and the force on the y-axis (fig.).

In the case of a spring the force-elongation (displacement) curve is a straight line as shown in the figure. The work done in producing an elongation ‘x’ in a spring is the area under the curve (which is the area of the triangle OAB) and is equal to ½ Fx = ½ (kx)x = ½ kx².

(2) A force is conservative if the work done by the force in moving an object depends only on the initial and final positions of the object and is independent of the path followed between these positions. The total work done by a conservative force on an object is zero when it moves round any closed path and returns to the initial position. Gravtational force and electrostatic force are examples of conservative forces.

If the work done by a force on an object moving between two positions depends on the path taken, the force is called non-conservative force. Friction and visous force are non-conservative forces.

(3) Kinetic energy (K) of a body of mass m moving with velocity v is given by

K = ½ mv²

Since momentum, p = mv, we have

K = p²/2m

[Remember this useful expression].

(4) Work-energy theorem states that the work done by a force acting on a body is equal to the change in the kinetic energy of the body. If W is the work done and K_i and K_f are the initial and final kinetic energies respectively, we have

W = K_f – K_i

(5) The gravitational potential energy of a body of mass m at a height h near the surface of the earth is mgh where g is the acceleration due to gravity at the place. Note that this energy is with respect to the reference level used for measuring the height and the value of h is negligible compared to the radius of the earth.

The change in the gravitational potential energy (∆U_g) of a mass m raised through a small height h can therefore be written as

∆U_g = mgh

You will find more details on gravitational potential energy in the posts dated 9^th, 12^th and 17^th May 2008 which you can access by clicking on the label ‘gravitation’ below this post.

(6) When a spring is compressed or elongated, the work done on the spring is stored as elastic potential energy in the spring. The elastic potential energy in the spring which is elongated or compressed through a distance ‘x’ is ½ kx².

(7) Law of conservation of energy states that energy can neither be created nor destroyed but can only be transferred from one form to another.

You should understand that the production of energy by annihilating mass in nuclear reactions does not violate the law of conservation of energy since mass itself is to be treated as a cocentrated form of energy in accordance with Einstein’s mass-energy relation, E = mc².

(8) Power is the time rate at which work is done or energy is transferred.

The average power P_av = W/t where W is the total work done in a time t.

Instantaneous power P = dW/dt where dW is the work done in a very small time dt at the instant t.

(9) The instantaneous power can be expressed as the scalar product of the force vector F and the velocity vector v as

P = F.v

This is easily obtained since dW = F.dr so that P = dW/dt = F.(dr/dt) = F.v

(10) Elastic collision is one in which momentum and kinetic energy are conserved. Inelastic collision is one in which kinetic energy is not conserved, but momentum is conserved.

You should note that momentum is conserved in elastic as well as inelastic collisions; but kinetic energy is conserved in the case of elastic collisions only.

In a completely inelastic collision the two colliding bodies move together after the collision.

If the mass m₁ moves with velocity v_1i in the positive x-direction and suffers a completely inelastic head on collision (collision in one dimension) with the mass m₂ at rest (fig.), the common velocity (v_f) with which the two masses will move is given by

m₁v_1i = (m₁ + m₂) v_f, on applying the law of conservation of momentum.

Therefore, v_f = m₁v_1i /(m₁ + m₂)

If the collision is elastic in the case of the above masses, we will have different

final velocities v_1f and v_2f for the two bodies after the collision, as given by the momentum conservationlaw:

m₁v_1i = m₁ v_1f + m₂ v_2f -------(i)

Since there are two unknowns (v_1f and v_2f) we require one more equation to solve for the unknowns and we have the kinetic energy equation,

½ m₁v_1i² =½ m₁ v_1f ²+ ½ m₂ v_2f ²

Or, m₁v_1i² = m₁ v_1f ²+ m₂ v_2f ²-------(ii)

Equations (i) and (ii) give

v_1f = (m₁– m₂) v_1i /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(m₁ + m₂)

If the the two bodies are of the same mass, the colliding mass m₁ will come to rest and will hand over its velocity to the other mass m₂ which was initially at rest. If m₂>> m₁, the velocity of m₁ will get reversed and m₂ will continue to remain at rest.

If m₂ also has an initial velocity v_2i you can incorporate its initial momentum and kinetic energy in the above equations and solve for the final velocities of the bodies. You will then get

v_1f = [2 m₂v_2i + (m₁– m₂) v_1i] /(m₁ + m₂) and

v_2f = [2 m₁v_1i + (m₂– m₁) v_2i] /(m₁ + m₂)

Try to derive these equations as an exercise.

In the next post we will discuss questions in this section. Meanwhile go through some useful multiple choice questions (with solution) here at physicsplus.

AP Physics B & C – Practice Questions (MCQ) on Conservation of Linear Momentum

In the post dted 24^th September 2008 we had discussed the essential points to be remembered under Newton’s laws of motion. It was then stated that we will discuss the conservation of momentum separately. Today we will discuss some typical multiple choice questions involving the law of conservation of linear momentum.

As you know, Newton’s second law and third law lead to the law of conservation of momentum: The total momentum of an isolated system of interacting particles is conserved.

By isolated system you mean that the system is not acted on by external forces. There may be internal forces by which pairs of particles in the system interact thereby changing their individual momentum. But since the mutual forces for each pair are equal and opposite, the changes in momentum cancel in pairs and the total momentum remains unchanged. A typical case usually used to illustrate the law of conservation of momentum is the recoil of a gun when a bullet is fired. Initially, the bullet and the gun are at rest and hence the momentum of the system (of bullet and gun) is zero. On firing, the bullet moves forward because of the internal forces developed in the system. Since the final momentum of the system has to be zero, the forward momentum of the bullet has to be equal in magnitude to the backward (recoil) momentum of the gun at the instant of firing. If m and M represent the masses of the bullet and the gun and v and V the magnitudes of their velocities respectively, we have

mv + MV = 0

This gives the recoil velocity of the gun: V = – mv/M

Now consider the following questions:

(1) A hose-pipe delivering water at the rate of 2.5 litre per second is held horizontally by a fire man. If the rate of flow of water is increased to 5 litre per second by increasing the speed of water, the fire man has to

(a) push forward with the same force as before

(b) push forward twice as hard

(c) push forward four times as hard

(d) push forward with half the force

(e) exert no horizontal force in both cases

Since the water jet has a forward momentum, the hose-pipe should have a backward momentum of the same magnitude (similar to the recoil momentum of a gun firing a bullet). The fire man has to exert a force in the direction of the water jet to hold the hose-pipe in position. When the rate of flow is doubled by increasing the speed of warer, the momentum imparted to the water jet per second is made four times sine the mass and velocity are doubled. Therefore, the fire man has to push forward four times as hard.

(2) A horizontal force F acts for a small time ∆t on an object of mass m₁ initially at rest on a smooth horizontal surface (fig.). After the force has ceased, the object gets attached to another stationary object of mass m₂ . What is the common velocity of the two objects moving together?

(a) Zero

(b) F∆t m₁/(m₁+ m₂)

(c) Fm₁/(m₁+ m₂)

(d) F∆t m₂/(m₁+ m₂)

(e) F∆t/(m₁+ m₂)

The impulse received by the object of mass m₁ is F∆t. The momentum of this object therefore becomes F∆t. [Note that impulse is equal to the change in momentum. Since the mass m₁ is initially at rest, its momentum becomes F∆t].

The initial momentum of the system (made of m₁ and m₂) is thus F∆t. The final momentum also is F∆t since the force has ceased. Therefore, we have

F∆t = (m₁+ m₂)v where ‘v’ is the common velocity of m₁ and m₂. This gives

v = F∆t/(m₁+ m₂)

(3) A shell fired from a cannon with velocity v at an angle of 45º with the horizontal explodes into two parts of equal mass at the highest point of its trajectory. One of the pieces retraces its path. What will be the speed of the other piece immediately after the explosion?

(a) 3v/2

(b) 3v/√2

(c) v/2

(d) v/√2

(e) 2v

At the highest point, the projectile (shell) has horizontal momentum only which is mvcosθ where m is the mass of the shell and θ is the angle of projection (45º here). Thus just before the explosion, the shell has momentum equal to mvcos45º = mv/√2

Since the total momentum of the two fragments immediately after the explosion must be equal to this, we have

mv/√2 = – (m/2) vcos 45º + (m/2)v’ where v’ is the speed of the second piece.

Note that the first piece retraces its path and that’s why we have written its velocity as – v cos 45º.

This gives

v/√2 = – v/(2√2) + v’/2 from which v’ = 3v/√2

(4) An object initially at rest explodes into 3 fragments with momenta 30 kg ms^–1, 40 kg ms^–1 and 50 kg ms^–1. What is the angle between the directions of flight of the fragments carrying momenta 30 kg ms^–1 and 40 kg ms^–1?

(a) Zero

(b) 30º

(c) 45º

(d) 90º

(e) 120º

Since the total momentum of the fragments after the explosion has to be zero, the momentum 50 kg ms^–1 of the 3^rd fragment has to be equal and opposite to the resultant of the momenta of the other two fragments. In other words, the resultant of the momenta 30 kg ms^–1and 40 kg ms^–1 must be equal to 50 kg ms^–1(represented by the vector OR in fig). Evidently, the angle between the momentum vectors OA and OB is 90º.

If you want you may write

50² = 30² + 40² + 2×30×40 cosθ

From this θ = 90º.

You will find a couple of solved questions in this section at physicsplus here as well as here