Physics written test Questions & Answers on “System of Particles – Centre of Mass – 2”.
1. The centre of mass for an object always lies inside the object.
a) True
b) False
Answer: b
Clarification:The centre of mass for an object need not necessarily lie inside the object. A ring or a C-section element are classic examples for this.
2. For which of the following does the centre of mass lie outside the body?
a) Pen
b) Dice
c) Rectangular tile
d) Bangle
Answer: d
Clarification: The centre of mass of a bangle lies at its geometric centre which does not lie on or in the object. Hence the centre of mass of a bangle lies outside the body.
3. The centre of mass of the body always lies on the line of symmetry for an object with uniform density.
a) True
b) False
Answer: a
Clarification: The centre of mass of the body always lies on the line of symmetry for an object with uniform density if there exists a line of symmetry for that geometry. Because, by definition, symmetry implies an equivalence of mass on either side of the symmetrical axis if the density is uniform.
4. Numerous particles in a system are placed at a distance of “D” from the origin. The position of the centre of mass will definitely be _____
a) Equal to D
b) Lesser than or equal to D
c) Greater than D
d) Greater than or equal to D
Answer: b
Clarification: Since all objects are placed at the same distance, they are either placed in a spherical configuration, circular-arc configuration or a linear configuration. In all of the cases, the centre of mass of the system will lie within the geometry of the shape. Hence the centre of mass of the system is at a position lesser than or equal to D.
5. An asteroid enters the atmosphere of the earth and breaks into two pieces. One of the pieces is larger than the other. Which of the following is true considering the centre of mass of both the pieces together?
a) The centre of mass shifts horizontally towards the larger piece
b) The centre of mass shifts horizontally towards the smaller piece
c) The centre of mass remains in the same trajectory as before breaking apart of the asteroid
d) Depends on the velocity of the asteroid
Answer: c
Clarification: Since internal forces do not change the position of the centre of mass of a system as their net force is zero, the centre of mass remains in the same trajectory as before breaking apart of the asteroid. However, it is an influential factor for the trajectories of the two pieces after breaking apart.
6. Consider a system on which there are external forces acting. If the vector sum of all these external forces is zero, then the centre of mass _____
a) must not move
b) may move
c) may accelerate
d) must accelerate
Answer: b
Clarification: Since the vector addition of all external forces is zero, we know that the acceleration vector is zero for the resultant force vector as mass cannot be zero. Zero acceleration can imply zero or non-zero constant velocity. Thus, the centre of mass may move.
7. For an object with the centre of mass at the origin, the x-coordinates of particles of the object _____
a) may be all positive
b) may be all negative
c) may be all non-negative
d) cannot be predicted
Answer: c
Clarification: For an object to have the x-coordinate of its centre of mass, there should be particles on both the negative and positive coordinates of the plane for the sum to be zero. However, for 1-dimensional particles that all lie on the y-axis symmetric about the x-axis, the x-coordinate is zero – which is neither positive nor negative, i.e., non-negative.
8. Two bodies of masses 5 kg and 3 kg are moving towards each other at velocities of 3 m/s and 5 m/s, respectively. What is the velocity of their centre of mass?
a) 0.25 m/s towards 3 kg
b) Upredictable
c) The centre of mass is stationary
d) 0.5 m/s towards 5 kg
Answer: c
Clarification: The velocity of the centre of mass = (m1*v1 + m2*v2)/(m1 + m2)
m1 = 5kg
m2 =3 kg
v1 = 3 m/s
v2 = -5 m/s [opposite direction)
Velocity of centre of mass = (5*3 – 3*5)/(5+3)
= 0.
9. The velocities of three particles of masses 10 kg, 20 kg and 30 kg are 10i, 10j and 10k m/s, respectively. What is the velocity of their centre of mass?
a) (i + 2j + 3k) m/s
b) 10 (i + j + k) m/s
c) (10i + 20j + 30k) m/s
d) (3i + 2j + k) m/s
Answer: a
Clarification: The velocity of the centre of mass = (m1*v1 + m2*v2 + m3*v3)/(m1 + m2 + m3)
= (10*10i + 20*10j + 30*10k) / (10 + 20 + 30)
= (100i + 200j + 300k) / 100
= (i + 2j + 3k) m/s.
10. A system consists of 2 particles of the same mass. Let one particle be at rest and another particle have an acceleration of “2a”. What would be the acceleration of the centre of mass of the system?
a) a/4
b) a/2
c) a
d) 2a
Answer: c
Clarification:The acceleration of the centre of mass = (m1*a1 + m2*a2)/(m1 + m2)
m1 = m2 = m
a1 = 0
a2 = 2a
The acceleration of the centre of mass = (m*0 + m*2a) / (m + m)
= 2ma/2m
= a.
11. Two particles A and B, initially at rest, start moving towards each other under a mutual force of attraction. When the speed of A is “10 m/s” and the speed of B is “800 m/s”, what is the speed of their centre of mass?
a) 0
b) 0.8v
c) 80v
d) 8v
Answer: a
Clarification: Since both A and B are moving towards each other due to mutual force of attraction, we can infer that the particles of the system are in motion due to internal forces and hence the state of centre of mass will remain unchanged, i.e., at rest.
13. A uniform free rectangular steel plate is heated from 273 to 373 kelvin. The initial area of the plate is 15 sq. cm. What is the shift of the centre of mass?
a)0 cm
b) 1 cm
c) 2 cm
d) 3 cm
Answer: a
Clarification: Since the plate is free, the net force on the plate is zero. So, there will be no shift in the centre of mass of the plate, i.e., the shift in the centre of mass of the plate will be 0 cm.