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**Quiz Description** :

**Name: **Center of Mass objective question answers test

**Subject: **Physics

**Topic: **Center of Mass

**Questions: **9 Objective type

**Time Allowed: **10 Minutes

**Important for: **11th & 12th School students, Engineering and Medical Entrance exams.

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- Question 1 of 9
##### 1. Question

1 pointsThe centre of mass of a system of particles does not depend upon

CorrectWe know that the resultant of all forces, on any system of particles, is zero. Therefore their centre of mass does not depend upon the forces acting on the particles.

IncorrectWe know that the resultant of all forces, on any system of particles, is zero. Therefore their centre of mass does not depend upon the forces acting on the particles.

- Question 2 of 9
##### 2. Question

1 pointsIf two bodies of masses 5Kg and 10Kg are moving towards each other without any external forces, then their centre of mass

CorrectWe have Mass of the first body (M

_{1}) = 5Kg and mass of the second body (M_{2}) = 20Kg. We know that as both the bodies are moving towards each other, without any external forces, therefore their centre of mass remains stationary.IncorrectWe have Mass of the first body (M

_{1}) = 5Kg and mass of the second body (M_{2}) = 20Kg. We know that as both the bodies are moving towards each other, without any external forces, therefore their centre of mass remains stationary. - Question 3 of 9
##### 3. Question

1 pointsA man hangs from a rope attached to a hot-air balloon. The mass of the man is greater than the mass of the balloon and its contents. The system is stationary in air. If the man now climbs up to the balloon using a rope, then centre of mass of the ‘man and balloon’ system will

CorrectWe know that when the man climbs up to the balloon using a rope, then no external force is involved.

Therefore centre of mass of the ‘man and balloon’ system will remain stationary.IncorrectWe know that when the man climbs up to the balloon using a rope, then no external force is involved.

Therefore centre of mass of the ‘man and balloon’ system will remain stationary. - Question 4 of 9
##### 4. Question

1 pointsA body A of mass M will falling vertically downwards under gravity breaks into two parts, a part B of mass M/3 and body C of mass 2 M/3. The centre of mass of the bodies B and C taken together

CorrectWe have Mass of body A (m

_{A}) = M; Mass of part B (m_{B}) = M/3 and mass of part C (m_{C}) = 2M/3.

We know that as the body A breaks into two parts while falling vertically downwards, under gravity, therefore there is no external force acting on the bodies. Thus centre of mass of bodies B and C taken together does not shift.IncorrectWe have Mass of body A (m

_{A}) = M; Mass of part B (m_{B}) = M/3 and mass of part C (m_{C}) = 2M/3.

We know that as the body A breaks into two parts while falling vertically downwards, under gravity, therefore there is no external force acting on the bodies. Thus centre of mass of bodies B and C taken together does not shift. - Question 5 of 9
##### 5. Question

1 pointsTwo identical particles move towards each other with velocity 2v and v respectively. The velocity of their centre of mass is

CorrectWe have, Mass of first particles (m

_{1}) = m; Mass of second particle (m_{2}) = m; Velocity of first particle (v_{1}) = 2v and velocity of second particle (v_{2}) = -v (Minus sing due to opposite direction).

We know that velocity of their centre of mass (v_{cm}) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+ m_{2}) = [(m × 2v) + m(-v)]/(m+m) = (2mv – mv)/ 2m = mv/2m = v/2.IncorrectWe have, Mass of first particles (m

_{1}) = m; Mass of second particle (m_{2}) = m; Velocity of first particle (v_{1}) = 2v and velocity of second particle (v_{2}) = -v (Minus sing due to opposite direction).

We know that velocity of their centre of mass (v_{cm}) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+ m_{2}) = [(m × 2v) + m(-v)]/(m+m) = (2mv – mv)/ 2m = mv/2m = v/2. - Question 6 of 9
##### 6. Question

1 pointsConsider a system of two particles having masses m

_{1}and m_{2}. If the particle of mass m_{1}is pushed towards the mass centre of particles through a distance d, by what distance would the particle of mass m_{2}move, so as to keep centre of the particles at the original position?CorrectGiven: Mass of first particle = m

_{1}; Mass of the second particle = m_{2}; Distance through which the first particle is pushed toward which the first particle is pushed towards mass centre (d_{1}) = d.

Let distance of first particle from mass centre = a and distance of second particle form mass centre = -b (Minus sign due to opposite side of second particle from mass centre)

We know that sum of moments of masses about mass centre is zero.

Therefore initial sum of moments of mass = m_{1}a + m_{2}(-b) = 0 or m_{1}a – m_{2}(b) = 0

Similarly, final sum of moments after the mass m_{1}is pushed = m_{1 }(a-d_{1}) + m_{2}(-b-d_{2}) = 0

d_{2}= m_{1}d / m_{2}

(Where d_{2}is the distance moved by second particle towards the mass centre and negative sign indicates direction of movement of second particle).IncorrectGiven: Mass of first particle = m

_{1}; Mass of the second particle = m_{2}; Distance through which the first particle is pushed toward which the first particle is pushed towards mass centre (d_{1}) = d.

Let distance of first particle from mass centre = a and distance of second particle form mass centre = -b (Minus sign due to opposite side of second particle from mass centre)

We know that sum of moments of masses about mass centre is zero.

Therefore initial sum of moments of mass = m_{1}a + m_{2}(-b) = 0 or m_{1}a – m_{2}(b) = 0

Similarly, final sum of moments after the mass m_{1}is pushed = m_{1 }(a-d_{1}) + m_{2}(-b-d_{2}) = 0

d_{2}= m_{1}d / m_{2}

(Where d_{2}is the distance moved by second particle towards the mass centre and negative sign indicates direction of movement of second particle). - Question 7 of 9
##### 7. Question

1 pointsTwo blocks of masses 10Kg and 4Kg are connected by a spring of negligible mass and placed on a friction less horizontal surface. An impulse gives a velocity of 14m-s

^{-1}to the heavier block in the direction of the lighter block. The velocity of the centre of mass isCorrectGiven that Mass of the first block (m

_{1}) = 10 Kg; Mass of second block (m_{2}) = 4 Kg; Velocity of first block due to impulse (v_{1}) = 14 m-s^{-1}and velocity of second block (v_{2}) = 0 (because the impulsive force acts only on the first block).

We know that the velocity of the centre of mass (v_{cm}) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+m_{2}) = [(10×14) + (4×0)]/(10+4) = 140/14 = 10 m-s^{-1}.IncorrectGiven that Mass of the first block (m

_{1}) = 10 Kg; Mass of second block (m_{2}) = 4 Kg; Velocity of first block due to impulse (v_{1}) = 14 m-s^{-1}and velocity of second block (v_{2}) = 0 (because the impulsive force acts only on the first block).

We know that the velocity of the centre of mass (v_{cm}) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+m_{2}) = [(10×14) + (4×0)]/(10+4) = 140/14 = 10 m-s^{-1}. - Question 8 of 9
##### 8. Question

1 pointsA system consists of two identical particles, one of the particle has an acceleration a and the other is at rest. The centre of mass has an acceleration of

CorrectGiven that Mass of the first particle (m

_{1}) = m; Mass of second particle (m_{2}) = m; Acceleration of first particle (a_{1}) = a and acceleration of second particle (a_{2}) = 0 (because it is at rest.)

We know that acceleration of the centre of mass (a) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+m_{2}) = (m×a+m×0)/(m+m) = ma/2m = a/2.IncorrectGiven that Mass of the first particle (m

_{1}) = m; Mass of second particle (m_{2}) = m; Acceleration of first particle (a_{1}) = a and acceleration of second particle (a_{2}) = 0 (because it is at rest.)

We know that acceleration of the centre of mass (a) = (m_{1}v_{1}+ m_{2}v_{2})/(m_{1}+m_{2}) = (m×a+m×0)/(m+m) = ma/2m = a/2. - Question 9 of 9
##### 9. Question

1 pointsThree identical metal balls, of same radius and mass are placed touching each other on a horizontal surface, such that an equilateral triangle is formed when centers of three balls are joined. The centre of mass of the system is located at

CorrectWe know that as all the metal balls are identical in mass and radius, therefore the center of mass of the system will be at the point of intersection of their medians.

IncorrectWe know that as all the metal balls are identical in mass and radius, therefore the center of mass of the system will be at the point of intersection of their medians.

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