Newton's Laws for System of Particles
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(Neglect friction everywhere and neglect the mass of the rope and pulley)
- g9
- g3
- g27
- g5
Internal forces are the forces which bodies exert on each other when the bodies are a part of the system.
- True
- False
- 7 m/s2
- 5 m/s2
- 3.77 m/s2
- 10.5 m/s2
- COM of bomb will remain at rest.
- COM of bomb will move along the surface with a constant velocity after the explosion.
- After explosion, all the fragments of bomb will fly in vertically upwards direction.
- Explosion of bomb will cause the centre of mass of bomb to accelerate.
- mw2L
- Mw2L2
- mw2L4
- zero
- −20 J
- −10 J
- −30 J
- −40 J
A pulley fixed to the ceiling carried a thread with bodies of masses m1 and m2 attached to its ends. The masses of the pulley and the thread are negligible and friction is absent. Find the acceleration of the center of mass of this system.
(g(m2+m1)2(m1−m2)2)
(m1+m2)g
(g(m1+m2)2(2m1m2))
((m2−m1)2g(m1+m2)2)
If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
Linear momentum of the system does not change in time.
Kinetic energy of the system does not change in time.
Angular momentum of the system does not change in time.
Potential energy of the system does not change in time.
- 20m
- 30m
- 10m
- 40m
- 2mL(m+M) and mL2(m+M)
- mL2(m+M) and 2mL(m+M)
- 2mL(m+M) and mL2(m+M)
- mL2(m+M) and mL(m+M)
A pulley fixed to the ceiling carried a thread with bodies of masses m1 and m2 attached to its ends. The masses of the pulley and the thread are negligible and friction is absent. Find the acceleration of the center of mass of this system.
(g(m2+m1)2(m1−m2)2)
(m1+m2)g
(g(m1+m2)2(2m1m2))
((m2−m1)2g(m1+m2)2)
- 25 N
- 20 N
- 50 N
- 5 N
- 2g
- g
- 2g3
- g2
Two blocks A and B each of equal masses 'm' are released from the top of a smooth fixed wedge as shown in the figure. Find the magnitude of the acceleration of the centre of mass of the two blocks.
g4
g2
g6
g
- 10 m/s, 14 m/s, 10 m/s
- 10 m/s, 4 m/s, 10 m/s
- 10 m/s, 4 m/s, -10 m/s
- 10 m/s, 10 m/s, 10 m/s
- 3y=x
- y=3x
- y=2x
- y=2x5