A 6 kg box is travelling on ice at a speed of 9 m/s when a 12 kg packet is gently placed on it. The velocity will now be
From Principal of conservation of momentum
m 1 v 1 = (m 1 + m 2 ) v
6 × 9 = ( 6 + 12) × v ⇒ v = 3 m/s
A ball, moving with a speed v towards north, collides with an identical ball, moving with a speed v towards east. After collision the two balls stick together and move towards north-east. The speed of the combination is
Taking both the balls as one system
mv i + mv j= 2m× v
so, v= v/2 i + v/2 j
so, |v|= v/√2
A bomb of mass M at rest explodes into three pieces, two of which of mass M/4 each, are thrown off in perpendicular directions with speeds of 3 m/s and 4 m/s. The third piece is thrown off with a speed
As the bomb was initially at rest and no external force acts on it total momentum of the bomb should remain constant.
so, (m/4) 3 i +(m/4) 4 j + (m/2) v1 = 0
v1 = 3/2 i + 4/2 j
|V1|= 2.5 m/s
Two bodies of masses 2 kg and 4 kg, initially at rest, start moving towards each other due to mutual gravitational attraction. At a certain instant their speeds are 2 m/s and 1 m/s respectively. The speed of their centre of mass at that instant is
As Centre of mass was initially at rest and no external force act on it, Center of mass will remain at rest.
Note: Gravitational force here is internal force acting between two objects.
A boat of mass 40 kg is at rest. A dog of mass 4 kg starts moving in the boat with velocity of 10 m/s. What is the speed of boat ?
\[ v_{cm}=\frac{(m_{1}v_{1}+ m_{2}v_{2})}{(m_{1}+m_{2})} \]
As Center of mass will remain at rest \[ (m_{1}v_{1}+ m_{2}v_{2}) = 0 \]
4 × 10 + 40 × v = 0
v = -1 m/s
The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg be placed so that the centre of mass of the system of all particles shifts to the point (1, 1, 1)
Center of mass of 1 kg, 2 kg and 3 kg is at (3, 3, 3) so, 6kg can be assumed to present at (3, 3, 3).
Assume 4 kg is placed at (x,y,z) Center of mass is at (1,1,1)
1= (4x+6×3)/10
x=-2
similarly y=-2 and z=-2
COM(-2,-2,-2)
A system consists of two identical particles. One particle is at rest and the other particle has an acceleration a. The centre of mass of the system has an acceleration
\[ a_{cm} = \frac{(m_{1}a_{1}+ m_{2}a_{2})}{(m_{1}+ m_{2})} \]
\[ a_{cm} = \frac{(m\times a+ m\times 0)}{(m+ m)}= \frac{a}{2} \]
Two persons of masses 55 kg and 65 kg are at the opposite ends of a boat. The length of the boat is 3.0 m and its mass is 100 kg. The 55 kg man walks upto the 65 kg man and sits with him. If the boat is in still water, the centre of mass of the system shifts by
As the boat was initially at rest and no external force acts on it centre of mass will remain at rest.
Center of mass is closer to
Center of Mass is the weighted mean of masses so it is closer to heavier mass.
Centre of mass is the point which divides the line joining the masses in
Centre of mass is the point which divides the line joining the masses in reverse ratio of masses.