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
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
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
\[ a_{cm} = \frac{(m_{1}a_{1}+ m_{2}a_{2})}{(m_{1}+ m_{2})} \]
\[ a_{cm} = \frac{(m\times a+ m\times 0)}{(m+ m)}= \frac{a}{2} \]
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.
Masses 2 kg and 5kg are located at origin and (7,0) respectively. Centre of mass of the arrangement is located at
Centre of mass is the point which divides the line joining the masses in reverse ratio of masses.
Here ratio of mass is 2:5 so distance from 2kg mass located at origin will be 5/7 of 7 cm so
\[ x_{cm}= \frac{2(0)+5(7)}{5+2}= 5 \]
Two particles \(A\) and \(B\) initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of \(A\) is \(v\) and speed of \(B\) is \(3v\), the speed of centre of mass is
Since there is no external force acting on the system, the acceleration of the centre of mass is zero. Since the system was initially at rest, the velocity of the centre of mass remains zero.
A bullet of mass \(m\) hits a block of mass \(M\) elastically. The transfer of energy is the maximum, when
In a one-dimensional elastic collision between a moving body of mass \(m\) and a stationary body of mass \(M\), maximum transfer of kinetic energy occurs when the masses are equal (\(m = M\)).