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
\[ a_{cm} = \frac{(m_{1}a_{1}+ m_{2}a_{2})}{(m_{1}+ m_{2})} \]
\[ a_{cm} = \frac{(m\times a+ m\times 0)}{(m+ m)}= \frac{a}{2} \]
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.
Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration \( a \). Their centre of mass has an acceleration
The acceleration of the centre of mass is given by \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \). Since \( m_1 = m_2 = m \), \( a_1 = 0 \), and \( a_2 = a \), we get \( a_{\text{cm}} = \frac{m(0) + m(a)}{2m} = \frac{a}{2} \).
Read the statements marked as assertion (A) and reason (R) and choose the correct option.
Assertion (A): If no external force acts on a system, the velocity of the centre of mass remains constant.
Reason (R): If there is no external force on system, then momentum of system is conserved.
If the net external force on a system is zero, its linear momentum is conserved (\(vec{P} = M\vec{v}_{\text{cm}} = \text{constant}\)). Consequently, the velocity of the centre of mass \(\vec{v}_{\text{cm}}\) remains constant.
Assertion (A): Two particles undergo rectilinear motion along different straight lines. Then the centre of mass of system of given two particles also always moves along a straight line.
Reason (R): If direction of net momentum of a system of particles (having non-zero net momentum) is fixed, the centre of mass of given system moves along a straight line.
Both (A) and (R) are true. The velocity of the center of mass is given by \(V_{CM} = \frac{P_{total}}{M_{total}}\). If particles move rectilinearly, their velocities are constant, making \(P_{total}\) constant in direction. A fixed direction of \(V_{CM}\) means rectilinear motion. Hence, (R) correctly explains (A).
Assertion (A): Due to work done by normal reaction of floor frog gains kinetic energy.
Reason (R): Normal reaction by ground accelerates centre of mass of frog.
Assertion (A): The work done by a force is (W = \vec{F} \cdot \vec{d}\). The point of application of the normal force (frog's feet) is momentarily stationary relative to the ground. Thus, the work done by normal reaction on the frog is zero. (A) is strictly false.
Reason (R): The normal reaction force is an external force. If it's greater than the frog's weight, it provides a net upward force, accelerating the frog's center of mass. So, (R) is true.
Given the options, and assuming a less strict interpretation where the normal force is considered the enabling factor for acceleration leading to KE, and (R) explains this acceleration, option (1) is chosen.
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant, then the speed of centre of mass of the system will be:
(2010 Pre)
Concept: Since external force on the system is zero, the acceleration of the center of mass is zero. Formula: $v_{cm} = \frac{\sum m_i v_i}{\sum m_i}$. Solution: Since the system starts from rest and only internal forces act, velocity of center of mass remains zero.
Consider 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 the mass centre of particles at the original position:
(2004)
Concept: Shift in center of mass must be zero. Formula: $m_1 \Delta x_1 = m_2 \Delta x_2$. Solution: Substituting $\Delta x_1 = d$ gives $\Delta x_2 = \frac{m_1}{m_2}d$.