Question 1:
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.
Question 2:
easyA 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
Question 3:
easyA 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
\[ a_{cm} = \frac{(m_{1}a_{1}+ m_{2}a_{2})}{(m_{1}+ m_{2})} \]
\[ a_{cm} = \frac{(m\times a+ m\times 0)}{(m+ m)}= \frac{a}{2} \]
Question 4:
moderateTwo 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.
Question 5:
easyTwo 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.
Question 6:
moderateA boat of length \( 12\text{ m} \) and mass \( 840\text{ kg} \) is floating without motion in still water. A man of mass \( 60\text{ kg} \) standing at one end of it walks to the other end of it and stops. The magnitude of displacement of the boat relative to the ground is:
Since no external horizontal force acts on the boat-man system, the center of mass does not move. The displacement of the boat is \( x = \frac{m L}{m + M} = \frac{60 \times 12}{60 + 840} = 0.8\text{ m} = 80\text{ cm} \).
Question 7:
moderateTwo bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:
The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).
Question 8:
easyConsider 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 \( a_{\text{cm}} = \frac{m(0) + m(a)}{2m} = \frac{a}{2} \).
Question 9:
easyRead 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.
Question 10:
easyAssertion (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).