
Two objects P and Q initially at rest move towards each other under the mutual force of attraction. At the instant when the velocity of P is v and that of Q is 2v, then find the velocity of the center of mass of the system.
A. v
B. 3v
C. 2v
D. 0
Answer
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Hint: Before we proceed with the problem, it is important to know about the velocity of the centre of mass. The velocity of the centre of mass for a group of objects is defined as the sum of the product of each object's mass and velocity, divided by the total mass.
Formula Used:
The formula to find the velocity of the centre of mass is given by,
\[{V_{CM}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\]
Where,
\[{m_1},{m_2}\] are the masses of two particles.
\[{v_1},{v_2}\] are the velocities of two particles.
Complete step by step solution:
Consider two objects P and Q that are initially at rest and they move towards each other under the mutual force of attraction. At the instant the velocity of P is v and that of Q is 2v and we need to find the velocity of the centre of mass of the system. In order to find the velocity of the centre of mass, we have,
\[{V_{CM}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\]
Since particles are starting from rest, in the initial case:
\[{V_{CM}} = 0\] and \[{v_1} = {v_2} = 0\]
As they start moving towards each other under the influence of mutual attraction, taking both particles as system attractive forces become internal for the system which cannot make the centre of mass.
\[{\overrightarrow F _{ext}} = 0\]
Now, for the system of particles, applying Newton's second law of motion we have,
\[{\overrightarrow F _{ext}} = {\overrightarrow a _{com}}\]
Hence \[{\overrightarrow a _{com}} = 0\]
\[{\overrightarrow v _{com}} = \text{constant}\]
Therefore the velocity of the centre of mass will not change.
\[ \therefore {v_i} = {v_f} = 0\]
Therefore, the velocity of the centre of mass of the system at any instant will be Zero.
Hence, option D is the correct answer.
Note:When there is no external force acting on the system, the velocity of the centre of mass remains the same before and after the collision.
Formula Used:
The formula to find the velocity of the centre of mass is given by,
\[{V_{CM}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\]
Where,
\[{m_1},{m_2}\] are the masses of two particles.
\[{v_1},{v_2}\] are the velocities of two particles.
Complete step by step solution:
Consider two objects P and Q that are initially at rest and they move towards each other under the mutual force of attraction. At the instant the velocity of P is v and that of Q is 2v and we need to find the velocity of the centre of mass of the system. In order to find the velocity of the centre of mass, we have,
\[{V_{CM}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\]
Since particles are starting from rest, in the initial case:
\[{V_{CM}} = 0\] and \[{v_1} = {v_2} = 0\]
As they start moving towards each other under the influence of mutual attraction, taking both particles as system attractive forces become internal for the system which cannot make the centre of mass.
\[{\overrightarrow F _{ext}} = 0\]
Now, for the system of particles, applying Newton's second law of motion we have,
\[{\overrightarrow F _{ext}} = {\overrightarrow a _{com}}\]
Hence \[{\overrightarrow a _{com}} = 0\]
\[{\overrightarrow v _{com}} = \text{constant}\]
Therefore the velocity of the centre of mass will not change.
\[ \therefore {v_i} = {v_f} = 0\]
Therefore, the velocity of the centre of mass of the system at any instant will be Zero.
Hence, option D is the correct answer.
Note:When there is no external force acting on the system, the velocity of the centre of mass remains the same before and after the collision.
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