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Question: Two particles of equal mass have velocities \[\overrightarrow {{V_1}} = 4\hat i\] and \[\overrightar...

Two particles of equal mass have velocities V1→=4i^\overrightarrow {{V_1}} = 4\hat i and V2→=4j^\overrightarrow {{V_2}} = 4\hat j. First particle has an acceleration a1=(5i+5j)ms−2{a_1} = \left( {5i + 5j} \right)m{s^{ - 2}} while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
A. Straight Line
B. Parabola
C. Circle
D. Ellipse

Explanation

Solution

Two particles are given with certain velocities. If the particles are moving with some velocity, then the center of mass also moves with some velocity. The velocity of the center of mass is equal to the sum of each particle's momentum divided by the total mass of the system. If the velocity of the system is changing with respect to time, then the system is accelerating. The acceleration of the center of mass is equal to the sum of the net force acting on each particle divided by the total mass of the system.

Formula Used:
The velocity of centre of mass is given by: Vcm→=mV1→+mV2→M\overrightarrow {{V_{cm}}} = \dfrac{{m\overrightarrow {{V_1}} + m\overrightarrow {{V_2}} }}{M}
The acceleration of centre of mass is given by: acm→=ma1→+ma2→M\overrightarrow {{a_{cm}}} = \dfrac{{m\overrightarrow {{a_1}} + m\overrightarrow {{a_2}} }}{M}
where, Vcm{V_{cm}} is the velocity of centre of mass, acm→\overrightarrow {{a_{cm}}} is the acceleration of centre of mass, mV1→m\overrightarrow {{V_1}} is the momentum of the first particle , mV2→m\overrightarrow {{V_2}} is the momentum of the second particle , MM is the total mass of the system, V1→\overrightarrow {{V_1}} is the velocity of the first particle, V2→\overrightarrow {{V_2}} is the velocity of the second particle, a1{a_1} is the acceleration of the first particle and a2{a_2} is the acceleration of the second particle.

Complete step by step answer:
Two particles are given having the same masses and moving with certain velocities and accelerations. The centre of mass of this system of particles will exist at the point where the whole mass of the system is concentrated. Let the mass of the given two particles be m1{m_1} and m2{m_2}. These two particles have equal masses. Therefore, m1=m2≈m{m_1} = {m_2} \approx m. The velocity of the first particle and the second particle is given as V1→=4i^\overrightarrow {{V_1}} = 4\hat i and V2→=4j^\overrightarrow {{V_2}} = 4\hat j respectively.

And, the acceleration is a1=(5i+5j)ms−2{a_1} = \left( {5i + 5j} \right)m{s^{ - 2}} and a2=0{a_2} = 0.Let the total mass of the system beMM. Then, M=m1+m2=m+m=2mM = {m_1} + {m_2} = m + m = 2m
The velocity of the centre of mass is equal to the sum of each particle's momentum divided by the total mass of the system. Momentum can be written as mass times the velocity of the particle.
Vcm→=mV1→+mV2→M\overrightarrow {{V_{cm}}} = \dfrac{{m\overrightarrow {{V_1}} + m\overrightarrow {{V_2}} }}{M}
Therefore,

\overrightarrow {{V_{cm}}} = \dfrac{{m\left( {4\hat i} \right) + m\left( {4\hat j} \right)}}{{2m}} \\\ \Rightarrow\overrightarrow {{V_{cm}}} = \dfrac{{4m\left( {\hat i + \hat j} \right)}}{{2m}} \\\ \Rightarrow\overrightarrow {{V_{cm}}} = 2\left( {\hat i + \hat j} \right)$$ The acceleration of center of mass $$\overrightarrow {{a_{cm}}} $$ is equal to the sum of the net force acting on each particle divided by the total mass of the system. The net force can also be written as the mass of each particle times its acceleration.

\overrightarrow {{a_{cm}}} = \dfrac{{m\overrightarrow {{a_1}} + m\overrightarrow {{a_2}} }}{M} \\
\Rightarrow\overrightarrow {{a_{cm}}} = \dfrac{{m\left( {5i + 5j} \right) + m\left( 0 \right)}}{{2m}} \\
\therefore\overrightarrow {{a_{cm}}} = \dfrac{5}{2}\left( {i + j} \right)If Ifiisthehorizontalcomponentandis the horizontal component andjisaverticalcomponent,thenis a vertical component, then\left( {i + j} \right)$$ is a straight line. Therefore, the centre of mass of the two particles moves in a straight line.

Hence, option A is the correct answer.

Note: The centre of mass of a system of two particles lies in between them on the line joining the particles. The location of the centre of mass of the body does not depend on the choice of the coordinate system. In the absence of external force, the velocity of the centre of the mass of the object remains constant. The nature of the motion of the centre of mass depends on the external force and is independent of the internal forces.