Two particles of equal masses have initial velocities 2- ms-1 and 2- ms-1- First particle has a constant acceleration of - ms-2 while the acceleration of the second particle is always zero- The centre of mass of the two particles moves in a-

Initial velocity of centre of mass (v_i)_com=m× 2î+m × 2ĵ2m=(î+ĵ) The acceleration of centre of mass is, a_com=m(î+ĵ)+m× 02m= ( î+ĵ2 ) Now, we see that v_i_com ...

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Byju's Answer

Standard XII

Physics

Magnitude and Direction of Acceleration

Two particles...

Question

Two particles of equal masses have initial velocities $2^i {ms}^{- 1} and 2^j {ms}^{- 1}$ . First particle has a constant acceleration of $(^i +^j) {ms}^{- 2}$ while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in a,

Circle

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parabola

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Ellipse

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Straight line

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Solution

The correct option is D Straight line
Initial velocity of centre of mass
$(_{i})_{c o m} = \frac{m \times 2^i + m \times 2^j}{2 m} = (^i +^j)$

The acceleration of centre of mass is,
${\to a}_{c o m} = \frac{m (^i +^j) + m \times 0}{2 m} = (\frac{^i +^j}{2})$
Now, we see that ${_{i}}_{c o m} \times {\to a}_{c o m} = 0$
So, velocity and acceleration have the same direction, and therefore, the path must be straight line.

Hence, $(D)$ is the correct answer.

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Two particles of equal masses have initial velocities 2^i ms−1 and 2^j ms−1. First particle has a constant acceleration of (^i+^j) ms−2 while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in a,

Two particles of equal masses have initial velocities $2^i {ms}^{- 1} and 2^j {ms}^{- 1}$ . First particle has a constant acceleration of $(^i +^j) {ms}^{- 2}$ while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in a,