Two particles of equal masses have velocity vecv_1 = 2hati m/s and vecv_2 = 2hatj m/s. The first particle has an acceleration veca_1 = (3hati + 3hatj) m/s2 while the -

Question

Two particles of equal masses have velocity ${\vec{v}}_{1} = 2 \hat{i}$ m/s and ${\vec{v}}_{2} = 2 \hat{j}$ m/s. The first particle has an acceleration ${\vec{a}}_{1} = (3 \hat{i} + 3 \hat{j})$ m/s² while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a

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circle
parabola
ellipse
straight line

MCQ

Solution

straight line

Explanation:

Initial velocity of C.M in x-direction

$μ_{x} = \frac{m_{1} μ_{x 1} + m_{2} μ_{x_{2}}}{m_{1} + m_{2}} = \frac{m (2 + 0)}{2 m}$ = 1

Acceleration of C.M in x-direction

$a_{x} = \frac{m_{1} a_{x_{1}} + m_{2} a_{x_{2}}}{m_{1} + m_{2}} = \frac{m (3 + 0)}{2 m} = \frac{3}{2}$

From, v = u + at, final velocity of C.M in x-direction is

v_x = u_x + a_xt

∴ v_x = $1 + \frac{3}{2}$ t

Initial velocity of C.M in y-direction

u_y = $\frac{m_{1} u_{y_{1}} + m_{2} u_{y_{2}}}{m_{1} + m_{2}} = \frac{m (0 + 2)}{2 m}$ = 1

Acceleration of C.M in y-direction

a_y = $\frac{m_{1} a_{y_{1}} + m_{2} a_{y_{2}}}{m_{1} + m_{2}} = \frac{m (3 + 0)}{2 m} = \frac{3}{2}$

Now, v_y = u_y + a_yt

∴ v_y = $1 + \frac{3}{2}$ t

C.M must be travelling in a straight line because it has the same velocity in both x and y directions.

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Linear Momentum of a System of Particles

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Two particles of equal masses have velocity vecv_1 = 2hati m/s and vecv_2 = 2hatj m/s. The first particle has an acceleration veca_1 = (3hati + 3hatj) m/s2 while the -

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Two particles of equal masses have velocity vecv_1 = 2hati m/s and vecv_2 = 2hatj m/s. The first particle has an acceleration veca_1 = (3hati + 3hatj) m/s2 while the -

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