Question

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 acceleration of the other particle is zero. The centre of mass of the two particles moves in a

विकल्प

• circle

• parabola

• ellipse

• straight line

Answer 1

straight line

Explanation:

Initial velocity of C.M in x-direction

`mu_x = (m_1 mu_(x1) + m_2 mu_(x_2))/(m_1 + m_2) = (m(2 + 0))/(2m)` = 1

Acceleration of C.M in x-direction

`a_x = (m_1a_(x_1) + m_2a_(x_2))/(m_1 + m_2) = (m(3 + 0))/(2m) = 3/2`

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

v_x = u_x + a_xt

∴ v_x = `1 + 3/2` t

Initial velocity of C.M in y-direction

u_y = `(m_1 u_(y_1) + m_2 u_(y_2))/(m_1 + m_2) = (m(0 + 2))/(2m)` = 1

Acceleration of C.M in y-direction

a_y = `(m_1 a_(y_1) + m_2 a_(y_2))/(m_1 + m_2) = (m(3 + 0))/(2m) = 3/2`

Now, v_y = u_y + a_yt

∴ v_y = `1 + 3/2` t

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