Concepts Of Physics MCQ Edition [Volume 1]

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

Three particles with masses of 1.0 kg , 2.0 kg and 3.0 kg are...

Three particles with masses of

1.0 \text{ kg}

2.0 \text{ kg}

and

3.0 \text{ kg}

are situated at the corners

A

B

and

C

respectively of an equilateral triangle

ABC

having an edge of

1 \text{ m}

. Find the coordinates of the centre of mass of the system, taking

A

as the origin and

B

along the positive

x

-axis.

Options

A. $\left(\dfrac{7}{12} \text{ m}, \dfrac{\sqrt{3}}{4} \text{ m}\right)$

Taking

A

as the origin and

B

along the positive

x

-axis, the coordinates of the vertices of the equilateral triangle of side

1 \text{ m}

are:

A = (0, 0)

B = (1, 0)

C = (1 \cos 60^{{}^{{}^{\circ}}}, 1 \sin 60^{{}^{{}^{\circ}}}) = \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)<br><br>The masses at these points are

m_A = 1.0

\text{ kg},

m_B = 2.0

\text{ kg}, and

m_C = 3.0

\text{ kg}.<br><br>The

-coordinate of the centre of mass is given by:<br>

x_{cm} =

\dfrac{m_A x_A + m_B x_B + m_C x_C}{m_A + m_B + m_C}<br><br>

x_{cm} =

\dfrac{1.0(0) + 2.0(1) + 3.0\left(\dfrac{1}{2}\right)}{1.0 + 2.0 + 3.0}<br><br>

x_{cm} =

\dfrac{0 + 2 + 1.5}{6} = \dfrac{3.5}{6} = \dfrac{7}{12} \text{ m}<br><br>The

-coordinate of the centre of mass is given by:<br>

y_{cm} =

\dfrac{m_A y_A + m_B y_B + m_C y_C}{m_A + m_B + m_C}<br><br>

y_{cm} =

\dfrac{1.0(0) + 2.0(0) + 3.0\left(\dfrac{\sqrt{3}}{2}\right)}{6}<br><br>

y_{cm} =

\dfrac{\dfrac{3\sqrt{3}}{2}}{6} = \dfrac{\sqrt{3}}{4} \text{ m}<br><br>The coordinates of the centre of mass are

\left(\dfrac{7}{12} \text{ m}, \dfrac{\sqrt{3}}{4} \text{ m}\right).