Two particles of equal masses have velocities v1→=4i^ms−1and v2→=4j^ms−1. First particle has an acceleration a1→=(2i^+2j^)ms−2while 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

Solution:

Given,
$\begin{array}{l} \vec{v_{1}} = 4 \hat{i} {ms}^{- 1}, \vec{v_{2}} = 4 \hat{j} {ms}^{- 1} \\ \vec{a_{1}} = (2 \hat{i} + 2 \hat{j}) {ms}^{- 1}, \vec{a_{2}} = 0 {ms}^{- 2} \end{array}$
$∴$ Velocity of centre of mass,
$\begin{array}{l} v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{(v_{1} + v_{2}) m}{2 m} \\ [∵ m_{1} = m_{2} = m] \\ \vec{v_{C M}} = \frac{4 \hat{i} + 4 \hat{j}}{2} = 2 (\hat{i} + \hat{j}) {ms}^{- 1} \end{array}$
Similarly, acceleration of centre of mass,
$a_{CM} = \frac{a_{1} + a_{2}}{2} = \frac{2 \hat{i} + 2 \hat{j} + 0}{2} = (\hat{i} + \hat{j}) {ms}^{- 2}$
Since, from above values, it can be seen that $\vec{v_{CM}}$ is parallel to $\vec{a_{CM}}$ , so the path will be a straight line.

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us