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Newton's Laws for System of Particles

Two particles...

Question

Two particles of equal mass have velocities $V_{1} = 4 ¯ i$ and $V_{2} = 4 ¯ j$ . First particle has an acceleration $a_{1} = (5 ¯ i + 5 ¯ j)^{m s^{- 2}}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Straight line

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Parabola

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Circle

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Ellipse

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Solution

The correct option is A Straight line
$X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$X_{1} = (4 t + \frac{1}{2} 5 t^{2})$
$X_{2} = 0$
$X_{c m} = \frac{4 t + \frac{1}{2} 5 t^{2}}{2}$
$Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$
$y_{1} = \frac{1}{2} 5 t^{2}$
$y_{2} = 4 t$
$Y_{c m} = \frac{m (\frac{5}{2} + 2) + m (4 t)}{m + m}$
$Y_{c m} = \frac{4 t + \frac{1}{2} 5 t^{2}}{2}$
$X_{c m} = Y_{c m}$
Hence centre of mass moves in a straight line

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Similar questions

{\to v}_{1} = (2^i) m/s

{\to v}_{2} = (2^j) m/s

{\to a}_{1} = (3^i + 3^j) {m/s}^{2}

{\to v}_{1} = (2^i) m/s

{\to v}_{2} = (2^j) m/s

{\to a}_{1} = (3^i + 3^j) {m/s}^{2}

{\to V}_{1} = 2^i

{\to V}_{2} = 3^j

{\to a}_{1} = 2^i + 2^j

2^i m s^{- 1} a n d 2^j m s^{- 1}

(^i +^j) m s^{- 1}

2^i {ms}^{- 1} and 2^j {ms}^{- 1}

(^i +^j) {ms}^{- 2}

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