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Normal Force

A particle of...

Question

A particle of mass $1 g$ having velocity $(3^i - 2^j)$ has a glued impact with another particle of
mass $2 g$ and velocity $(4^i + 6^k)$ . Magnitude of velocity of the formed particle is

5.46 {m s}^{- 1}

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6.44 {m s}^{- 1}

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4.64 {m s}^{- 1}

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Solution

The correct option is A $5.46 {m s}^{- 1}$
The linear momentum should be conserved so $m_{1} V_{1} + m_{2} V_{2} = (m_{1} + m_{2}) V$
or $1 (3^i - 2^j) + 2 (4^i + 6^k) = (1 + 2) V$
so the velocity of formed particle will be $V = \frac{1}{3} (11^i - 2^j + 12^k)$
whose magnitude is $V = \frac{1}{3} \sqrt{121 + 4 + 144} = 5.46 m / s$

Suggest Corrections

Similar questions

1 g

_{1} = 3^i - 2^j m s^{- 1}

2 g

_{2} = 4^j - 6^k m s^{- 1}

1

_{2} = (3^i - 2^j)

m s^{- 1}

2

_{2} = (4^j - 6^k) m s^{- 1}

m

(3^i + 2^j) m s^{- 1}

M

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

1 kg

{\to v}_{1} = (2 t)^i

2 kg

{\to v}_{2} = (t^{2})^j

\begin{matrix} Column I & Column II (a) & Net force on centre of mass at 2 s & (p) & \frac{20}{9} unit (b) & Velocity of centre of mass at 2 s & (q) & \sqrt{68} unit (c) & Displacement of centre of mass in 2 s & (r) & \frac{\sqrt{80}}{3} unit (s) & None \end{matrix}

(2 m, 4 m, 6 m)

(6 m, 2 m, 8 m)

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

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

t = 0

t = 1 s

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