First slide

Law of conservation of momentum and it's applications

Question

A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other, one with a velocity $2 \hat{i} m / s$ and the other with a velocity $3 \hat{j} m / s$ . If the explosion takes place in 10^-5 sec, the average force acting on the third piece in newton is:

Moderate

A

$(2 \hat{i} + 3 \hat{j}) \times 10^{- 5}$
B

$- (2 \hat{i} + 3 \hat{j}) \times 10^{5}$
C

$(3 \hat{j} - 2 \hat{i}) \times 10^{5}$
D

$(2 \hat{i} - 3 \hat{j}) \times 10^{- 5}$

Solution

Mass of each piece (m) = 1 kg. Initial momentum = 0. Final momentum $= p_{1} + p_{2} + p_{3}$ .
From the principle of conservation of momentum, we have $p_{1} + p_{2} + p_{3} = 0$
or
$\begin{array}{l} p_{3} = - (p_{1} + p_{2}) = - ({mv}_{1} + {mv}_{2}) \\ = - m (v_{1} + v_{2}) \\ = - 1 kg \times (2 \hat{i} + 3 \hat{j}) {ms}^{- 1} \end{array} = - (2 \hat{i} + 3 \hat{j}) {kgms}^{- 1}$
Force
$\begin{array}{r} F = \frac{p_{3}}{t} = \frac{- (2 \hat{i} + 3 \hat{j}) {kgms}^{- 1}}{10^{- 5} s} \\ = - (2 \hat{i} + 3 \hat{j}) \times 10^{5} newton \end{array}$

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