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

− ( 2 i ^ + 3 j ^ ) × 10 − 5

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

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

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

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

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

answer is B.

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Detailed Solution

Since the body explodes into three equal parts, therefore $m_{1} = m_{2} = m_{3} =$ $\frac{m}{3}$ =1 kg
Let the velocity of the third part be $\vec{v}$ .According to the principle of conservation of linear momentum,
Momentum of system before explosion = Momentum of system after explosion
Or, $m \vec{v} = m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}} + m_{3} \vec{v_{3}}$
Or, $3 \times 0 = 1 \times 2 \hat{i} + 1 \times 3 \hat{j} + 1 \times \vec{v}$
Or, $\vec{v} = - (2 \hat{i} + 3 \hat{j})$ m/s
Average force acting on the third particle is
$\vec{F} = \frac{m \vec{v}}{t} = \frac{- 1 \times (2 \hat{i} + 3 \hat{j})}{10^{- 5}} = - (2 \hat{i} + 3 \hat{j}) \times 10^{5} N$