First slide

Collisions

Question

A stationery 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 $3 \hat{i}$ m/s.If the explosion takes place in $10^{- 5}$ s, the average force acting on the third piece in newtons 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

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 $m v = m_{1} v_{1} + m_{2} v_{2} + m_{3} v_{3}$
Or $3 \times 0 = 1 \times 2 \hat{i} + 1 x 3 \hat{j} + 1 \times \vec{v}$
Or v= - ( $2 \hat{i} + 3 \hat{j}$ )m/s
Average force acting on the third particle is
$\vec{F} = \frac{m \vec{v}}{t} = \frac{- 1 \times (2 \hat{i} + 3 \hat{j})}{10^{- 5}} = - (2 \hat{i} + 3 \hat{j}) \times 10^{5} N$

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