First slide

Third law

Question

A stationary body of mass 3kg explodes into three equal pieces. Two of the pieces fly off at 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 $10^{- 5} s e c$ , the average force acting on the third piece in newton is

Moderate

A

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

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

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

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

Solution

$a c c o r d i n g t o c o n s e r v a t i o n o f l i n e a r m o m e n t u m 0 = \frac{m}{3} 2 \hat{i} + \frac{m}{3} 2 \hat{j} + \frac{m}{3} v \frac{m}{3} v = - (\frac{m}{3} 2 \hat{i} + \frac{m}{3} 2 \hat{j}) f o r c e o n t h i r d p i e c e i s \frac{m}{3} \frac{v}{t} = - (2 \hat{i} + 2 \hat{j}) 10^{5}$

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