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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 velocity $2 \hat i \;m/s$ and the other with velocity $3 \hat j\; m/s$. If the explosion takes place in $10^{-5}s,$ the average force acting on the third piece in Newtons is

\[\begin {array} {1 1} (a)\;(2 \hat i+ 3 \hat j) \times 10^{-5} & \quad (b)\;-(2 \hat i+ 3 \hat j) \times 10^{+5} \\ (c)\;(3 \hat i- 2 \hat j) \times 10^{5} & \quad  (d)\;(2 \hat i- 3 \hat j) \times 10^{-5} \end {array}\]

1 Answer

Force= rate of change of momentum by momentum conservation
$1(2 \hat i)+1(3 \hat j)+1(\bar v)=0$
$\Delta p= -[2 \hat i+3 \hat j]$
$F= \large\frac{\Delta p}{\Delta t}$
$\quad=-[2 \hat i+3 \hat j] \times 10^5\;N$
answered Nov 23, 2013 by meena.p

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