A rocket is projected straight up and explodes into three equally massive fragments just as it reaches the top of its flight. One of the fragments is observed to come straight down in a time $t$ , while the other two land at a time $t_{2,}$ after the burst. If the initial velocity of first segments is $v_{1} = \frac{g (t_{2}^{2} - t_{1}^{2})}{n t_{1} + t_{2}}$ , find $n$ .

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Solution

The velocity and momenta of the rocket are zero when it reaches the top of its flight.

$\Rightarrow$ After burst (by conservation of momentum)

$m_{1} v_{1} + m_{2} v_{2} + m_{3} v_{3} = 0$

As $m_{1} = m_{2} = m_{3}, v_{1} + v_{2} + v_{3} = 0$

As the second and third fragments land at the same time, the vertical components of $v_{2}$ and $v_{3}$ are the same.

As $v_{1}$ is vertically downward the vertical components of $v_{2}$ and $v_{3}$ are each - $\frac{v_{1}}{2}$

$\Rightarrow$ For the first and second fragments, we have

$h = v_{1} t_{1} + \frac{g t_{1}^{2}}{2}$

$h = - \frac{v_{1} t_{2}}{2} + \frac{g t_{2}^{2}}{2}$

on subtracting

$0 = v_{1} t_{1} + \frac{v_{1} t_{2}}{2} + \frac{g t_{1}^{2}}{2} - \frac{g t_{2}^{2}}{2}$

$0 = v_{1} (t_{1} + \frac{t_{2}}{2}) - \frac{g}{2} (t_{2}^{2} - t_{1}^{2})$

$\Rightarrow v_{1} (t_{1} + \frac{t_{2}}{2}) = \frac{g}{2} (t_{2}^{2} - t_{1}^{2})$

$\Rightarrow v_{1} = \frac{g (t_{2}^{2} - t_{1}^{2})}{2 t_{1} + t_{2}}$

$\Rightarrow$ n = 2

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