Question

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 $t_1$, while the other two land together at a time $t_2$, after the burst. If the initial velocity of first segment is
$v_1=\frac{g(t_2^2-t_1^2)}{n{t}_1+t_2}$, find $n$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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

Let the mass of the fragments are $m_1,m_2$ and $m_3$ and their vertical component of the velocity are $v_1,v_2$ and $v_3$ respectively.

By applying momentum conservation in vertical direction.
Initial momentum = Final momentum
$0=m_1{v}_1+m_2{v}_2+m_3{v}_3$

$\Rightarrow v_1+v_2+v_3=0$ (Given $m_1=m_2=m_3$)
As the second and third fragments land at the same time, the velocity $v_2$ and $v_3$ are the same.

$\Rightarrow v_1+v_2+v_2=0$ ($v_2=v_3$)
$\Rightarrow v_2=v_3=-\frac{v_1}{2}$

Let the rocket explodes at maximum height $h$ and for the first and second fragments, we have
$h=v_1{t}_1+\frac{g{t}_1^2}{2}$ ....(i)
$h=-\frac{v_1{t}_2}{2}+\frac{g{t}_2^2}{2}$ ...(ii)
On subtracting equation (ii) from (i).
$0=v_1{t}_1+\frac{v_1{t}_2}{2}+\frac{g{t}_1^2}{2}-\frac{g{t}_2^2}{2}$
$\Rightarrow 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)$
$\Rightarrow v_1=\frac{g(t_2^2-t_1^2)}{2t_1+t_2}$
$\Rightarrow n=2$

Hence, option (b) is correct.