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Relative Motion in 1D

PSLV launches...

Question

PSLV launches a rocket in a gravity free space with constant acceleration $5 {m/s}^{2}$ along +x direction (see figure). The length of the chamber inside the rocket is $6 m$ . At time $t = 0$ , a ball is thrown from the left end of the chamber in +x direction with a speed $0.7 m/s$ relative to the rocket. At the same time, another ball is thrown in opposite direction to the previous ball with a speed $0.3 m/s$ relative to the rocket. The time in seconds when the two balls hit each other is

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Solution

Given that
$u_{A} = 0.7 m/s; u_{B} = 0.3 m/s$
$a = 5 {m/s}^{2}; l = 6 m$

Let $t$ be the time at which both balls hit each other.

At time $t$ ,
Relative distance covered by ball A $S_{r e l} = 6 m$
Relative velocity of ball A w.r.t ball b $u_{r e l} = 0.7 + 0.3 = 1 m/s$
Relative acceleration, $a_{r e l} = 0$
[since both the balls are inside rocket chamber, moving with the same acceleration]

Using formula,
$S_{r e} = u_{r e l} \times t$
$\Rightarrow t = \frac{S_{r e l}}{u_{r e l}} - - - - - (1)$
$\Rightarrow t = \frac{6}{1} = 6 s$
is the time when the two balls hit each other.

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