Question

A rocket is moving in a gravity space with a constant acceleration of \[2\,m/{s^2}\], along +x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 m/s relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 m/s from its right end relative to the rocket. The time in seconds when the two balls hit each other is

A. $2$
B. $3$
C. $4$
D. $5$

Answer 1

Hint: Determine the maximum distance covered by the ball thrown from the left end using a kinematic equation. Use the kinematic equation to express the distance moved by the ball thrown from the right end. Then determine the time taken by this ball to reach the total distance by rearranging this equation.

Formula used:
\[{v^2} = {u^2} + 2as\]
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
\[s = ut + \dfrac{1}{2}a{t^2}\]
Here, t is the time.

Complete step by step answer:
We assume the +x direction is the positive direction for velocity and acceleration while the –x direction is the negative direction for velocity and acceleration. We have given that the velocity of ball A is \[{u_A} = 0.2\,m/s\] and velocity of ball B is \[{u_B} = 0.3\,m/s\].

Let’s determine the maximum distance covered by the ball thrown from the left end of the rocket using the kinematic equation as follows,
\[v_B^2 = u_B^2 + 2{a_B}s\]
Here, \[{v_B}\] is the final velocity of ball B, \[{a_B}\] is the acceleration of ball B relative to the rocket and s is the maximum distance covered by ball B.

After the impact, the final velocity of the ball B becomes zero. Therefore, we can write the above equation as,
\[v_B^2 = u_B^2 + 2{a_B}s\]
\[ \Rightarrow s = \dfrac{{u_B^2}}{{2{a_B}}}\]

Substituting 0.3 m/s for \[{u_B}\] and \[2\,m/{s^2}\] for \[{a_B}\] in the above equation, we get,
\[s = \dfrac{{{{\left( {0.3} \right)}^2}}}{{2\left( 2 \right)}}\]
\[ \Rightarrow s = 0.022\,{\text{m}}\]
Therefore, the collision of the two balls will happen near to the left wall of the rocket.

So, we can assume that the distance travelled by the ball A is almost 4 m. Let’s determine the time required to reach 4 m for the ball A using the kinematic equation as follows,
\[ - s = - {u_A}t - \dfrac{1}{2}{a_A}{t^2}\]

The negative signs in the above equation imply that the motion of ball A is along the negative x-direction.

Substituting 4 m for s, 0.2 m/s for \[{u_A}\] and \[2\,m/{s^2}\] for \[{a_B}\] in the above equation, we get,
\[ - 4 = - \left( {0.2} \right)t - \dfrac{1}{2}\left( 2 \right){t^2}\]
\[ \Rightarrow {t^2} + 0.2t - 4 = 0\]

Solving the above equation using the formula, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get,
\[t = \dfrac{{ - 0.2 \pm \sqrt {{{\left( {0.2} \right)}^2} - 4\left( 1 \right)\left( { - 4} \right)} }}{2}\]
\[ \Rightarrow t = 1.9\,{\text{s}}\]
\[ \therefore t \approx 2\,{\text{sec}}\]

So, the correct answer is option A.

Note: While calculating the distance moved by ball B, we have neglected the negative sign for distance and have taken the magnitude. To solve these types of equations, the direction of motion is very important. You should always choose the positive and negative direction of motion of the body. In this question, we have chosen +x-direction as the positive direction and therefore, the velocity and acceleration are positive along +x-direction.