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QJEEAdv14Ph1f8d: JEE Advanced 2014 Physics - Class 11 JEEAdv14Ph1f8d Solution: A rocket is moving in a gravity free space with a constant…

A rocket is moving in a gravity free space with a constant acceleration of 2 ms–2 along + x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 ms–1 relative to the rocket. At the same time, another ball is thrown in - x direction with a speed of 0.2 ms–1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is JEE Advanced 2014 Paper 1 Offline Physics - Motion Question 9 English

Step-by-Step Verified Solution

JEE Advanced 2014 Paper 1 Offline Physics - Motion Question 9 English Explanation

Consider the motion of two balls with respect to rocket.

Distance travelled by ball A from left end of the chamber is

=u22a=(0.3)22×2=0.094≈0.02= {{{u^2}} \over {2a}} = {{{{(0.3)}^2}} \over {2 \times 2}} = {{0.09} \over 4} \approx 0.02 m

So, collision of two balls will take place very near to left end of the chamber.

For ball B

S=ut12at2S = ut{1 \over 2}a{t^2}

−4=−0.2×t−12×2×t2- 4 = - 0.2 \times t - {1 \over 2} \times 2 \times {t^2}

t2+0.2t−4=0{t^2} + 0.2t - 4 = 0

t=−0.2±(0.2)2−4(1)(−4)2=−0.2±0.04+162t = {{ - 0.2 \pm \sqrt {{{(0.2)}^2} - 4(1)( - 4)} } \over 2} = {{ - 0.2 \pm \sqrt {0.04 + 16} } \over 2}

t = 1.9 s, −- 2.1 s

Since t can't be negative

∴\therefore t = 1.9 s

Nearest integer is 2 s.

Also, from

S=ut+12at2S = ut + {1 \over 2}a{t^2}

SA=0.3t+12(−2)t2=0.3t−t2{S_A} = 0.3t + {1 \over 2}( - 2){t^2} = 0.3t - {t^2}

SB=0.2t+12(2)t2=0.2t+t2{S_B} = 0.2t + {1 \over 2}(2){t^2} = 0.2t + {t^2}

∵\because SA+SB=4{S_A} + {S_B} = 4

⇒\Rightarrow 0.5 t = 4 or t = 8 s

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