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Relative Motion in 1D

A tram moves ...

Question

A tram moves with a velocity $v$ . A particle moving horizontally with a speed $u$ enters through corner $B$ perpendicular to $v$ . this particle strikes the diagonally opposite corner $A$ . If the dimensions of tram are $16 \times 2.4 \times 3.2 m^{3}$ , the value of $v$ is

3 m s^{- 1}

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20 m s^{- 1}

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15 m s^{- 1}

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30 m s^{- 1}

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Solution

The correct option is D $20 m s^{- 1}$
Consider the vertical motion of the particle after entering the compartment . Let it reach the floor in time t .
$3.2 = \frac{1}{2} (10) t^{2} \to t = 0.8 s e c$
Due to the velocity component $u$ it covers 2.4 m in this time .
Hence , $u = \frac{2.4}{0.8} = 3 m / s$
Also , the distance moved by the train in this time is the length of the train with speed $v$
Hence , $v = \frac{16}{0.8} = 20 m / s$

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A tram moves with a velocity v. A particle moving horizontally with a speed u enters through corner B perpendicular to v. this particle strikes the diagonally opposite corner A. If the dimensions of tram are 16×2.4×3.2m3, the value of v is

A tram moves with a velocity $v$ . A particle moving horizontally with a speed $u$ enters through corner $B$ perpendicular to $v$ . this particle strikes the diagonally opposite corner $A$ . If the dimensions of tram are $16 \times 2.4 \times 3.2 m^{3}$ , the value of $v$ is