Question

Two parallel rail tracks run north to south. The train $A$ moves north with a speed of $54 km h^{- 1}$ and the train $B$ moves south with a speed of $90 km h^{- 1}$ . A monkey running on the roof of the train $A$ against its motion (with its velocity of $18 km h^{- 1}$ with respect to the train $A$ ). What is the velocity of the monkey, $(a)$ as observed by an observer on the ground? $(b)$ as observed by an observer moving in train $B$ ?

Accepted Answer

Let's take the north direction as positive.

Then the velocity of the train $A$ is $54 km h^{- 1}$ towards the north,

${\vec{v}}_{A} = 54 km h^{- 1} = 54 \times \frac{5}{18} = 15 m s^{- 1}$ .

The velocity of the train $B$ is $90 km h^{- 1}$ towards the south,

${\vec{v}}_{B} = - 90 km h^{- 1} = - 90 \times \frac{5}{18} = - 25 m s^{- 1}$ .

The velocity of the monkey with respect to train $A$ is $18 km h^{- 1}$ towards the south,

${\vec{v}}_{MA} = - 18 km h^{- 1} = - 18 \times \frac{5}{18} = - 5 m s^{- 1}$ .

$(a)$ The velocity of a monkey as observed by an observer on the ground is $v_{monkey}$ ,

${\vec{v}}_{MA} = {\vec{v}}_{monkey} - {\vec{v}}_{A}$

$\Rightarrow {\vec{v}}_{monkey} = {\vec{v}}_{MA} + {\vec{v}}_{A}$

$\Rightarrow {\vec{v}}_{monkey} = (- 5) + 15 = 10 m s^{- 1}$ .

For an observer standing on the ground, the monkey will appear to move with $10 m s^{- 1}$ towards the north.

$(b)$ The velocity of a monkey as observed by an observer moving in train $B$ is,

${\vec{v}}_{M - B} = {\vec{v}}_{monkey} - {\vec{v}}_{B} = 10 - (- 25)$

$\Rightarrow {\vec{v}}_{M - B} = 35 m s^{- 1}$ .

For an observer moving in the train $B$ the monkey will appear to move with $35 m s^{- 1}$ towards the north.

B M Sharma Solutions for Chapter: Kinematics I, Exercise 38: ILLUSTRATION 4.35

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 38: ILLUSTRATION 4.35

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 38: ILLUSTRATION 4.35 with Hints & Solutions

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