Question

A train is moving with a constant speed of 10 m/s in a circle of radius $\frac{16}{\pi }$ m. The plane of the circle lies in horizontal. x-y plane. At time t=0, train is at point $P(\frac{16}{\pi },0,0)$ and moving in counter-clockwise direction. At this instant a stone is thrown from the train with speed 10 m/s relative to train towards negative x-axis at an angle of ${37}^{\circ }$ with vertical z-axis.

Find the velocity of particle relative to train at the highest point of its trajectory.

• A. $(-3\hat{i}+4\hat{j})m{s}^{-1}$
• B. $(-3\hat{i}+5\hat{j})m{s}^{-1}$
• C. $(-6\hat{i}+8\hat{j})m{s}^{-1}$
• D. $(4\hat{i}+10\hat{j})m{s}^{-1}$

Answer 1

Answer: (D)

$(4\hat{i}+10\hat{j})m{s}^{-1}$

The stone has three components of velocity.

The velocity of stone $v_s=v_{s/train}+v_{train}$
$v_{s/train}=-10cos{37}^{\circ }\hat{i}+10sin{37}^{\circ }\hat{k}m/s$ ;$v_{train}=10\hat{j}$
$v_s=-10cos{37}^{\circ }\hat{i}+10\hat{j}+10sin{37}^{\circ }\hat{k}m/s$

$\overrightarrow{v_s}=-6\hat{i}+10\hat{j}+8\hat{k}$
At the highest point of projectile the vertical velocity will be zero

Velocity along z-direction is 8m/s, thus, time to reach max height is

$T=\frac{v_z}{g}=\frac{8}{10}=0.8s$

Velocities along x and y directions won't change as there is no aceleretion in these directions.At max height velocity of the particle is

$\overrightarrow{v_s}=-6\hat{i}+10\hat{j}$

Till this instant distance travelled by the train is $=10\times 0.8=8m$

$\frac{8}{R}=\frac{8}{\frac{16}{\pi }}=\frac{\pi }{2}\Rightarrow$ angular displacement of the train is ${90}^0$

$\overrightarrow{v_t}=-10\hat{i}\overrightarrow{v_{p/t}}=\overrightarrow{v_s}-\overrightarrow{v_t}=4\hat{i}+10\hat{j}$