Question

Two stones are thrown simultaneously from the same point with equal speed \[{{V}_{0}}\]at an angle \[30{}^\circ \] and \[60{}^\circ \] respectively with the positive axis. The velocity with which the stones move relative to each other in air is.
\[A)2{{v}_{0}}\cos 15{}^\circ \]
\[B)2{{v}_{0}}\sin 15{}^\circ \]
\[\text{C)2}{{\text{v}}_{\text{0}}}\text{tan15 }\!\!{}^\circ\!\!\text{ }\]
\[\text{D)2}{{\text{v}}_{\text{0}}}\text{cot15 }\!\!{}^\circ\!\!\text{ }\]

Answer 1

Hint: Here, two stones are thrown at different angles. The stones are moving in the same direction. Then, the relative velocity is the difference of the velocities of stone 1 and 2. This resultant velocity can be calculated by finding the x component of both stones velocity, and y component of both stones velocity. And then using the x and y component of the resultant velocity, we can find its magnitude.

Formula used:
\[{{v}_{x}}=v\text{ cos}\theta \]
\[{{v}_{y}}=v\text{ sin}\theta \]
\[\left| {\vec{v}} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}\]
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
\[\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B\]
\[\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A\sin B\]
\[\left( 1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2} \right)\]

Complete answer:
Given,
Stone 1 is thrown at an angle of\[30{}^\circ \], with velocity\[{{v}_{0}}\]. Then,
X-axis component of the velocity is,\[{{\left( {{v}_{x}} \right)}_{1}}={{v}_{0}}\cos \theta ={{v}_{0}}\cos 30{}^\circ \]
Y-axis component of the velocity is,\[{{\left( {{v}_{y}} \right)}_{1}}={{v}_{0}}\sin \theta ={{v}_{0}}\sin 30{}^\circ \]
Stone 2 is thrown at an angle of \[60{}^\circ \], with velocity\[{{v}_{0}}\] Then,
 X-axis component of the velocity is, \[{{\left( {{v}_{x}} \right)}_{2}}={{v}_{0}}\cos \theta ={{v}_{0}}\cos 60{}^\circ \]
Y-axis component of the velocity is, \[{{\left( {{v}_{y}} \right)}_{2}}={{v}_{0}}\sin \theta ={{v}_{0}}\sin 60{}^\circ \]
Since the two stones move in the same direction, the relative velocity is the difference of the two velocities.
Then,
Relative velocity of two stones with respect to x axis,
\[{{\left( \text{Relative velocity} \right)}_{x}}\text{ = }{{\text{v}}_{0}}\cos 60-{{\text{v}}_{0}}\cos 30={{\text{v}}_{0}}\left( \cos 60-\cos 30 \right)\]
Relative velocity of two stones with respect to y axis,
\[{{\left( \text{Relative velocity} \right)}_{y}}\text{ = }{{\text{v}}_{0}}\sin 60-{{\text{v}}_{0}}\sin 30={{\text{v}}_{0}}\left( \sin 60-\sin 30 \right)\]
Then,
Resultant relative velocity,
\[{{\left( Velocity \right)}_{\text{resultant}}}\text{= }{{\text{v}}_{0}}\left( \cos 60-\cos 30 \right)+{{\text{v}}_{0}}\left( \sin 60-\sin 30 \right)\]
We have,
Magnitude of a vector \[\vec{v}\] is given by,
\[\left| {\vec{v}} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}\]
Where,
\[{{v}_{x}}=v\text{ cos}\theta \], \[{{v}_{y}}=v\text{ sin}\theta \] and, \[\theta \]is the angle between \[{{\text{v}}_{\text{x}}}\text{ and v}\]
Then,
\[\left| {{\left( Velocity \right)}_{\text{resultant}}} \right|=\sqrt{{{v}_{0}}^{2}{{\left( \cos 60-\cos 30 \right)}^{2}}+{{v}_{0}}^{2}{{\left( \sin 60-\sin 30 \right)}^{2}}}\]
\[={{v}_{0}}\sqrt{\text{co}{{\text{s}}^{\text{2}}}\text{60+si}{{\text{n}}^{\text{2}}}\text{60+co}{{\text{s}}^{\text{2}}}\text{30+si}{{\text{n}}^{\text{2}}}\text{30-2cos60cos30-2sin60sin30}}\]
We have,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\],
\[\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B\]
\[\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A\sin B\]
Then,
\[\left| {{\left( Velocity \right)}_{\text{resultant}}} \right|={{v}_{0}}\sqrt{1+1-2\left( \cos \left( 60-30 \right) \right)}={{v}_{0}}\sqrt{2-2\cos 30}\]
\[={{v}_{0}}\sqrt{2\left( 1-\cos 30 \right)}\]
\[={{v}_{0}}\sqrt{2\times 2{{\sin }^{2}}15}\] \[\left( 1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2} \right)\]
\[={{v}_{0}}\sqrt{4{{\sin }^{2}}15}\]
\[=2{{v}_{0}}\sin 15{}^\circ \]
Therefore, the velocity with which the stones move relative to each other in air is \[2{{v}_{0}}\sin 15{}^\circ \]

Answer is option B.

Note:
The velocity of a body with respect to the velocity of another body is known as the relative velocity of the first body with respect to the second.
If \[{{V}_{A}}\] and \[{{V}_{B}}\] represent the velocities of two bodies A and B respectively, then the relative velocity of A with respect to B is represented by \[{{V}_{AB}}\].Similarly, the relative velocity of B with respect to A is represented by \[{{V}_{BA}}\].
If the two objects are moving in the same direction, the relative velocity is given by,
\[{{V}_{AB}}={{V}_{A}}-{{V}_{B}}\], \[{{V}_{BA}}={{V}_{B}}-{{V}_{A}}\]
And if the two objects are moving in the opposite direction, then their relative velocity is given by,
\[{{V}_{AB}}={{V}_{BA}}={{V}_{A}}+{{V}_{B}}\]