Question

Take the z=axis as vertical and the x-y plane as horizontal. A particle ‘A’ is projected with velocity $4\sqrt{2}m{s}^{-1}$ making an angle ${45}^{\circ }$ to the horizontal. Particle B is projected at $5m{s}^{-1}$ an angle $\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$ to y axis in y-z plane then velocity of B wrt A.
A) Has its initial magnitude$5m{s}^{-1}$.
B) Magnitude will change with time.
C) Lies in the xy plane.
D) Will initially make an angle$\left(\theta +{\displaystyle \frac{\pi }{2}}\right)$.

Answer 1

Hint: The x,y and z are the three directions which are mutually perpendicular to each other, the particle A is in plane x-z and the particle B is in the plane y-z. The relative velocity is the difference between the velocities of the two bodies.

Formula used:
The relative velocity of the two particles is given by,
$\Rightarrow v_B-v_A$
Where the velocity of particle B is $v_B$ and the velocity of particle A is$v_A$.

Complete step by step solution:
In this problem it is given that a particle ‘A’ is projected with velocity $4\sqrt{2}m{s}^{-1}$ making an angle ${45}^{\circ }$ to the horizontal. Particle B is projected at $5m{s}^{-1}$ an angle $\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$ to y axis in y-z plane then we need to find the velocity of B wrt A.
According to the condition the figure will be.

The velocity of particle A is $v_A=4\hat{i}+4\hat{k}$ and the velocity of particle B is equal to $v_B=3\hat{j}+4\hat{k}$.
The relative velocity of the particle B with respect to particle A is equal to,
$\Rightarrow v_B-v_A=\left(3\hat{j}+4\hat{k}\right)-\left(4\hat{i}+4\hat{k}\right)$
$\Rightarrow v_B-v_A=3\hat{j}+4\hat{k}-4\hat{i}-4\hat{k}$
$\Rightarrow v_B-v_A=3\hat{j}+4\hat{k}-4\hat{i}-4\hat{k}$
$\Rightarrow v_B-v_A=3\hat{j}-4\hat{i}$
The magnitude of the velocity is equal to,
$\Rightarrow v_B-v_A=\sqrt{3^2+4^2}$
$\Rightarrow v_B-v_A=\sqrt{9+16}$
$\Rightarrow v_B-v_A=\sqrt{25}$
$\Rightarrow v_B-v_A=5m{s}^{-1}$
Which means option A is correct. The magnitude of the acceleration will not change with time and therefore the option B is wrong.
The relative velocity is $v_B-v_A=3\hat{j}-4\hat{i}$ which means it is in the x-y plane, the option C is also correct. The angle of the initially equal to,
$\Rightarrow \theta ={\tan }^{-1}\left(-{\displaystyle \frac{4}{3}}\right)$
As the angle is negative and therefore we add ${\displaystyle \frac{\pi }{2}}$ so the angle becomes $\theta +{\displaystyle \frac{\pi }{2}}$, so the option D is correct.

The wrong option is option B, so the answer for this problem is option B.

Note: The students are advised to remember the formula of the relative velocity and also the diagram of the velocity of the particle A and particle B should be drawn very carefully as the answer is very dependent on the diagram.