Two spheres A and B of masses m_1 and m_2 respectively collide. A...

Two spheres

A

and

B

of masses

m_1

and

m_2

respectively collide.

A

is at rest initially and

B

is moving with velocity

y

along

x

-axis. After collision

B

has a velocity

\frac{v}{2}

in a direction perpendicular to the original direction. The mass

A

moves after collision in the direction

Options

Asame as that of $B$
Bopposite to that of $B$
C\theta=\tan ^{-1} $\left(\frac{1}{2}\right)$ to the $x$ -axis
D\theta=\tan ^{-1} $\left(\frac{-1}{2}\right)$ to the $x$ -axis

Correct answer

C. \theta=\tan ^{-1} $\left(\frac{1}{2}\right)$ to the $x$ -axis

Step-by-step solution

\text{Here, } \mathbf{p}_i=m_2 v \mathbf{i}+m_2 \times 0 \mathbf{p}_f=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_2 Law

of conservation of momentum\begin{gathered}\mathbf{p}_i=\mathbf{p}_f \\ m_2 v \mathbf{i}= m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_1 \\ \mathbf{v}_1= \frac{m_2}{m_1} v \mathbf{i}+\frac{m_2}{m_1} \frac{v}{2} \mathbf{j}\end{gathered} From

this equation we can find

\begin{aligned}\tan \theta & =\frac{1}{2} \\ \theta & =\tan ^{-1}\left(\frac{1}{2}\right) \text{ to the } x \text{-axis. } \end{aligned}

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