Question

Two sphere $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $v$ 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.

• A. $\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$ to the x-axis
• B. $\theta ={\tan }^{-1}(-\frac{1}{2})$ to the y-axis
• C. Same as that of $B$
• D. Opposite to that of $B$

Answer 1

The correct option is A $\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$ to the x-axis

There is no external force acting on the spheres. So linear momentum will be conserved.

Before the collision, In direction $x$, Linear momentum $=m_{2}v$ ....1

In direction $y$, Linear momentum$=0$

After the collision, spheres moves as shown in figure. Let velocity of sphere $A$ is $v_1$

In direction $x$, Linear momentum $=m_1{v}_{1}cos\left(\theta \right)$ ...2

In direction $y$, Linear momentum $=\frac{m_{2}v}{2}-m_1{v}_{1}sin\left(\theta \right)$ .......3

Linear momentum will be conserved,

From equation 1 and 2, $\Rightarrow m_1{v}_{1}cos\left(\theta \right)=m_{2}v$ ...4

From the equation 2, $\Rightarrow \frac{m_{2}v}{2}-m_1{v}_{1}sin\left(\theta \right)=0$ $\Rightarrow m_1{v}_{1}sin\left(\theta \right)=\frac{m_{2}v}{2}$ ......5

Dividing equation 5 by 4, $\Rightarrow tan\left(\theta \right)=\frac{1}{2}$ $\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{2}\right)$