A ball of mass 1kg , moving with a velocity of 0.4m / s collides with another stationary ball. After the collision, the first ball moves with a velocity of 0.3m / s in a direction making an angle of of 90° with its initial direction. The momentum of second ball after collision will be (in kg-m/s)

The correct option is 0.5.

Step-by-step Explanation:

We can solve this problem using the law of conservation of linear momentum. According to this law, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

Let us define the coordinate system such that the initial direction of motion of the first ball is along the positive x-axis ($\hat{i}$ direction), and the perpendicular direction is along the positive y-axis ($\hat{j}$ direction).

1. Initial Momentum of the System:
The first ball has mass $m_1=1\text{kg}$ and initial velocity ${\overrightarrow{v}}_{1i}=0.4\hat{i}\text{m/s}$.
The second ball is initially stationary, so its velocity is ${\overrightarrow{v}}_{2i}=0$.
Therefore, the initial momentum of the first ball (${\overrightarrow{p}}_{1i}$) is:
${\overrightarrow{p}}_{1i}=m_1{\overrightarrow{v}}_{1i}=1\times 0.4\hat{i}=0.4\hat{i}\text{kg-m/s}$
The initial momentum of the second ball is ${\overrightarrow{p}}_{2i}=0$.
Thus, the total initial momentum of the system (${\overrightarrow{P}}_i$) is:
${\overrightarrow{P}}_i=0.4\hat{i}\text{kg-m/s}$

2. Final Momentum of the First Ball:
After the collision, the first ball moves at an angle of 90° relative to its initial direction with a speed of 0.3 m/s. We can represent this direction along the positive y-axis ($\hat{j}$ direction).
The final velocity of the first ball is ${\overrightarrow{v}}_{1f}=0.3\hat{j}\text{m/s}$.
So, the final momentum of the first ball (${\overrightarrow{p}}_{1f}$) is:
${\overrightarrow{p}}_{1f}=m_1{\overrightarrow{v}}_{1f}=1\times 0.3\hat{j}=0.3\hat{j}\text{kg-m/s}$

3. Conservation of Momentum:
Let the final momentum of the second ball be ${\overrightarrow{p}}_{2f}$.
By conservation of momentum:
${\overrightarrow{P}}_i={\overrightarrow{p}}_{1f}+{\overrightarrow{p}}_{2f}$
Substituting the known values:
$0.4\hat{i}=0.3\hat{j}+{\overrightarrow{p}}_{2f}$
Solving for ${\overrightarrow{p}}_{2f}$ gives:
${\overrightarrow{p}}_{2f}=0.4\hat{i}-0.3\hat{j}$

4. Magnitude of the Momentum of the Second Ball:
The magnitude of the momentum vector is calculated as:
$\left|{\overrightarrow{p}}_{2f}\right|=\sqrt{{\left(0.4\right)}^2+{(-0.3)}^2}$
$\left|{\overrightarrow{p}}_{2f}\right|=\sqrt{0.16+0.09}$
$|{\overrightarrow{p}}_{2f}$

The magnitude of the momentum of the second ball after the collision is 0.5 kg-m/s.