A particle of mass `2kg` moving with a velocity `5hatim//s` collides head-on with another particle of mass `3kg` moving with a velocity `-2hatim//s`. After the collision the first particle has speed of `1.6m//s` in negative x-direction, Find
(a) velocity of the centre of mass after the collision,
(b) velocity of the second particle after the collision.
(c) coefficient of restitution.

Explanation

To solve the problem step by step, we will tackle each part of the question systematically.

### Given Data:
- Mass of particle 1, \( m_1 = 2 \, \text{kg} \)
- Initial velocity of particle 1, \( u_1 = 5 \, \text{m/s} \) (in positive x-direction)
- Mass of particle 2, \( m_2 = 3 \, \text{kg} \)
- Initial velocity of particle 2, \( u_2 = -2 \, \text{m/s} \) (in negative x-direction)
- Final velocity of particle 1 after collision, \( v_1 = -1.6 \, \text{m/s} \) (in negative x-direction)

### Part (a): Velocity of the Centre of Mass After the Collision

The formula for the velocity of the center of mass \( V_{cm} \) after the collision is given by:

\[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]

We need to find \( v_2 \) (the final velocity of particle 2) first, which we will do in Part (b).

### Part (b): Velocity of the Second Particle After the Collision

Using the conservation of momentum:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the known values:

\[ 2 \times 5 + 3 \times (-2) = 2 \times (-1.6) + 3 v_2 \]

Calculating the left side:

\[ 10 - 6 = -3.2 + 3 v_2 \]

This simplifies to:

\[ 4 = -3.2 + 3 v_2 \]

Rearranging gives:

\[ 3 v_2 = 4 + 3.2 \]
\[ 3 v_2 = 7.2 \]
\[ v_2 = \frac{7.2}{3} = 2.4 \, \text{m/s} \]

### Part (a) Continued: Velocity of the Centre of Mass After the Collision

Now that we have \( v_2 = 2.4 \, \text{m/s} \), we can calculate \( V_{cm} \):

\[ V_{cm} = \frac{2 \times (-1.6) + 3 \times 2.4}{2 + 3} \]

Calculating the numerator:

\[ = \frac{-3.2 + 7.2}{5} \]
\[ = \frac{4.0}{5} = 0.8 \, \text{m/s} \]

### Part (c): Coefficient of Restitution

The coefficient of restitution \( e \) is given by:

\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]

Substituting the known values:

\[ e = \frac{2.4 - (-1.6)}{5 - (-2)} \]
\[ = \frac{2.4 + 1.6}{5 + 2} \]
\[ = \frac{4.0}{7} \approx 0.571 \]

### Summary of Results:
- (a) Velocity of the centre of mass after the collision: \( 0.8 \, \text{m/s} \)
- (b) Velocity of the second particle after the collision: \( 2.4 \, \text{m/s} \)
- (c) Coefficient of restitution: \( \approx 0.571 \)