Two bodies of equal masses moving along the same straight line with velocities +3m/s and -5m/s resp collide elastically. Their velocities after collision will be reapectively-1) 0.3m/s and -0.5m/s2) -0.4m/s and0.3m/s3) -5m/s and +3m/s4) -0.3m/s and 0.5m/s

To determine the velocities of two bodies after an elastic collision, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's break this down step by step.

Understanding the Problem

We have two bodies with equal masses, moving towards each other with velocities of +3 m/s and -5 m/s. In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum before the collision equals the total momentum after the collision, and the same applies to kinetic energy.

Step 1: Calculate Initial Momentum

The initial momentum (p_initial) of the system can be calculated using the formula:

• p_initial = m1 * v1 + m2 * v2

Here, m1 and m2 are the masses of the two bodies (which are equal), and v1 and v2 are their respective velocities. Since the masses are equal, we can denote them as 'm'. Thus, we have:

• p_initial = m * (+3) + m * (-5) = m(3 - 5) = -2m

Step 2: Calculate Initial Kinetic Energy

The initial kinetic energy (KE_initial) is given by:

• KE_initial = 0.5 * m1 * v1² + 0.5 * m2 * v2²

Substituting the values, we get:

• KE_initial = 0.5 * m * (3²) + 0.5 * m * (-5)² = 0.5m(9 + 25) = 17m

Step 3: Set Up Equations for Final Velocities

Let the final velocities after the collision be v1' and v2'. According to the conservation of momentum:

• -2m = m * v1' + m * v2'

This simplifies to:

• -2 = v1' + v2'

For conservation of kinetic energy, we have:

• 17m = 0.5 * m * (v1'² + v2')

Dividing through by 0.5m gives:

• 34 = v1'² + v2'²

Step 4: Solve the Equations

Now we have a system of two equations:

• 1. v1' + v2' = -2
• 2. v1'² + v2'² = 34

From the first equation, we can express v2' in terms of v1':

• v2' = -2 - v1'

Substituting this into the second equation gives:

• v1'² + (-2 - v1')² = 34

Expanding this, we get:

• v1'² + (4 + 4v1' + v1'²) = 34

This simplifies to:

• 2v1'² + 4v1' + 4 - 34 = 0

Or:

• 2v1'² + 4v1' - 30 = 0

Dividing the entire equation by 2 gives:

• v1'² + 2v1' - 15 = 0

Now we can factor this quadratic equation:

• (v1' + 5)(v1' - 3) = 0

This gives us two possible solutions for v1': v1' = -5 or v1' = 3. However, since we are looking for the velocities after the collision, we can use the first equation to find v2' for each case.

Final Velocities

1. If v1' = -5, then:

• v2' = -2 - (-5) = 3

2. If v1' = 3, then:

• v2' = -2 - 3 = -5

However, these results indicate that the bodies simply exchanged their velocities, which is a characteristic of elastic collisions when masses are equal.

Choosing the Correct Option

Now, looking at the options provided:

• 1) 0.3 m/s and -0.5 m/s
• 2) -0.4 m/s and 0.3 m/s
• 3) -5 m/s and +3 m/s
• 4) -0.3 m/s and 0.5 m/s

None of these options match the calculated velocities of -5 m/s and +3 m/s, indicating that the bodies simply exchanged their velocities. Therefore, the correct answer is that the velocities after the collision will be -5 m/s and +3 m/s, which corresponds to option 3.