Refer to Fig. Suppose the initial velocity of the 2.4-kg block is reversed; it is headed directly toward the 1.6-kg block. a) What would be the velocity of the 1.6-kg block after the collision?

b) Is the colission elastic?

To analyze the scenario you've presented, we need to apply the principles of conservation of momentum and the characteristics of elastic collisions. Let's break this down step by step.

Understanding the Collision

In this situation, we have two blocks: a 2.4-kg block moving toward a stationary 1.6-kg block. When they collide, we need to determine the final velocity of the 1.6-kg block and whether the collision is elastic.

Step 1: Conservation of Momentum

The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. The formula for momentum (p) is:

• p = mass × velocity

Let's denote:

• m1 = 2.4 kg (mass of the first block)
• u1 = initial velocity of the first block (let's assume it is negative since it is moving towards the second block)
• m2 = 1.6 kg (mass of the second block)
• u2 = initial velocity of the second block = 0 m/s (since it is stationary)
• v1 = final velocity of the first block after the collision
• v2 = final velocity of the second block after the collision

The equation for conservation of momentum before and after the collision can be expressed as:

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

Step 2: Solving for Final Velocities

Assuming a perfectly elastic collision, we also need to consider the conservation of kinetic energy. The kinetic energy (KE) before and after the collision can be expressed as:

• KE_initial = 0.5 * m1 * u1² + 0.5 * m2 * u2²
• KE_final = 0.5 * m1 * v1² + 0.5 * m2 * v2²

For an elastic collision, KE_initial = KE_final. However, to simplify our calculations, we can use the following equations derived from both conservation laws:

• v1 = ((m1 - m2) / (m1 + m2)) * u1 + ((2 * m2) / (m1 + m2)) * u2
• v2 = ((2 * m1) / (m1 + m2)) * u1 + ((m2 - m1) / (m1 + m2)) * u2

Substituting u2 = 0 into these equations gives us:

• v1 = ((2.4 - 1.6) / (2.4 + 1.6)) * u1
• v2 = ((2 * 2.4) / (2.4 + 1.6)) * u1

Step 3: Calculating the Final Velocities

Now, let's calculate:

• v1 = (0.8 / 4) * u1 = 0.2 * u1
• v2 = (4.8 / 4) * u1 = 1.2 * u1

Thus, the final velocity of the 1.6-kg block (v2) after the collision is 1.2 times the initial velocity of the 2.4-kg block, but in the opposite direction since it was initially stationary.

Examining Elasticity of the Collision

To determine if the collision is elastic, we need to check if kinetic energy is conserved. If the calculated final velocities lead to the same total kinetic energy as before the collision, then it is elastic.

Using the kinetic energy formulas:

• KE_initial = 0.5 * 2.4 * u1²
• KE_final = 0.5 * 2.4 * (0.2 * u1)² + 0.5 * 1.6 * (1.2 * u1)²

After calculating both sides, if they are equal, the collision is elastic. If not, it is inelastic.

In summary, the final velocity of the 1.6-kg block after the collision is 1.2 times the initial velocity of the 2.4-kg block, and whether the collision is elastic can be confirmed by checking the conservation of kinetic energy. This approach allows us to understand the dynamics of the collision thoroughly.