To solve this problem, we need to use two key principles:

1. Conservation of mechanical energy for object $m_1$ as it swings down (from height $h_{1i}$ to the point of collision).
2. Conservation of momentum and kinetic energy for the elastic collision between $m_1$ and $m_2$.
3. Once we know the final velocity of $m_1$ after the collision, we can use the conservation of energy again to find the final height.

Step 1: Mechanical Energy Conservation before the collision (for $m_1$)

The total mechanical energy of $m_1$ at the initial height $h_{1i}$ includes both potential energy and kinetic energy: $E_{1i} = m_1 g h_{1i} + \frac{1}{2} m_1 v_{1i}^2$ where:

• $m_1 = 1.60 \, \text{kg}$
• $h_{1i} = 0.285 \, \text{m}$
• $v_{1i} = 4.00 \, \text{m/s}$
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)

Let's calculate the total energy before the collision: $E_{1i} = (1.60)(9.81)(0.285) + \frac{1}{2} (1.60)(4.00)^2$ $E_{1i} = 4.47 \, \text{J} + 12.8 \, \text{J} = 17.27 \, \text{J}$

This is the total energy of $m_1$ before the collision.

Step 2: Conservation of Momentum in the Elastic Collision

For an elastic collision between $m_1$ and $m_2$, the velocities of both objects are governed by the following equations: $v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2}$ $v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1v_{1i}}{m_1 + m_2}$ Since $v_{2i} = 0$ (initial velocity of $m_2$), the equations simplify to: $v_{1f} = \frac{(m_1 - m_2)v_{1i}}{m_1 + m_2}$ $v_{2f} = \frac{2m_1v_{1i}}{m_1 + m_2}$

Substitute the known values: $v_{1f} = \frac{(1.60 - 4.50)(4.00)}{1.60 + 4.50}$ $v_{1f} = \frac{-2.90(4.00)}{6.10}$ $v_{1f} = -1.90 \, \text{m/s}$

This is the velocity of $m_1$ after the collision (the negative sign indicates it's in the opposite direction).

Step 3: Use Energy Conservation to Find Final Height

After the collision, $m_1$ swings upward, converting its kinetic energy back into potential energy. At the highest point, all kinetic energy will have been converted to potential energy, so: $\frac{1}{2} m_1 v_{1f}^2 = m_1 g h_{1f}$

Canceling $m_1$ from both sides and solving for $h_{1f}$: $\frac{1}{2} v_{1f}^2 = g h_{1f}$ $h_{1f} = \frac{v_{1f}^2}{2g}$

Substitute the known values: $h_{1f} = \frac{(-1.90)^2}{2(9.81)}$ $h_{1f} = \frac{3.61}{19.62}$ $h_{1f} \approx 0.184 \, \text{m}$

Final Answer:

The final height of object $m_1$ is approximately 0.184 meters.

Would you like any more details or have any questions?

Here are 5 related questions you might find interesting:

1. How would the result change if the collision were inelastic?
2. What is the velocity of object $m_2$ after the collision?
3. How does the ratio of masses affect the velocities after the collision?
4. What is the significance of elastic collisions in physics?
5. How would the outcome differ if $m_1$ had started with no initial speed?

Tip: Elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve only momentum.