This is a classical physics problem involving energy conservation and an elastic collision. Here's a step-by-step breakdown of how to approach it:

Part (a): Determine the speeds of the two blocks after the collision.

Let's denote:

• $m_1$ as the mass of the first block (the one sliding down the incline),
• $m_2$ as the mass of the second block (the one initially at rest),
• $v_1$ as the velocity of block $m_1$ just before the collision,
• $v_2$ as the velocity of block $m_2$ just before the collision (which is $0$ since it's at rest),
• $v_1'$ as the velocity of block $m_1$ after the collision,
• $v_2'$ as the velocity of block $m_2$ after the collision.

Step 1: Find the speed of $m_1$ just before the collision

The block $m_1$ slides down the incline without friction, so we can use conservation of mechanical energy.

The potential energy at the top is entirely converted into kinetic energy at the bottom:

$m_1 g h = \frac{1}{2} m_1 v_1^2$

Solve for $v_1$:

$v_1 = \sqrt{2gh}$

Given:

• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $h = 3.60 \, \text{m}$ (height of the incline),

$v_1 = \sqrt{2 \times 9.8 \times 3.60} = \sqrt{70.56} \approx 8.40 \, \text{m/s}$

So, the speed of $m_1$ just before the collision is approximately $8.40 \, \text{m/s}$.

Step 2: Apply the equations for an elastic collision

For a perfectly elastic collision, two key principles are used:

1. Conservation of momentum: $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$ Since $v_2 = 0$ (block 2 is initially at rest), the equation simplifies to: $m_1 v_1 = m_1 v_1' + m_2 v_2'$

2. Conservation of kinetic energy: $\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 {v_1'}^2 + \frac{1}{2} m_2 {v_2'}^2$

Solving for the final velocities:

In an elastic collision, there's a known formula for the final velocities of two colliding objects:

$v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1$

$v_2' = \frac{2 m_1}{m_1 + m_2} v_1$

We can substitute the known values of $v_1$ once we know the masses $m_1$ and $m_2$.

Part (b): How far back up the incline the smaller mass will go

After the collision, block $m_1$ will travel back up the incline. We can use conservation of energy again. The kinetic energy of block $m_1$ after the collision will convert back into potential energy as it moves up the incline.

$\frac{1}{2} m_1 {v_1'}^2 = m_1 g h_{\text{max}}$

Solve for $h_{\text{max}}$:

$h_{\text{max}} = \frac{{v_1'}^2}{2g}$

Once $v_1'$ is calculated from part (a), we can find how high the block rises.

Final Steps:

• To solve for the final velocities and height, we'd need the values of $m_1$ and $m_2$.
• If these values are given, we can plug them into the equations above for specific results.

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Would you like to provide values for the masses, or do you need further details on the general case?