We will solve this problem using the principles of conservation of momentum and work-energy theorem.

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Step 1: Find the Initial Velocities of the Pieces

From the work-energy theorem, the work done by friction is equal to the initial kinetic energy of each piece:

$W = \Delta KE$

Since each piece comes to a stop, its final kinetic energy is zero, so:

$f_k d = \frac{1}{2} m v^2$

The force of kinetic friction is given by:

$f_k = \mu m g$

Thus, the work done by friction is:

$(\mu m g) d = \frac{1}{2} m v^2$

Canceling $m$ from both sides:

$\mu g d = \frac{1}{2} v^2$

Solving for $v$:

$v = \sqrt{2 \mu g d}$

Using $g = 9.81$ m/s²:

For piece L ($m_L = 3.4$ kg, $\mu_L = 0.54$, $d_L = 0.39$):

$v_L = \sqrt{2 (0.54) (9.81) (0.39)}$

$v_L \approx \sqrt{4.222}$

$v_L \approx 2.05 \text{ m/s}$

For piece R ($\mu_R = 0.47$, $d_R = 0.31$):

$v_R = \sqrt{2 (0.47) (9.81) (0.31)}$

$v_R \approx \sqrt{2.873}$

$v_R \approx 1.70 \text{ m/s}$

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Step 2: Use Conservation of Momentum

Before the explosion, the block is stationary, so the total momentum is zero. After the explosion:

$m_L v_L + m_R v_R = 0$

$(3.4)(2.05) + m_R (-1.70) = 0$

$6.97 - 1.70 m_R = 0$

$m_R = \frac{6.97}{1.70}$

$m_R \approx 4.1 \text{ kg}$

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Final Answer:

The mass of piece $R$ is 4.1 kg.

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Would you like me to go into more detail or clarify any steps?

Related Questions:

1. How would the answer change if the friction coefficients were different?
2. What would happen if the floor were not frictionless before the pieces entered the rough region?
3. How much energy was lost to friction for each piece?
4. What was the total kinetic energy immediately after the explosion?
5. How would the result differ if the explosion occurred in space (zero gravity)?

Tip:

Whenever solving physics problems involving explosions or collisions, always start by checking if momentum is conserved before moving to energy considerations! 🚀