1. Defining Readings and Masses

The balance is tared (calibrated) to read zero for the empty pan.

• Final Reading ($m_2 = 2.0$ kg): Since the system settles here, this is the mass of the added object (the sandbag). $$ m_{bag} = 2.0 \text{ kg} $$
• Total Oscillating Mass ($M_{tot}$): The collision involves both the bag and the pan. $$ M_{tot} = M_{pan} + m_{bag} $$
• Peak Reading ($m_1 = 6.0$ kg): This corresponds to the maximum force exerted by the spring above the empty pan weight. $$ F_{spring, max} = (M_{pan} + m_1)g $$

2. Oscillation Parameters

The oscillation takes place around the new equilibrium (Pan + Bag).

Amplitude ($A$): At the lowest point (max reading), the spring force balances the total weight plus the restoring force $kA$. $$ F_{spring, max} = M_{tot}g + kA $$ Substituting the reading definition: $$ (M_{pan} + m_1)g = (M_{pan} + m_{bag})g + kA $$ $$ m_1 g = m_{bag} g + kA $$ $$ kA = (m_1 – m_{bag})g $$ $$ A = \frac{(6.0 – 2.0)10}{1500} = \frac{40}{1500} = \frac{4}{150} \text{ m} $$

Initial Position Shift ($x_0$): The impact happens at the level of the empty pan equilibrium. The distance between the empty equilibrium and the new equilibrium is determined by the added weight of the bag. $$ k x_0 = m_{bag} g \implies x_0 = \frac{2.0 \times 10}{1500} = \frac{20}{1500} = \frac{2}{150} \text{ m} $$

3. Collision and Energy Conservation

Step A: Inelastic Collision
Bag ($m_{bag}$) falls height $h=0.04$ m. Velocity $v = \sqrt{2gh}$. The collision with the pan ($M_{pan}$) is perfectly inelastic. $$ m_{bag} \sqrt{2gh} = (M_{pan} + m_{bag}) v’ $$ $$ v’ = \frac{m_{bag} \sqrt{2gh}}{M_{tot}} $$

Step B: Energy Balance
Consider the energy relative to the new equilibrium position. Just after impact, the system is at position $x_0$ (above equilibrium) with velocity $v’$. It reaches max extension $A$ (where velocity is 0).

$$ \text{KE}_{initial} + \text{PE}_{spring, initial} = \text{PE}_{spring, max} $$ $$ \frac{1}{2} M_{tot} (v’)^2 + \frac{1}{2} k x_0^2 = \frac{1}{2} k A^2 $$ Multiply by 2 and substitute $v’$: $$ M_{tot} \left( \frac{m_{bag} \sqrt{2gh}}{M_{tot}} \right)^2 + k x_0^2 = k A^2 $$ $$ \frac{m_{bag}^2 (2gh)}{M_{tot}} = k (A^2 – x_0^2) $$

4. Solving for $M_{pan}$

Calculate the term $(A^2 – x_0^2)$:

$$ A^2 – x_0^2 = \left(\frac{4}{150}\right)^2 – \left(\frac{2}{150}\right)^2 = \frac{16 – 4}{22500} = \frac{12}{22500} $$

Right Hand Side ($k(A^2 – x_0^2)$):

$$ 1500 \times \frac{12}{22500} = \frac{1500 \times 12}{15 \times 1500} = \frac{12}{15} = 0.8 \text{ Joules} $$

Left Hand Side:

$$ \frac{m_{bag}^2 (2gh)}{M_{tot}} = \frac{2.0^2 \times 2 \times 10 \times 0.04}{M_{tot}} = \frac{4 \times 0.8}{M_{tot}} = \frac{3.2}{M_{tot}} $$

Equating LHS and RHS:

$$ \frac{3.2}{M_{tot}} = 0.8 $$ $$ M_{tot} = \frac{3.2}{0.8} = 4.0 \text{ kg} $$

Since $M_{tot} = M_{pan} + m_{bag}$ and $m_{bag} = 2.0$ kg:

$$ M_{pan} = 4.0 – 2.0 = 2.0 \text{ kg} $$

The mass of the pan is 2.0 kg.