91Ó°ÊÓ

First, we need to apply the conservation of momentum to the collision between the stone and the block. The total momentum before the collision is equal to the total momentum after the collision. The initial momentum is the momentum of the stone since the block is initially at rest. Therefore, we have:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]where \(m_1 = 3.00\, \mathrm{kg}\), \(v_{1i} = 8.00\, \mathrm{m/s}\), and \(v_{1f} = -2.00\, \mathrm{m/s}\) for the stone. The block is initially at rest, so \(v_{2i} = 0\). Substituting the known values gives:\[ 3.00\, \mathrm{kg} \times 8.00\, \mathrm{m/s} = 3.00\, \mathrm{kg} \times (-2.00\, \mathrm{m/s}) + 15.0\, \mathrm{kg} \times v_{2f} \]\[ 24.0 = -6.0 + 15.0 \times v_{2f} \]\[ 30.0 = 15.0 \times v_{2f} \]\[ v_{2f} = 2.00\, \mathrm{m/s} \]The speed of the block immediately after the collision is \(2.00\, \mathrm{m/s}\).

After the collision, the block moves with a speed of \(2.00\, \mathrm{m/s}\) and compresses the spring until it comes to a stop, entirely converting its kinetic energy to the potential energy stored in the spring. We apply conservation of energy:\[ \text{Initial kinetic energy of the block} = \text{Potential energy of the spring} \]\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]where \(m = 15.0\, \mathrm{kg}\), \(v = 2.00\, \mathrm{m/s}\), \(k = 500.0\, \mathrm{N/m}\), and \(x\) is the maximum compression of the spring. Solving for \(x\):\[ \frac{1}{2} \times 15.0 \times (2.00)^2 = \frac{1}{2} \times 500.0 \times x^2 \]\[ 30.0 = 250.0 \times x^2 \]\[ x^2 = \frac{30.0}{250.0} \]\[ x^2 = 0.12 \]\[ x = \sqrt{0.12} \approx 0.346\, \mathrm{m} \]The maximum compression of the spring is approximately \(0.346\, \mathrm{m}\).