91Ó°ÊÓ

First, let's apply the conservation of momentum principle to the bullet-block system. The total initial momentum is the momentum of the bullet before it collides with the block, and the total final momentum is the combined momentum of the bullet and the block after the collision when they are moving together. Let \(v_b\) be the initial speed of the bullet, \(m_b = 2.75 \mathrm{~g}\) be the mass of the bullet, \(m_s = 1.50 \mathrm{~kg}\) be the mass of the block, and \(v_{bs}\) be the final speed of the bullet-block system. Then: Initial_total_momentum = Final_total_momentum \(m_b \cdot v_b = (m_b + m_s) \cdot v_{bs}\) Now, we need to determine the potential energy stored in the spring when it's compressed at its maximum point, 4.30 cm. The potential energy of a compressed spring is given by the equation: \(PE_{spring} = \frac{1}{2} k x^2\) where \(k = 850 \mathrm{~N} / \mathrm{m}\) is the spring constant and \(x = 4.30 \mathrm{~cm} = 0.043 \mathrm{~m}\) is the maximum compression of the spring. By plugging these values, we can find the potential energy: \(PE_{spring} = \frac{1}{2} \cdot 850 \mathrm{~N/m} \cdot (0.043 \mathrm{~m})^2 = 0.7868 \mathrm{~J}\) When the spring is compressed at its maximum point, the total kinetic energy of the bullet-block system has been converted into potential energy. Hence: \(PE_{spring} = KE_{bullet} + KE_{block}\) Assuming that all the kinetic energy is carried by the bullet-block system, since the block is initially at rest: \(p = \frac{1}{2} (m_b + m_s) v_{bs}^2\) We now have two equations with two unknowns (\(v_{bs}\) and \(v_b\)): (1) \(m_b \cdot v_b = (m_b + m_s) \cdot v_{bs}\) (2) \(0.7868 \mathrm{~J} = \frac{1}{2} (m_b + m_s) v_{bs}^2\) We first solve the first equation for \(v_{bs}\): \(v_{bs} = \frac{m_b}{m_b + m_s} v_b\) Plugging this expression into the second equation, we get: \(0.7868 \mathrm{~J} = \frac{1}{2} (m_b + m_s) \left( \frac{m_b}{m_b + m_s} v_b \right)^2\) Now, we isolate the variable \(v_b\) and solve the equation for it: \(v_b = \sqrt{\frac{2 \cdot 0.7868 \mathrm{~J} (m_b + m_s)^2}{m_b^2}} = 155.07 \mathrm{~m/s}\) So, the initial speed of the bullet is 155.07 m/s.

Since the spring force is a conservative force, we can use the conservation of mechanical energy principle. At the maximum compression, the kinetic energy is fully converted into the potential energy stored in the spring, which can be expressed as: \(KE_{initial} = PE_{spring}\) Thus, the initial kinetic energy of the bullet-block system is 0.7868 J. Since the mass of the system is the combined mass of the bullet and the block, the initial velocity of the bullet-block system can be found through: \(0.7868 \mathrm{~J} = \frac{1}{2} (m_b + m_s) v_{bs}^2\) \(v_{bs} = \sqrt{\frac{2 \cdot 0.7868 \mathrm{~J}}{(m_b + m_s)}} = 1.013 \mathrm{~m/s}\) Now, we will use the spring's force constant and Hooke's Law to find the period of oscillation: \(T = 2 \pi \sqrt{\frac{m_{system}}{k}}\) where \(m_{system} = m_b + m_s\) and \(k = 850 \mathrm{~N} / \mathrm{m}\). Therefore, \(T = 2 \pi \sqrt{\frac{m_b + m_s}{850 \mathrm{~N/m}}} = 0.249 \mathrm{~s}\) The time it takes the bullet-block system to come to rest is half the period of oscillation, as it returns to its initial position when the spring's potential energy is completely converted back into kinetic energy. \(t_{rest} = \frac{T}{2} = 0.1245 \mathrm{~s}\) So, it takes 0.1245 seconds for the bullet-block system to come to rest.