Let's solve part **(a)** of the problem step by step, following your requested structure. --- **Given** - Mass of tray, \( m_t = 1.50 \) kg - Mass of metal ball, \( m_b = 0.275 \) kg - Spring constant, \( k = 185 \) N/m - Displacement from equilibrium, \( y_A = 15.0 \) cm = 0.150 m - The system is released from rest at point A. --- ### Step 1: **Determine the condition for the ball to leave the tray** #### Explanation The ball will leave the tray when the normal force between the tray and the ball becomes zero. This happens when the acceleration of the tray exceeds the acceleration due to gravity in the upward direction (since the tray is accelerating upward faster than gravity is pulling the ball downward). The equation of motion for the tray-ball system is: \[ (m_t + m_b) a = -k y \] where \( y \) is the displacement from equilibrium (downward positive). The acceleration of the tray is: \[ a = -\frac{k}{m_t + m_b} y \] The ball loses contact when the upward acceleration of the tray equals \( g \): \[ \left| a \right| = g \] \[ \frac{k}{m_t + m_b} y = g \] \[ y = \frac{g (m_t + m_b)}{k} \] --- ### Step 2: **Calculate the height above point A where the ball leaves the tray** #### Explanation Point A is 0.150 m below the equilibrium point, and the ball leaves at a displacement \( y \) above this point. The difference in height from point A: \[ \Delta y = y_A - y \] Insert values: \[ y = \frac{9.8 \times (1.50 + 0.275)}{185} \] \[ y = \frac{9.8 \times 1.775}{185} \] \[ y = \frac{17.395}{185} \] \[ y = 0.0941 \text{ m} \] Now, calculate the height above point A: \[ \Delta y = 0.150 - 0.0941 = 0.0559 \text{ m} \] --- ### **Summary** **Definition of Parameters:** - \( m_t \): Mass of tray - \( m_b \): Mass of ball - \( k \): Spring constant - \( y_A \): Initial displacement below equilibrium - \( g \): Acceleration due to gravity **Final Answer for (a):** **The tray will be 0.0559 m (5.59 cm) above point A when the metal ball leaves the tray.** --- **Quality Review on 6 Parameters** - **Relevancy:** The solution directly addresses the condition when the ball leaves the tray. - **Completeness:** Key equations and each calculation step are shown. - **Accuracy:** All values and physical principles (Newton's laws, SHM) are correctly used. - **Clarity:** Each step explains the reasoning and calculation. - **Structure:** The steps progress logically from physics condition to calculation. - **Voice:** Language is clear and instructive, avoiding non-human tone.

Let's proceed to solve parts **(b)** and **(c)** step by step, following the structured approach with explanations. --- ### **Part (b): How much time elapses between releasing the system at point A and the ball leaving the tray?** #### **Step 1: Determine the initial conditions and the oscillation parameters** **Explanation:** The system is released from rest at a displacement \( y_A = 0.150 \) m downward from equilibrium. Since the system behaves like a simple harmonic oscillator (SHO), the motion of the tray (and the ball when in contact) can be described by SHO equations. The angular frequency: \[ \omega = \sqrt{\frac{k}{m_t + m_b}} \] Insert values: \[ \omega = \sqrt{\frac{185}{1.50 + 0.275}} = \sqrt{\frac{185}{1.775}} \approx \sqrt{104.225} \approx 10.21 \text{ rad/sec} \] **Initial condition:** - Initial displacement: \( y(0) = y_A = 0.150 \) m - Initial velocity: \( v(0) = 0 \) (system released from rest) The displacement as a function of time: \[ y(t) = y_A \cos(\omega t) \] because initial velocity is zero. --- ### **Step 2: Find the time when the ball leaves the tray** **Explanation:** The ball leaves when the acceleration of the tray exceeds gravity: \[ a(t) = -\omega^2 y(t) \] The system leaves contact when: \[ |a(t)| = g \] which occurs at: \[ \omega^2 y(t) = g \] \[ y(t) = \frac{g}{\omega^2} \] Calculate: \[ y(t) = \frac{9.8}{(10.21)^2} = \frac{9.8}{104.24} \approx 0.094 \text{ m} \] We already know: \[ y(t) = y_A \cos(\omega t) = 0.150 \cos(\omega t) \] Set equal to 0.094 m: \[ 0.150 \cos(\omega t) = 0.094 \] \[ \cos(\omega t) = \frac{0.094}{0.150} \approx 0.627 \] Calculate \( \omega t \): \[ \omega t = \arccos(0.627) \approx 0.887 \text{ radians} \] Solve for \( t \): \[ t = \frac{0.887}{\omega} = \frac{0.887}{10.21} \approx 0.087 \text{ seconds} \] --- ### **Final answer for (b):** **Approximately 0.087 seconds elapse between releasing the system at point A and the ball leaving the tray.** --- ### **Part (c): How fast is the ball moving just as it leaves the tray?** #### **Step 1: Find the velocity of the tray at the moment the ball leaves** **Explanation:** The velocity of the tray is the time derivative of displacement: \[ v(t) = -y_A \omega \sin(\omega t) \] Using \( y_A = 0.150 \) m, \( \omega = 10.21 \) rad/sec, and \( \omega t = 0.887 \): \[ v(t) = -0.150 \times 10.21 \times \sin(0.887) \] Calculate \( \sin(0.887) \): \[ \sin(0.887) \approx 0.777 \] Now: \[ v(t) = -0.150 \times 10.21 \times 0.777 \approx -1.19 \text{ m/sec} \] The negative sign indicates direction (upward or downward), but speed is magnitude: \[ |v| \approx 1.19 \text{ m/sec} \] --- ### **Step 2: Velocity of the ball at the moment of leaving** **Explanation:** When the ball leaves, it has the same velocity as the tray (assuming no slip or additional forces). Therefore, the velocity of the ball just as it leaves the tray is approximately: \[ \boxed{ \text{Speed of the ball} \approx 1.19 \text{ m/sec} } \] --- ### **Final summaries:** - **(a)** The tray is **0.0559 m (5.59 cm)** above point A when the ball leaves. - **(b)** The time elapsed from release at point A until the ball leaves is approximately **0.087 seconds**. - **(c)** The speed of the ball as it leaves the tray is approximately **1.19 m/sec**. --- **Note:** These calculations assume ideal conditions, such as no damping and perfect energy transfer, which are typical in physics problems of this nature.

Yes, the answer for part **(a)** is correct. The tray will be approximately **0.0559 meters (5.59 centimeters)** above point A when the metal ball leaves the tray. This conclusion is based on equating the upward acceleration of the tray to gravity, which determines the point at which the ball loses contact. The calculations correctly incorporate the parameters provided and utilize the physics principles of simple harmonic motion and contact loss conditions.