Question

Maximum energy stored in the spring in the subsequent motion will be:

• A. $5v_0^2$
• B. $15v_0^2$
• C. zero
• D. $10v_0^2$

Answer 1

Answer: (B)

15v₀²

Answer 2

The problem involves two blocks connected by a spring on a smooth horizontal surface. This is a system where both linear momentum and mechanical energy are conserved. The motion can be effectively analyzed by considering the center of mass frame.

1. Calculate the velocity of the center of mass ($V_{CM}$): Let the initial velocity of the 2 kg block ($m_1$) be $u_1 = +4v_0$ (taking right as positive). Let the initial velocity of the 3 kg block ($m_2$) be $u_2 = -v_0$ (to the left).

The velocity of the center of mass is constant because there are no external horizontal forces. $V_{CM} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$ $V_{CM} = \frac{(2 \text{ kg})(4v_0) + (3 \text{ kg})(-v_0)}{2 \text{ kg} + 3 \text{ kg}}$ $V_{CM} = \frac{8v_0 - 3v_0}{5} = \frac{5v_0}{5} = v_0$ So, $V_{CM} = v_0$ (to the right).

2. Maximum energy stored in the spring: The maximum energy stored in the spring occurs when the relative velocity between the two blocks is momentarily zero. At this instant, both blocks move with the velocity of the center of mass ($V_{CM}$). This also corresponds to the state of maximum compression or extension of the spring. The potential energy stored in the spring is the kinetic energy of the system in the center of mass frame. Initial velocities in the center of mass frame: $u_{1,CM} = u_1 - V_{CM} = 4v_0 - v_0 = 3v_0$ $u_{2,CM} = u_2 - V_{CM} = -v_0 - v_0 = -2v_0$ The initial relative velocity is $u_{rel} = u_1 - u_2 = 4v_0 - (-v_0) = 5v_0$. The reduced mass of the system is $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{2 \times 3}{2+3} = \frac{6}{5} \text{ kg}$. The initial kinetic energy in the center of mass frame is: $KE_{CM, initial} = \frac{1}{2}\mu u_{rel}^2$ $KE_{CM, initial} = \frac{1}{2} \left(\frac{6}{5}\right) (5v_0)^2$ $KE_{CM, initial} = \frac{1}{2} \left(\frac{6}{5}\right) (25v_0^2) = \frac{6 \times 25}{10} v_0^2 = 3 \times 5 v_0^2 = 15v_0^2$ At maximum compression/extension, all this kinetic energy in the CM frame is converted into potential energy of the spring. Thus, the maximum energy stored in the spring is $U_{max} = 15v_0^2$.