Question

In the shown figure, spring is light and surface is smooth. Initially, when spring is relaxed, the point mass (1) is given velocity $v_0$. If $U_{max}$ is the maximum potential energy stored in the spring and $v$ is velocity of mass (1) at this moment, then

• A. A. $U_{max} = \frac{mv_0^2}{4}$
• B. B. $U_{max} = \frac{mv_0^2}{2}$
• C. C. $v = \frac{v_0}{2}$
• D. D. $v = v_0$

Answer 1

Answer: (A), (C)

A and C

Answer 2

The system's linear momentum is conserved because there are no external horizontal forces. Initial momentum: $P_i = m v_0$. At maximum potential energy, both masses move with the same velocity $v$. Final momentum: $P_f = mv + mv = 2mv$. Conservation of momentum: $m v_0 = 2mv \implies v = \frac{v_0}{2}$.

Mechanical energy is also conserved. Initial energy: $E_i = \frac{1}{2} m v_0^2$. At maximum potential energy, kinetic energy is $K_f = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = mv^2 = m (\frac{v_0}{2})^2 = \frac{1}{4} m v_0^2$. The total energy is $E_f = K_f + U_{max}$. Conservation of energy: $E_i = E_f \implies \frac{1}{2} m v_0^2 = \frac{1}{4} m v_0^2 + U_{max}$. Thus, $U_{max} = \frac{1}{4} m v_0^2$.