Question

If the collision occurs at time \(𝑡_0 = 0\), the value of \(\frac{𝑣_cm}{(\alpha\omega)}\) will be __________

Answer 1

Correct Answer: 0.75

Velocity and Center of Mass Calculation After Collision 

At \( t_0 = 0 \), the velocities of the particles are:

\[ v_1 = a \omega \cos 0 = a \omega, \quad v_2 = -a \omega \]

After collision, velocity exchange occurs. The center of mass velocity is:

\[ v_{\text{cm}} = \frac{m \cdot a \omega + m \cdot (-a \omega)}{2m} \] Simplifying: \[ v_{\text{cm}} = \frac{m \cdot a \omega - m \cdot a \omega}{2m} = 0 \]

Quick Tip:

\[ \frac{v_{\text{cm}}}{a \omega} = 0.75 \]

Answer 2

Collision Analysis in a Spring-Mass System 

If the collision occurs at \( t_0 = \frac{\pi}{2\omega} \), the system undergoes the following transitions:

Before Collision:

\[ 3m \rightarrow 2 \rightarrow u = 0 \rightarrow 1 \rightarrow u = 0 \] The mass \( 3m \) moves with initial velocity \( u_0 \), while the other masses remain stationary.

After Collision:

\[ 3m \, u = 0 \quad 2 \rightarrow u_0 \quad 1 \rightarrow u = 0 \] The second mass now moves with velocity \( u_0 \), and the third remains at rest.

Final Situation:

\[ 3m \, u = 0 \quad 2 \rightarrow u_0 \quad 1 \rightarrow \frac{u_0}{2} \] After the system stabilizes, the second and third masses share the velocity.

Applying Conservation of Mechanical Energy:

\[ \frac{1}{2} K (2a)^2 + \frac{1}{2} m u_0^2 = \frac{1}{2} K (2b)^2 + \frac{1}{2} (2m) u_0^2 \] where: \[ K = m \omega^2, \quad u_0 = a \omega / 2 \]

Solving for the Ratio:

\[ \frac{4b^2}{a^2} = 17 \quad \Rightarrow \quad \frac{4b^2}{a^2} = 4.25 \]

Final Answer:

The ratio \( \frac{4b^2}{a^2} = 4.25 \), indicating the equilibrium conditions of the system after the collision.