Elastic Collision of Moving Mass with Pivoted Rigid Bar

To find the final angular velocity $\omega$ of the bar and the speed $v$ of the small mass after the collision, we apply the principles of conservation of angular momentum and conservation of kinetic energy.

1. Conservation of Angular Momentum

Since there are no external torques acting on the system about the pivot point, the angular momentum about the pivot is conserved during the collision.

The moment of inertia of the bar about its pivoted end is given by: $I = \frac{1}{3} M L^2$

Taking the direction of initial angular momentum as positive:

• Initial angular momentum about the pivot ($L_i$): $L_i = m u \left(\frac{L}{2}\right)$

• Final angular momentum about the pivot ($L_f$): $L_f = -m v \left(\frac{L}{2}\right) + I \omega$

Equating $L_i$ and $L_f$: $m u \left(\frac{L}{2}\right) = -m v \left(\frac{L}{2}\right) + \frac{1}{3} M L^2 \omega$

$m (u + v) \left(\frac{L}{2}\right) = \frac{1}{3} M L^2 \omega$

Solving for $\omega$: $\omega = \frac{3 m (u + v)}{2 M L} \quad \text{--- (Equation 1)}$

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2. Conservation of Kinetic Energy

Since the collision is perfectly elastic, the total kinetic energy before and after the collision is equal: $\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$

$m(u^2 - v^2) = I \omega^2$

Substitute $I = \frac{1}{3} M L^2$ into the energy equation: $m (u - v)(u + v) = \frac{1}{3} M L^2 \omega^2 \quad \text{--- (Equation 2)}$

Substitute $\omega$ from Equation (1) into Equation (2): $m (u - v)(u + v) = \frac{1}{3} M L^2 \left[ \frac{3 m (u + v)}{2 M L} \right]^2$

$m (u - v)(u + v) = \frac{3 m^2 (u + v)^2}{4 M}$

Since $(u + v) \neq 0$, we can cancel $m(u + v)$ from both sides: $u - v = \frac{3 m}{4 M} (u + v)$

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3. Numerical Calculation

Given values:

• $M = 1.00 \text{ kg}$
• $L = 0.20 \text{ m}$
• $m = 0.10 \text{ kg}$
• $u = 5.00 \text{ m s}^{-1}$

Calculate the factor $\frac{3 m}{4 M}$: $\frac{3 m}{4 M} = \frac{3 \times 0.10}{4 \times 1.00} = 0.075$

Now, substitute this into the relation between $u$ and $v$: $u - v = 0.075(u + v)$

$u(1 - 0.075) = v(1 + 0.075)$

$0.925 u = 1.075 v$

$v = \frac{0.925}{1.075} u = \frac{37}{43} \times 5.00 \approx 4.3023 \text{ m s}^{-1} \approx 4.30 \text{ m s}^{-1}$

Now, substitute $v$ back into Equation (1) to find $\omega$: $\omega = \frac{3 \times 0.10}{2 \times 1.00 \times 0.20} (5.00 + 4.3023)$

$\omega = 0.75 \times 9.3023 \approx 6.9767 \text{ rad s}^{-1} \approx 6.98 \text{ rad s}^{-1}$

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Conclusion

• $\omega = 6.98 \text{ rad s}^{-1}$
• $v = 4.30 \text{ m s}^{-1}$

Thus, Option A is the correct statement.