QJEEM24Phbbbf: JEE Main 2024 Physics - Class 11 JEEM24Phbbbf Solution: A simple pendulum of length has a wooden bob of mass . It is…

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}}$

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $v'$ . The final momentum $p_{\text{final}}$ is the combined mass of the bullet and bob times the common velocity :

$p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'$

According to the principle of conservation of linear momentum,

$p_{\text{initial}} = p_{\text{final}}$

$m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'$

Plugging in the values :

$(10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v'$

Solving for $v'$ :

$v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s}$

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $h$ that the bob reaches. Using the principle of conservation of energy :

$\text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}}$

$\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh$

Isolating $h$ , we get :

$h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g}$

Substituting the values for $v'$ and $g$ :

$h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m}$

Looking at the given options, the result most closely matches Option A :

$\boxed{0.20 \text{ m}}$

Therefore, the height to which the bob rises before swinging back is approximately $0.20 \text{ m}$ .