Mix Examples-Work, Energy, Power and Collision Questions in English

(B) The loss in kinetic energy during a perfectly inelastic collision is given by: $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (v_1 + v_2)^2$.
Substituting the given values: $m_1 = 15 \ kg$,$m_2 = 25 \ kg$,$v_1 = 10 \ m/s$,$v_2 = 30 \ m/s$.
$\Delta K = \frac{1}{2} \left( \frac{15 \times 25}{15 + 25} \right) (10 + 30)^2 = \frac{1}{2} \left( \frac{375}{40} \right) (40)^2 = \frac{1}{2} \times 375 \times 40 = 7500 \ J$.
This energy is converted into heat: $Q = (m_1 + m_2) S \Delta T$.
Given $S = 31 \ cal/kg \cdot ^{\circ}C = 31 \times 4.2 \ J/kg \cdot ^{\circ}C = 130.2 \ J/kg \cdot ^{\circ}C$.
$7500 = (15 + 25) \times 130.2 \times \Delta T$.
$7500 = 40 \times 130.2 \times \Delta T$.
$\Delta T = \frac{7500}{5208} \approx 1.44^{\circ}C$.

(B) $1$. Let $l = 1 \ m$ be the length of the rod. The velocity $V_A$ of bob $A$ at the lowest point is given by the conservation of mechanical energy:
$mgl(1 - \cos \theta) = \frac{1}{2} m V_A^2$
$V_A = \sqrt{2gl(1 - \cos 60^{\circ})} = \sqrt{2 \times 10 \times 1 \times (1 - 0.5)} = \sqrt{10} \ m/s$.
$2$. Since the collision between identical bobs $A$ and $B$ is elastic,the velocities are exchanged. Thus,after the collision,bob $B$ moves with velocity $V_B = V_A = \sqrt{10} \ m/s$.
$3$. For bob $B$ to just complete a vertical circular path of radius $R$,the minimum velocity at the bottom of the circular track must be $V_{min} = \sqrt{5gR}$.
$4$. Equating the velocities: $\sqrt{10} = \sqrt{5gR} \implies 10 = 5 \times 10 \times R \implies 10 = 50R \implies R = \frac{10}{50} = \frac{1}{5} \ m$.