In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 k

Question

In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg , moving in opposite directions with speeds of $10 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively, strike each other and stick together. The rise in temperature (in ${ }^{\circ} \mathrm{C}$ ), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material $31 \mathrm{cal} / \mathrm{kg} .{ }^{\circ} \mathrm{C}$ and $1 \mathrm{cal}=4.2 \mathrm{~J}$ )

Professor PC Thomas Classes · Answer

Applying the principle of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$. Let the $25 	ext{ kg}$ mass move in the positive direction. Momentum $P = 25(30) + 15(-10) = 750 - 150 = 600 	ext{ kg}\cdot	ext{m/s}$. The common velocity is $v = \frac{600}{15 + 25} = 15 	ext{ m/s}$. The loss in kinetic energy is $\Delta K = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \frac{1}{2}(m_1+m_2)v^2 = \frac{1}{2}(15)(100) + \frac{1}{2}(25)(900) - \frac{1}{2}(40)(225) = 750 + 11250 - 4500 = 7500 	ext{ J}$. This energy converts to heat $H = \Delta K$. The specific heat is $s = 31 	ext{ cal/kg}\cdot^\circ	ext{C} = 31 	imes 4.2 = 130.2 	ext{ J/kg}\cdot^\circ	ext{C}$. Using $H = ms\Delta T$, we get $\Delta T = \frac{7500}{40 	imes 130.2} = \frac{7500}{5208} \approx 1.44^\circ	ext{C}$.