In a perfectly inelastic collision, two spher | vedclass

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(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

Vedclass · Accepted Answer

$1.44$

Vedclass · Answer

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

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