Related Formula
Δ K = (1)/(2) m₁m₂m₁ + m₂ urel² (1 - e²)$$\Delta K = \frac{1}{2} \frac{m_{1}m_{2}}{m_{1} + m_{2}} u_{\text{rel}}^{2} (1 - e^{2})$$
Q = msΔ T$Q = ms\Delta T$
Step 1: Loss of Kinetic Energy
For a perfectly inelastic collision, e = 0$e = 0$.
Relative speed urel = v₁ + v₂$u_{\text{rel}} = v_{1} + v_{2}$ (since moving in opposite directions).
urel = 10 + 30 = 40 m/s$u_{\text{rel}} = 10 + 30 = 40 \text{ m/s}$.
Δ K = (1)/(2)((15 × 25)/(15 + 25))(40)²$$\Delta K = \frac{1}{2}\left(\frac{15 \times 25}{15 + 25}\right)(40)^{2}$$
Δ K = (1)/(2)((375)/(40))(1600)$$\Delta K = \frac{1}{2}\left(\frac{375}{40}\right)(1600)$$
Δ K = (1)/(2) × 375 × 40 = 7500 J$$\Delta K = \frac{1}{2} \times 375 \times 40 = 7500 \text{ J}$$
Step 2: Heat Conversion
This lost energy turns into heat Q$Q$. The combined mass is m = 15 + 25 = 40 kg$m = 15 + 25 = 40 \text{ kg}$.
Specific heat s = 31 cal/kg·°C = 31 × 4.2 J/kg·°C = 130.2 J/kg·°C$s = 31 \text{ cal/kg}\cdot^{\circ}\text{C} = 31 \times 4.2 \text{ J/kg}\cdot^{\circ}\text{C} = 130.2 \text{ J/kg}\cdot^{\circ}\text{C}$.
Q = mtotal · s · Δ T$$Q = m_{\text{total}} \cdot s \cdot \Delta T$$
7500 = 40 × 130.2 × Δ T$$7500 = 40 \times 130.2 \times \Delta T$$
Step 3: Calculate Temperature Rise
Δ T = (7500)/(40 × 130.2)$$\Delta T = \frac{7500}{40 \times 130.2}$$
Δ T = (750)/(520.8) ≈ 1.44°C$$\Delta T = \frac{750}{520.8} \approx 1.44^{\circ}\text{C}$$
Pattern Recognition
Sees: "inelastic collision" + "rise in temperature" → Lost kinetic energy = Heat gained. Use reduced mass relative velocity shortcut to find Δ K$\Delta K$ instantly instead of conserving momentum first.
Chapter Mix
Class 11 Physics: Work, Energy and Power
Class 11 Physics: Thermal Properties of Matter