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QPage 47: Q15: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q15 Solution: The blocks of masses 10 kg and 20 kg moving at speeds of and…

The blocks of masses 10 kg and 20 kg moving at speeds of 10 m s−110 \text{ m s}^{-1} and 20 m s−120 \text{ m s}^{-1} respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.

Step-by-Step Verified Solution

**Answer:** Thermal energy developed = 3000 J. **Step-by-Step Explanation:** Given: m1=10m_1 = 10 kg, u1=10u_1 = 10 m/s; m2=20m_2 = 20 kg, u2=20u_2 = 20 m/s (opposite directions). Completely inelastic collision. Conservation of momentum:
m2u2−m1u1=(m1+m2)vm_2u_2 - m_1u_1 = (m_1 + m_2)v
400−100=30v⇒v=10 m/s400 - 100 = 30v \Rightarrow v = 10 \text{ m/s}
Initial KE:
KEi=12×10×102+12×20×202=500+4000=4500 JKE_i = \frac{1}{2} \times 10 \times 10^2 + \frac{1}{2} \times 20 \times 20^2 = 500 + 4000 = 4500 \text{ J}
Final KE:
KEf=12×30×102=1500 JKE_f = \frac{1}{2} \times 30 \times 10^2 = 1500 \text{ J}
Thermal energy = 4500−1500=30004500 - 1500 = 3000 J.[citation:2]

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