NEET-XII-Physics
25: Calorimetry
- #15The blocks of masses 10 kg and 20 kg moving at speeds of 10 m s-1 and 20 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.Ans : Given:
Mass of the first block, m1 = 10 kg
Mass of the second block, m2 = 20 kg
Initial velocity of the first block, u1 = 10 m/s
Initial velocity of the second block, u2 = 20 m/s
Let the velocity of the blocks after collision be v.
Applying conservation of momentum, we get
m2u2 - m1u1 = (m1 + m2)v
⇒ 20 × 20 - 10 × 10 = (10 + 20)v
⇒ 400 - 100 = 30 v
⇒ 300 = 30 v
⇒ v = 10 m/s
Initial kinetic energy is given by
`` {K}_{\,\mathrm{\,i\,}}=\frac{1}{2}{m}_{1}{u}_{1}^{2}+\frac{1}{2}{m}_{2}{u}_{2}^{2}``
`` {K}_{\,\mathrm{\,i\,}}=\frac{1}{2}\times 10\times {\left(10\right)}^{2}+\frac{1}{2}\times 20\times {\left(20\right)}^{2}``
`` {K}_{\,\mathrm{\,i\,}}=500+4000=4500``
Final kinetic energy is given by
`` {K}_{\,\mathrm{\,f\,}}\mathit{=}\frac{1}{2}\left({m}_{1}\mathit{+}{m}_{2}\right){v}^{2}``
`` {K}_{\,\mathrm{\,f\,}}=\frac{1}{2}\left(10+20\right){\left(10\right)}^{2}``
`` {K}_{\,\mathrm{\,f\,}}=\left(\frac{30}{2}\right)\times 100=1500``
∴ Total change in KE = 4500 J - 1500 J = 3000 J
Thermal energy developed in the process = 3000 J
Page No 47: