Question

The 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.

Answer 1

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
$K_{\mathrm{i}}=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$
$$\begin{gathered} K_{\mathrm{i}}=\frac{1}{2}\times 10\times {\left(10\right)}^2+\frac{1}{2}\times 20\times {\left(20\right)}^2 \\ {K}_{\mathrm{i}}=500+4000=4500 \end{gathered}$$

Final kinetic energy is given by
$K_{\mathrm{f}}\mathit{=}\frac{1}{2}\left(m_1\mathit{+}{m}_2\right)v^2$
$$\begin{gathered} K_{\mathrm{f}}=\frac{1}{2}\left(10+20\right){\left(10\right)}^2 \\ {K}_{\mathrm{f}}=\left(\frac{30}{2}\right)\times 100=1500 \end{gathered}$$

∴ Total change in KE = 4500 J − 1500 J = 3000 J
Thermal energy developed in the process = 3000 J