Question

Two 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

$m_1$ = 10 kg, $V_1$ = 10 m/s

$m_2$ = 20 kg, $V_2$ = 20 m/s

Here, $m_2{V}_2-m_1{V}_1=(m_1+m_2)V$

$\Rightarrow 20\times 20-10\times 10$ = (10+20) V

$\Rightarrow$ 400-100 = 30 V

$\Rightarrow$ 300 = 30 V

$\Rightarrow$ V = 10 m/s

Initial kinetic energy

= $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$

= $\frac{1}{2}\times 10\times (10{)}^2+\frac{1}{2}\times 20\times (20{)}^2$

= 500 + 4000 = 4500

Final kinetic energy = $\frac{1}{2}(m_1+m_2)V^2$

=$\frac{1}{2}(10+20)(10{)}^2$

= $\left(\frac{30}{2}\right)\times 100$ = 1500

$\therefore$ The total change in K.E

= 4500- 1500 =3000 J

$\therefore$ The thermal energy developed in the process = 3000 J