A bullet of mass 4.2 10⁻² kg , moving at a speed of 300 ms ⁻¹ , gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be

Step-by-step solution

Let mass of bullet = m Mass of block = M Velocity of bullet =v=300 m / s Velocity of combined system M+m=V Here, fhom momentum conservation M+m m V=vv= 300 42 10⁻⁴ 4.2 10⁻⁴+9 (4.2 10⁻⁴ ) 30 ms Now, heat produced = Loss in kinetic energy of bullet array l = 1 2 m²- 1 2 (M+m) V² = 1 2 4.2 10⁻⁴(300)²- 1 2 (4.2 10⁻² . +9 4.2 1)(30)² =63 270 =1701 J = 1701 4.2 Cal array =405 Cal