Question

A bullet was moving with 200 m/s speed. On its way, it stopped penetrating 7 cm into a jelly block. The mass of the bullet is 10 g, its temperature is 60 degree celcius and specific heat is 200J/kg/K. The mass of the jelly block is 1 kg, temperature is 25 degree celcius and specific heat is 4000 J/kg/K. After the bullet has stopped what will be the temperature of the whole system?
Please show how to solve it if possible

Answer 1

From the conservation of energy:
(mv^2)/2=mc_1 (T-T_1 )+Mc_2 (T-T_2 )
0.01/2 〖200〗^2=(0.01)(200)(T-60)+(1)(4000)(T-25)
T=25.675℃