Answer to Question #79365 in Physics for Sayem

Question

A bullet was moving with 200 m/s speed. On its way, it stopped after penetrating 7 cm into a jelly block. The mass of the bullet is 10 g, its temperature is 60 degree Celcius and the specific heat is 200 J / kg / K. Also, mass of the jelly block is 1 kg, its temperature 25 degree celcius, specific heat is 4000 J / kg / k. What will the final temperature of the system after the bullet is stopped?

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℃