Question

An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 2.4 times the mass of piece B. The energy of 6400 J is released in the explosion. Part A Determine the kinetic energy of piece A after the explosion. Express your answer to two significant figures and include the appropriate units. Part B Determine the kinetic energy of piece B after the explosion. Express your answer to two significant figures and include the appropriate units.

Answer 1

Initial momentum = 0      Since it is initially at rest.

Mass of A = 2.4 x mass of B

            M = 2.4 m

From law of conservation of momentum , MV + mv = 0

2.4 mV + mv = 0

          2.4 mV = - mv

             2.4 V = - v

                   v = -2.4 V

Kinetic energy of B is K = (1/2) mv ²

Kinetic energy of A is K ' = ( 1/2) MV ²

                                    = (1/2)x2.4 m x (-v/2.4) ²

                                    = (1/2.4) (1/2)mv ²

                                    = K /2.4

Total kinetic energy of A and B is = K + (K/2.4)

                                                 = K + 0.4166 K

                                                  = 1.4166 K

According to problem , 1.4166 K = 6400 J

                                             K = 6400 J / 1.4166

                                                 = 4517.64 J

                                                 = 4500 J

Kinetic energy of A is = K /2.4

                                 = 1882.3 J

                                  = 1900 J

Kinetic energy of B = 4500 J