In: Physics
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.
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