A bomb of mass 9kg explodes into two pieces of mass 3kg and 6kg . The velocity of 3kg mass is 16 m/s.The K.E Zof 6 kg particle is
Answers
The original mass of the bomb was 9 kg .
The initial velocity of the bomb with that mass was 0 m/s since it was at rest position initially .
m = 9 kg .
u = 0 m/s .
Now the bomb is split into two masses of 3 kg and 6 kg .
m1 = 3 kg .
m2 = 6 kg .
Now it is given that the final velocity of the body with mass 3 kg is 16 m/s .
Let the final velocity of the body having mass of 6 kg be x .
By the law of conservation of energy we know that :
Initial momentum = final momentum .
⇒ 9 kg × 0 m/s = ( 3 kg × 16 m/s ) + 6 x kg
⇒ 0 = 48 kg m/s + 6 x kg .
⇒ 6 x kg = - 48 kg m/s
⇒ x = - 48/6
⇒ x = - 8
The final velocity is - 8 m/s .
The negative sign denotes that the piece had travelled the opposite direction to that of the 3 kg particle .
We know that the kinetic energy of a body is given by :
K = 1/2 m v²
⇒ K = 1/2 × 6 kg × ( - 8 m/s )²
⇒ K = 3 kg × 64 m²/s²
⇒ K = 192 kg m²/s²
⇒ K = 192 J
The kinetic energy is 192 J .