A fireworks rocket is fired vertically upward. At its maximum height of 70.0 m , it explodes and breaks into two pieces, one with mass mA = 1.30 kg and the other with mass mB = 0.300 kg . In the explosion, 950 J of chemical energy is converted to kinetic energy of the two fragments. What is the speed of each fragment just after the explosion?

The speed of each fragment is as follows

mA = 16.55 m/s

mB = 71.73 m/s.

Explanation:

To solve the question, we note that

Maximum height of fireworks rocket = 70.0 m

mA =  1.30 kg

mB =  0.300 kg

Energy converted = 950 J

Since the available energy to be converted = 950 J

we have Kinetic energy, K.E. = 950 J

However K.E. =  and the rocket breaks into two pieces mA and mB which constitute the mass in the K.E. equation.

That is K.E. = ×mA×v² +  

To find v and v, we note that from the principle of conservation of linear momentum we have at the time of explosion;

mA×v = mA×v

1.30 kg×v = 0.300 kg×v

That is

v = ×v

Therefore

K.E. = ×mA×v² +  

950 J = ×1.3 kg×v² +  

=   ×1.3 kg×(×v)² +  

= ×v² +  ×v

∴  v² 950 J × =5145.83 m²/s²

v = = 71.73 m/s

Therefore, since v = ×v

v = × 71.73 m/s = 16.55 m/s.