An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other. If 7200 J wer?

An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other. If 7200 J were released in the explosion, how much kinetic energy did each piece acquire?

1st this is a momentum problem
Momentum = mass * velocity
Let mass of P1 = m1
Mass of P2 = 1.5* m1
Momentum of P1 = m1 * v1
Momentum of P2 = 1.5m1 * v2

Momentum is always conserved

Initial momentum = 0, since the object was at rest before the explosion
Final momentum = m1 * v1 + 1.5m1 * v2
m1 * v1 + 1.5m1 * v2 = 0

m1 * v1 = -1.5m1 * v2 = 0
divide both sides b m1

v1 = -1.5v2

Kinetic energy = ½ * mass * velocity^2

Mass of P1 = m1
Velocity of P1 = v1
KE of P1 = ½ * m1 * v1^2

Now we need the mass and velocity of P2 in terms of the mass and velocity of P1

Mass of P2 = 1.5m1
Velocity of P2 = -v1/1.5 = -?v1
KE of P2 = ½ * 1.5m1 * (-?v1)^2 =
KE of P2 = ½ * 1.5m1 * -4/9v1^2
KE of P2 = ? * m1 * v1^2

KE1 + KE2 =7200 J

Let?s see what the ratio of KE1 to KE2 is?
KE1/KE2 = ½ /? = 1.5
This means KE1 = 1.5 * KE2
1.5 KE2 + KE2 =7200 J
2.5 KE2 = 7200J
KE2 = 2880 J
KE1 = 7200 ? 2880 = 4320 J

Check:
4320 / 2880 = 1.5

Well, what do you know about that!!! The only numbers given in this problem are 1.5 and 7200

The KE equation = ½ * mass * velocity^2
The mass of P2 = 1.5* m1
v1 = -1.5v2
v2 = v1 ÷ -1.5
The velocity of P2 = v1 ÷ -1.5
The velocity is squared, so velocity^2 of P2 = (v1 ÷ -1.5)^2
The velocity^2 of P2 = v1^2 ÷ (-1.5^2)

The mass of P2 = 1.5 times the mass of P1
The velocity of 2 = 1 ÷ 1.5^2 time the velocity of P1

So, the KE of P2 = 1.5 * 1/(1.5^2) * KE of P1 = 1/1.5 * KE of P1

So, when a particle explodes into 2 particles; and the mass of particle #1 is x times the mass of Particle #2; the KE of the particle #1 will be  1/x times the KE of the particle.