How much kinetic energy did each piece acquire after an explosion of an object?

Question

An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other. If 7100 J were released in the explosion, how much kinetic energy did each piece acquire?

NOOO idea... help, please??? Thanks in advance for your time and help.

Answer 1

Total energy of the system must be 7100 J.
Before the explosion, it was 7100 J of potential energy
After the explosion, it's 7100 J of kinetic (assuming none of the energy was lost as waste heat, or waste light, or other sources)

The two pieces of the object have mass 0.4x kg and 0.6x kg (the original body had mass 1.0x kg, assuming none of it was lost in the explosion). The two pieces have a speed after the explosion which is unknown, but we can call the smaller piece v1, and the larger piece v2.

Kinetic Energy = 1/2 m v^2

Total momentum must be conserved for the entire system. (Law of conservation of momentum) Since there are only two pieces, and the original system was at rest (we can assume this, since even if it's not true, the original system's momentum will just be added to the two pieces later) we therefore know that they flew apart in exactly opposite directions, and will have momentum equal to each other.

The formula for momentum is P = mv. So we can set m1v1 = m2v2

So here are the equations:
0.4x (v1) = 0.6x (v2)
7100J = 1/2 (0.4) (v1)^2 + 1/2 (0.6) (v2)^2

Knowing this, we can use algebra on the first equation to turn it into:

v1 = (0.6x / 0.4x) (v2)
the x cancels out, and we get:
v1 = 1.5 v2

Plug that into the first equation and we get:
7100J = 1/2 (0.4x) (1.5 v2)^2 + 1/2 (0.6x) (v2)^2
7100 J = 1/2 (0.4) (x) (2.25) v2^2 + 1/2 (0.6) (x) v2^2
7100 J = 0.45x v2^2 + 0.3x v2^2

Since we don't know x, we can't go any further with the calculations, but we CAN compare part one of the equation (the smaller piece) to part two (the bigger)

Smaller piece gets 0.45x v2^2 energy.
Bigger piece gets 0.3x v2^2 energy.

From this, we can tell what PERCENT of the energy went to the smaller piece.
Percent to smaller = 0.45x v2^2 / (0.3x v2^2 + 0.45x v2^2)
= 0.45 / (0.3 + 0.45)
= 0.45 / 0.75
= 0.6, or 60%

Therefore, the smaller piece got 60% of the total energy. And we know the total energy was 7100 J!

(60%) (7100 J) = 4260 J

The other piece got whatever's left over (40%, or 2840 J)

Answer 2

the total kenetic energy of the system is the energy of the explosion

1) 1/2 * MONE * VONE^^2 + 1/2 * MTWO * VTWO^^2 = 7100

Also the momentum of the system is conserved

2) MONE * VONE + MTWO * VTWO = 0

Also

3) MTWO = 3/2 MONE

Three equtions in 4 unknowns

But we do not have to solve for the mass of each piece, only the kenetic energy of each Piece

Let KEONE = the kenetic energy of MONE

then rewrite eq 1) as

4) KEONE + 1/2 MTWO VTWO ^^2 = 7100

subsittute eq 3) into eq 4)

5) MONE * VONE + 3/2 MONE * VTWO = 0

simplify

6) VONE = -3/2 VTWO

substitute into eq 4)

7) KEONE + 1/2 * MTWO * (-3/2 VONE) ^^2 = 7100

simplify

8) KEONE + 9/4 MTWO * (VONE)^^2 = 7100

substitute in eq 3)

9) KEONE + 9/4 * 3/2 * MONE * (VONE) ^^2 = 7100

rearrange to isolate KEONE

10) KEONE + 27/4 KEONE = 7100

solve for KEONE

11) KEONE = 7100 * 4/31 = 931

rewrite eq 4) as

12) KEONE + KETWO = 7100

solve for KETWO

13) KETWO = 7100 - 931 = 6069

Answer 3

I'll write m1 for the mass of the larger object and m2 for the mass of the smaller, similarly v1 and v2 for their respective velocities after the explosion and u1 and u2 for the initial velocities (which are 0).

We are given m1 = 1.5 m2.

From conservation of momentum we know
m1u1 + m2u2 = m1v1 + m2v2
and thus
0 = (1.5 m2) v1 + m2 v2 = m2 (1.5 v1 + v2)
So v2 = -1.5 v1.

The kinetic energy of the larger piece is KE1 = 1/2 m1 v1^2 and the kinetic energy of the smaller piece is KE2 = 1/2 m2 v2^2. So
KE2 / KE1 = (m2/m1) (v2/v1)^2 = (1/1.5) (-1.5)^2 = 1.5.
So KE2 = 1.5 KE1 and KE1 + KE2 = 7100 J.
This gives us 2.5 KE1 = 7100J and so KE1 = 2840 J, KE2 = 4260 J. So the larger piece has 2840 J and the smaller piece has 4260 J.