An internal explosion breaks an object, initially at rest, into two pieces, one of which has 2 times the mass of the other. If 22500 J is released in the explosion, how much kinetic energy does each piece acquire?
The lesser mass will have twice the velocity.
KE= 1/2mv^2 +1/2 2M (v/2)^2=22500
1/2mv^2+ 1/4mv^2=
so the smaller piece has twice as much energy than the larger
so smaller has 2/3 22500 Joules
It is not necessary that all of the energy of the explosion be converted to kinetic energy of the two pieces. There will usually also be heating of the pieces, and the release of gas.
IF 22500 J is released in the explosion as kinetic energy of the only two pieces, then the approach of BobPursley will yield the answer.
To determine the kinetic energy acquired by each piece, we need to use the principle of conservation of momentum. The momentum before the explosion is equal to the momentum after the explosion since there are no external forces acting on the system.
Let's denote the mass of the smaller piece as m and the mass of the larger piece as 2m. Since the object is initially at rest, the total initial momentum is 0.
Let the velocity of the smaller piece after the explosion be v1 and the velocity of the larger piece be v2. According to the law of conservation of momentum:
(momentum before explosion) = (momentum after explosion)
0 = (m x v1) + (2m x v2)
Now, we need to express the velocities in terms of the given information. The kinetic energy acquired by an object can be calculated using the equation:
Kinetic energy = (1/2) x (mass) x (velocity)^2
Since we want to find the kinetic energy acquired by each piece, we can set up two equations:
Kinetic energy of smaller piece = (1/2) x m x v1^2
Kinetic energy of larger piece = (1/2) x 2m x v2^2
Since the total energy released in the explosion is 22,500 J, we have:
Kinetic energy of smaller piece + Kinetic energy of larger piece = 22,500 J
Substituting the expressions for kinetic energy, we get:
(1/2) x m x v1^2 + (1/2) x 2m x v2^2 = 22,500 J
We can simplify this equation by canceling out the common term of m:
(1/2) x v1^2 + v2^2 = 22,500 J
Now we have two equations:
(m x v1) + (2m x v2) = 0
(1/2) x v1^2 + (1/2) x 2v2^2 = 22,500 J
Simplifying the first equation:
m x v1 + 2m x v2 = 0
v1 + 2v2 = 0
We can solve this equation for v1 in terms of v2:
v1 = -2v2
Substituting this value into the second equation:
(1/2) x (-2v2)^2 + (1/2) x 2v2^2 = 22,500 J
Simplifying the equation:
(1/2) x (4v2^2) + (1/2) x 2v2^2 = 22,500 J
2v2^2 + 2v2^2 = 22,500 J
4v2^2 = 22,500 J
v2^2 = 22,500 J / 4
v2^2 = 5,625 J
v2 = √(5,625 J)
v2 ≈ 75 J
Using the relationship v1 = -2v2:
v1 = -2 x 75 J
v1 = -150 J
Now that we have the velocities v1 and v2, we can substitute these values into the equations for kinetic energy:
Kinetic energy of smaller piece = (1/2) x m x v1^2
Kinetic energy of larger piece = (1/2) x 2m x v2^2
Plugging in the values:
Kinetic energy of smaller piece = (1/2) x m x (-150 J)^2
Kinetic energy of smaller piece = (1/2) x m x 22,500 J
Kinetic energy of smaller piece = 11,250 J x m
Kinetic energy of larger piece = (1/2) x 2m x (75 J)^2
Kinetic energy of larger piece = (1/2) x 2m x 5,625 J
Kinetic energy of larger piece = 5,625 J x m
Therefore, the smaller piece acquires 11,250 J x m of kinetic energy, and the larger piece acquires 5,625 J x m of kinetic energy.