Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart and they then fly off in opposite directions, free of the spring. The mass of A is 3.00 times the mass of B, and the energy stored in the spring was 65 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B?
My wrong answers were 10.8333 and 54.1667
total momentum still = 0
A goes at speed u left
B goes at speed v to right
3 m u = m v
so
v = 3 u
(1/2) (3m)u^2 + (1/2)mv^2 = 65
m [ 3 u^2 + v^2 ] = 130
m[ 3u^2 + 9u^2 ] = 130
12 m u^2 = 130
m u^2 = 10.8
so
(1/2) m u^2 = 5.4 Joules
(1/2)(3mu^2) = 16.25 Joules = Ke of A
(1/2) mv^2 = (1/2) m (9 u^2 ) = 9*5.4 = 48.75 Joules = Ke of B
To determine the kinetic energies of particle A and particle B, we need to understand the conservation of energy principle. According to this principle, the total mechanical energy of a system (in this case, the spring and the two particles) remains constant if no external forces are acting on it.
The total mechanical energy in this system is the sum of the potential energy stored in the spring and the kinetic energy of particle A and particle B. Initially, all of the energy is stored in the compressed spring. Then, when the spring is released, all of the energy is transferred between the particles as they move apart.
Let's denote the mass of particle A as mA and the mass of particle B as mB. Given that the mass of particle A is 3.00 times the mass of particle B, we can write mA = 3mB.
The energy stored in the spring is given as 65 J. This is equal to the initial potential energy of the spring. At this point, the kinetic energy of particle A and particle B is zero.
When the spring is released, the potential energy is fully converted into kinetic energy. Since the system is isolated, the total mechanical energy before and after the transfer of energy remains the same. Therefore, we can write the equation:
Potential energy before = Kinetic energy of A + Kinetic energy of B
65 J = KE(A) + KE(B)
Since particle A and particle B fly off in opposite directions, their velocities will have the same magnitude but opposite signs. The kinetic energy (KE) can be calculated using the formula:
KE = 0.5 * mass * velocity^2
Let's denote the velocity of particle A as vA and the velocity of particle B as vB.
Substituting the known values, we have:
65 J = 0.5 * mA * vA^2 + 0.5 * mB * vB^2
Substituting mA = 3mB, we can rewrite the equation as:
65 J = 0.5 * (3mB) * vA^2 + 0.5 * mB * vB^2
To solve this equation, we need an additional relationship between vA and vB. One possibility is the conservation of momentum, which states that the total momentum before and after the transfer remains the same if no external forces are acting. The momentum (p) is defined as mass (m) multiplied by velocity (v):
Total momentum before = Total momentum after
mA * vA = mB * vB
Substituting mA = 3mB, we have:
3mB * vA = mB * vB
Dividing both sides by mB, we get:
3vA = vB
Now we can express vB in terms of vA:
vB = 3vA
Substituting vB = 3vA into the equation 65 J = 0.5 * (3mB) * vA^2 + 0.5 * mB * vB^2, we have:
65 J = 0.5 * (3mB) * vA^2 + 0.5 * mB * (3vA)^2
Simplifying the equation, we get:
65 J = 1.5 * mB * vA^2 + 4.5 * mB * vA^2
65 J = 6 * mB * vA^2
Dividing both sides by 6mB, we have:
10.8333 J = vA^2
Taking the square root of both sides, we find:
vA = √10.8333 ≈ 3.29 m/s
Knowing vA, we can find vB using the relationship vB = 3vA:
vB = 3 * 3.29 ≈ 9.87 m/s
Now we can calculate the kinetic energies:
Kinetic energy of A = 0.5 * mA * vA^2
= 0.5 * 3mB * (3.29^2)
≈ 53.15 J
Kinetic energy of B = 0.5 * mB * vB^2
= 0.5 * mB * (9.87^2)
≈ 484.23 J
Therefore, the correct answers are:
(a) The kinetic energy of particle A is approximately 53.15 J.
(b) The kinetic energy of particle B is approximately 484.23 J.