An unstable nucleus with a mass of 18.0 ✕ 10-27 kg initially at rest disintegrates into three particles. One of the particles, of mass 4.1 ✕ 10-27 kg, moves along the positive y-axis with a speed of 8.0 ✕ 106 m/s. Another particle, of mass 7.7 ✕ 10-27 kg, moves along the positive x-axis with a speed of 4.0 ✕ 106 m/s. (Assume that mass is conserved.)
(a) Determine the third particle's speed.
(b) What is the direction of motion?
(from the positive x-axis)
To solve this problem, we will use the principles of conservation of momentum and conservation of kinetic energy.
(a) Determine the third particle's speed:
We need to find the speed of the third particle. Let's assume its mass is m and its speed is v.
Conservation of momentum tells us that the total momentum before the disintegration is equal to the total momentum after the disintegration.
Before disintegration (initially at rest) :
Total momentum = 0 kg.m/s
After disintegration:
The first particle, with a mass of 4.1 ✕ 10-27 kg, moves along the positive y-axis with a speed of 8.0 ✕ 106 m/s.
Momentum of the first particle = mass x velocity = (4.1 ✕ 10-27 kg) x (8.0 ✕ 106 m/s)
The second particle, with a mass of 7.7 ✕ 10-27 kg, moves along the positive x-axis with a speed of 4.0 ✕ 106 m/s.
Momentum of the second particle = mass x velocity = (7.7 ✕ 10-27 kg) x (4.0 ✕ 106 m/s)
The third particle, with a mass of m and a speed of v.
Momentum of the third particle = mass x velocity = m x v
According to conservation of momentum:
(4.1 ✕ 10-27 kg) x (8.0 ✕ 106 m/s) + (7.7 ✕ 10-27 kg) x (4.0 ✕ 106 m/s) + m x v = 0 kg.m/s
Simplifying the equation:
(32.8 ✕ 10-21 kg.m/s) + (30.8 ✕ 10-21 kg.m/s) + m x v = 0 kg.m/s
63.6 ✕ 10-21 kg.m/s + m x v = 0 kg.m/s
Since the initial total momentum is zero, the total momentum after disintegration must also be zero. Therefore, we can solve for the third particle's speed:
m x v = -63.6 ✕ 10-21 kg.m/s
v = -63.6 ✕ 10-21 kg.m/s / m
(b) Determine the direction of motion:
To find the direction of the third particle's motion, we need to consider the directions of the other two particles. The first particle moves along the positive y-axis, and the second particle moves along the positive x-axis.
By convention, we take the direction of the positive x-axis as the reference. Since the second particle moves in the positive x-axis direction, the third particle must move in the opposite direction.
Therefore, the direction of motion of the third particle is opposite to the positive x-axis.
Note: It's important to remember that we cannot determine the exact value of m or v without additional information. We can only provide a relationship between them through the conservation of momentum equation.
To solve this problem, we can apply the principle of conservation of momentum and conservation of kinetic energy.
(a) First, let's calculate the initial momentum and kinetic energy of the system.
The initial momentum is given by the sum of the momenta of the particles, and since the unstable nucleus is initially at rest, its momentum is zero. Therefore, we have:
Initial momentum = 0 + (mass of particle 1 × velocity of particle 1) + (mass of particle 2 × velocity of particle 2)
Initial momentum = (4.1 ✕ 10^-27 kg) × (8.0 ✕ 10^6 m/s) + (7.7 ✕ 10^-27 kg) × (4.0 ✕ 10^6 m/s)
Calculating this expression, we find the initial momentum to be:
Initial momentum = 32.8 ✕ 10^-21 kg⋅m/s + 30.8 ✕ 10^-21 kg⋅m/s
Initial momentum = 63.6 ✕ 10^-21 kg⋅m/s
Next, the initial kinetic energy of the system is given by the sum of the kinetic energies of the particles. Using the formula for kinetic energy (KE = 0.5 × mass × velocity^2), we have:
Initial kinetic energy = 0.5 × (mass of particle 1 × velocity of particle 1)^2 + 0.5 × (mass of particle 2 × velocity of particle 2)^2
Initial kinetic energy = 0.5 × (4.1 ✕ 10^-27 kg) × (8.0 ✕ 10^6 m/s)^2 + 0.5 × (7.7 ✕ 10^-27 kg) × (4.0 ✕ 10^6 m/s)^2
Calculating this expression, we find the initial kinetic energy to be:
Initial kinetic energy = 0.5 × 4.1^2 × 8.0^2 + 0.5 × 7.7^2 × 4.0^2 ✕ 10^-45 J
Initial kinetic energy = 264.32 ✕ 10^-45 J + 118.72 ✕ 10^-45 J
Initial kinetic energy = 383.04 ✕ 10^-45 J
Now, let's apply the principle of conservation of momentum and conservation of kinetic energy to determine the unknowns.
Conservation of momentum:
Final momentum = 0 + (mass of particle 1 × velocity of particle 1) + (mass of particle 2 × velocity of particle 2) + (mass of particle 3 × velocity of particle 3)
Since the unstable nucleus initially at rest, its momentum is zero. Therefore, we have:
Final momentum = 0 + (4.1 ✕ 10^-27 kg) × (8.0 ✕ 10^6 m/s) + (7.7 ✕ 10^-27 kg) × (4.0 ✕ 10^6 m/s) + (mass of particle 3 × velocity of particle 3)
The final momentum must be equal to the initial momentum:
(4.1 ✕ 10^-27 kg) × (8.0 ✕ 10^6 m/s) + (7.7 ✕ 10^-27 kg) × (4.0 ✕ 10^6 m/s) + (mass of particle 3 × velocity of particle 3) = 63.6 ✕ 10^-21 kg⋅m/s
Now we can solve this equation for the velocity of the third particle.
Conservation of kinetic energy:
Since mass is conserved, the initial kinetic energy of the system must equal the final kinetic energy. Therefore, we have:
Initial kinetic energy = Final kinetic energy
0.5 × (mass of particle 1 × velocity of particle 1)^2 + 0.5 × (mass of particle 2 × velocity of particle 2)^2 = 0.5 × (mass of particle 3 × velocity of particle 3)^2
Substituting the known values, we can solve this equation for the velocity of the third particle.
(b) To determine the direction of motion of the third particle, we need to consider the direction of the velocities of the other two particles. Since one particle moves along the positive y-axis and the other moves along the positive x-axis, we can assume that the third particle will move in a direction such that the sum of their components of motion cancels out to zero.