In an elastic collision between a proton and a helium atom, the total kinetic energy is conserved. To find the initial and final speeds of the proton, we can use the principles of conservation of momentum and kinetic energy.
Let's assume the initial speed of the proton is v1, and the initial speed of the helium atom is v2. Since the helium atom is initially at rest, v2 equals zero.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision. The momentum of an object is given by the product of its mass and velocity.
Momentum before collision:
p1 = m1 * v1 (proton)
p2 = m2 * v2 (helium, initially at rest)
Since the helium atom is initially at rest, p2 equals zero. Therefore, p1 = m1 * v1.
Momentum after collision:
p1' = m1 * v1' (proton)
p2' = m2 * v2' (helium)
Since the collision is elastic, the total momentum before and after the collision is the same. So, we have p1 = p1'. Therefore, m1 * v1 = m1 * v1'.
From this equation, we can find the final speed of the proton, v1'.
2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Kinetic energy before collision:
KE1 = (1/2) * m1 * v1^2 (proton)
KE2 = (1/2) * m2 * v2^2 (helium, initially at rest)
Since the helium atom is initially at rest, KE2 equals zero. Therefore, KE1 = (1/2) * m1 * v1^2.
Kinetic energy after collision:
KE1' = (1/2) * m1 * v1'^2 (proton)
KE2' = (1/2) * m2 * v2'^2 (helium)
Since the collision is elastic, the total kinetic energy before and after the collision is the same. So, we have KE1 = KE1'. Therefore, (1/2) * m1 * v1^2 = (1/2) * m1 * v1'^2.
From this equation, we can find the final speed of the proton, v1'.
To determine the fraction of the initial energy transferred to the helium atom, we can compare the kinetic energy of the helium atom before and after the collision.
Fraction of initial energy transferred to the helium atom:
E_transferred = (KE2' - KE2) / KE1
By substituting the equations for KE2' and KE2, we have:
E_transferred = (1/2) * m2 * v2'^2 / (1/2) * m1 * v1^2
Since v2 equals zero (initially at rest), the fraction of initial energy transferred simplifies to:
E_transferred = 0 / (1/2) * m1 * v1^2 = 0
Therefore, no energy is transferred to the helium atom in an elastic collision with a proton.
In summary:
- The initial speed of the proton is v1.
- The initial speed of the helium atom is zero.
- The final speed of the proton, v1', can be found by solving m1 * v1 = m1 * v1'.
- The final speed of the helium atom, v2', is zero.
- No fraction of the initial energy is transferred to the helium atom.