A proton of mass m undergoes a head on collision with a stationary atom of mass 20m. If the initial speed of the proton is 485 m/s, find the speed of the proton after the collision.
momentum conserved:
m*485=Vatom*20m+Vproton*m
so two unknowns, we need another equation. So we use use conservation of energy
1/2 m 485^2=1/2 *20m*Vatom^2 + 1/2 *m*Vpro^2
use the first equation to solve for Vatom in terms of Vproton, put that expression into the second equation (a bit of algebra ensures) and solve for Vproton
To find the speed of the proton after the collision, we can apply the principle of conservation of momentum and kinetic energy.
1. Conservation of momentum:
In a collision, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can express it as:
momentumbefore = momentumafter
Momentum (p) is calculated as the product of mass (m) and velocity (v):
momentum = mass * velocity
Considering the collision is a head-on collision, we can write the momentum equation as:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
In this case, m1 is the mass of the proton, v1_initial is the initial velocity of the proton, m2 is the mass of the stationary atom, v2_initial is the initial velocity of the stationary atom (which is 0 since it's stationary), v1_final is the final velocity of the proton after the collision, and v2_final is the final velocity of the stationary atom after the collision.
2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, we can express it as:
KE_before = KE_after
The kinetic energy (KE) of an object is calculated as half of the mass multiplied by the square of the velocity:
KE = (1/2) * mass * velocity^2
Since the atom is initially at rest, its initial kinetic energy is 0. Thus, we only consider the kinetic energy of the proton before and after the collision:
(1/2) * m1 * v1_initial^2 = (1/2) * m1 * v1_final^2 + 0
Now we have two equations (momentum and kinetic energy) and two unknowns (v1_final and v2_final), so we can solve for v1_final.
Let's plug in the given values:
m1 = m (mass of the proton)
v1_initial = 485 m/s
m2 = 20m (mass of the atom)
v2_initial = 0 m/s
m * 485 + 20m * 0 = m * v1_final + 20m * v2_final
(m * 485)^2 = (m * v1_final)^2
Simplifying the equations, we have:
485m = v1_final + 20v2_final (equation 1)
(485)^2 = v1_final^2 (equation 2)
Now we can solve these two equations simultaneously to find v1_final.
From equation 2, v1_final = 485 m/s (since 485^2 = v1_final^2)
Plugging this value back into equation 1, we have:
485m = 485 + 20v2_final
Solving for v2_final:
20v2_final = 485m - 485
v2_final = (485m - 485) / 20
Thus, the speed of the proton after the collision is v1_final = 485 m/s, and the speed of the stationary atom after the collision is v2_final = (485m - 485) / 20.