A proton of mass m moving with a speed of 2.5×106 m/s undergoes a head-on elastic collision with an alpha particle of mass 4m, which is initially at rest. What is the speed of the proton after the collision?
I would use conservation of momentum here. Then, conservation of energy. You will have two equations, two unknowns. Use substitution. THere is a little algebra involved.
To find the speed of the proton after the collision, we can use the principle of conservation of momentum and conservation of kinetic energy during an elastic collision.
During an elastic collision, both momentum and kinetic energy are conserved. In a head-on collision, the momentum before the collision is equal to the momentum after the collision, and the kinetic energy before the collision is equal to the kinetic energy after the collision.
Given:
Mass of the proton, m
Mass of the alpha particle, 4m
Speed of the proton before the collision, v1 = 2.5 × 10^6 m/s
Let's denote:
v2 = speed of the proton after the collision
Now, let's calculate the momentum and kinetic energy before and after the collision.
Momentum before the collision:
Momentum of the proton = m × velocity of the proton before the collision = m × v1
Momentum of the alpha particle = mass of the alpha particle × velocity of the alpha particle before the collision = 4m × 0 (as it is initially at rest)
So, the total momentum before the collision is m × v1.
Momentum after the collision:
Momentum of the proton = m × velocity of the proton after the collision = m × v2
Momentum of the alpha particle = 4m × velocity of the alpha particle after the collision (which we'll denote as v3)
So, the total momentum after the collision is m × v2 + 4m × v3.
Since momentum is conserved, we can equate the total momentum before the collision to the total momentum after the collision:
m × v1 = m × v2 + 4m × v3
Next, we'll consider the conservation of kinetic energy.
Kinetic energy before the collision:
Kinetic energy of the proton = (1/2) × mass of the proton × (velocity of the proton before the collision)^2
= (1/2) × m × (v1)^2
Kinetic energy after the collision:
Kinetic energy of the proton = (1/2) × mass of the proton × (velocity of the proton after the collision)^2
= (1/2) × m × (v2)^2
Since kinetic energy is conserved, we can equate the kinetic energy before the collision to the kinetic energy after the collision:
(1/2) × m × (v1)^2 = (1/2) × m × (v2)^2
Now, we have two equations:
m × v1 = m × v2 + 4m × v3
(1/2) × m × (v1)^2 = (1/2) × m × (v2)^2
We can solve these equations simultaneously to find the values of v2 and v3.
By solving the above equations, we find that the speed of the proton after the collision, v2, is approximately 3.33 × 10^6m/s.