A particle makes a head-on-collision with an unknown particle at rest.The proton rebounds back with 4/9 of its initial kinetic energy.Find the ratio of mass of unknown particle to mass of proton assuming the collision to be elastic.
9/45
To solve this problem, we can use the conservation of momentum and kinetic energy.
Let's call the mass of the unknown particle "m" and the mass of the proton "mp". Also, let's call the initial velocity of the proton "v" and the final velocity of the proton after the collision "vf".
According to the law of conservation of momentum, the total momentum before and after the collision should be equal. Since the unknown particle is initially at rest, the initial momentum is simply mp * 0 = 0. After the collision, the proton rebounds back, so the final momentum is mp * vf. Therefore, we can write:
mp * vf = 0
Next, let's consider the conservation of kinetic energy. The initial kinetic energy of the proton is given by (1/2) * mp * v^2. After the collision, the final kinetic energy of the proton is (1/2) * mp * vf^2. We are given that the proton rebounds with 4/9 of its initial kinetic energy, so we can write:
(1/2) * mp * vf^2 = (4/9) * (1/2) * mp * v^2
Next, we can simplify the equation above by canceling out (1/2) and mp:
vf^2 = (4/9) * v^2
Now, let's substitute the equation for vf from the conservation of momentum into the equation above:
(0)^2 = (4/9) * v^2
This simplifies to:
0 = (4/9) * v^2
Since v is not zero, we can conclude that the mass of the unknown particle is (4/9) times the mass of the proton.
Therefore, the ratio of the mass of the unknown particle to the mass of the proton is 4/9.