A proton traveling at 10^7m/s collides with a stationary particles and bounces back at 2*10^6m/s.if the particle moves forward at 3*10^6m/s.find the mass of the stationary particles.(m1=1.67*10^-27kg)
To find the mass of the stationary particle, we can use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. In a collision, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.
Let's denote the mass of the proton as m2 and the mass of the stationary particle as M. We are given the following information:
Initial velocity of the proton (u1) = 10^7 m/s
Final velocity of the proton (v1) = -2*10^6 m/s (Since it bounces back, the velocity is negative)
Velocity of the particle after the collision (v2) = 3*10^6 m/s
Using the principle of conservation of momentum, we can form the equation:
m1u1 + m2u2 = m1v1 + m2v2
Substituting the given values:
(1.67*10^-27 kg)(10^7 m/s) + m2(0 m/s) = (1.67*10^-27 kg)(-2*10^6 m/s) + m2(3*10^6 m/s)
We assume that the stationary particle is initially at rest, so its initial velocity u2 is 0 m/s.
Simplifying the equation:
(1.67*10^-27 kg)(10^7 m/s) = (1.67*10^-27 kg)(-2*10^6 m/s) + m2(3*10^6 m/s)
1670 kg m/s = -334 kg m/s + m2(3*10^6 m/s)
Now let's solve for m2:
2004 kg m/s = m2(3*10^6 m/s)
m2 = (2004 kg m/s) / (3*10^6 m/s)
m2 ≈ 0.000668 kg ≈ 6.68 * 10^-4 kg
Therefore, the mass of the stationary particle is approximately 6.68 * 10^-4 kg.