a particle having a speed of 1.0x10^6m/s collisdes with a stationary proton which gains an initial speed of 1.60x10^6 in the direction in which the particle is travelling. what is the speed of the particle immediately after collision? how much energy is gained by the proton in the collision. it is know that this collision is perfectly elastic. expalin what this means ( Ma= 6.64x10^-22 kg, Mp= 1.66x10^-27)

momentum is conserved , and in this case, energy also (perfectly elastic)

you're given the masses and initial velocity of the incident particle

the particle transfers energy (and momentum) to the proton

Given:

M1 = 6.64*10^-22, V1 = 1.0*10^6 m/s,
M2 = 1.66*10^-27, V2 = 0,
V3 = Velocity of M1(particle) after the collision,
V4 = 1.60*10^6 m/s = Velocity of M2(proton) after collision,

Momentum before = Momentum after,
A. M!*V1 + M2*V2 = M1*V3 + M2*V4,
6.64*10^-22 * 1.0*10^6 + M2*0 = 6.64*10^-22 * V3 +1.66*10^-27 * 1.6*10^6
6.64*10^-16 = 6.64*10^-22*V3 + 2.66*10^-21,
3.984*10^-37 = 6.64*10^-22*V3,
V3 = 6.0*10^-16 m/s.

B. KE = 0.5*M2*V4^2 = Energy gained by proton.

To find the speed of the particle immediately after the collision, we can use the law of conservation of momentum. According to this law, the total momentum before the collision must equal the total momentum after the collision.

Let's assign variables to the given information:
v1 = speed of the particle before the collision = 1.0 x 10^6 m/s
v2 = speed of the proton after the collision = 1.60 x 10^6 m/s
m1 = mass of the particle = 6.64 x 10^-22 kg
m2 = mass of the proton = 1.66 x 10^-27 kg

Using the conservation of momentum equation:
m1v1 + m2(0) = m1v1' + m2v2

Since the particle is initially moving, m2(0) (the momentum of the proton before the collision) can be ignored.

Substituting the known values:
(6.64 x 10^-22 kg)(1.0 x 10^6 m/s) = (6.64 x 10^-22 kg)v1' + (1.66 x 10^-27 kg)(1.60 x 10^6 m/s)

Simplifying the equation gives:
6.64 x 10^-16 kg·m/s = (6.64 x 10^-22 kg)v1' + 2.66 x 10^-21 kg·m/s

Now we can solve for v1':

(6.64 x 10^-22 kg)v1' = 6.64 x 10^-16 kg·m/s - 2.66 x 10^-21 kg·m/s
v1' = (6.64 x 10^-16 kg·m/s - 2.66 x 10^-21 kg·m/s) / (6.64 x 10^-22 kg)

Calculating this gives:
v1' ≈ 10^6 m/s

Therefore, the speed of the particle immediately after the collision is approximately 1.0 x 10^6 m/s.

To calculate the energy gained by the proton in the collision, we can use the equation for kinetic energy:
KE = 1/2 mv^2

Using the given values:
m = 1.66 x 10^-27 kg
v = 1.60 x 10^6 m/s

Plugging these values into the equation:
KE = 1/2 (1.66 x 10^-27 kg) (1.60 x 10^6 m/s)^2

Simplifying and calculating gives:
KE ≈ 2.13 x 10^-14 J

Therefore, the proton gains approximately 2.13 x 10^-14 Joules of energy in the collision.

A perfectly elastic collision is one in which both momentum and kinetic energy are conserved. In this type of collision, the total momentum before the collision is equal to the total momentum after the collision, and the total kinetic energy is also conserved.

In the given problem, the collision between the particle and the proton is stated to be perfectly elastic. This means that the total momentum of the system (particle + proton) is conserved, and the total kinetic energy is also conserved. The resulting speeds of the particle and the proton after the collision are determined by these conservation principles.