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
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)
2 answers
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.
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.
1 answer