標題:
who can help me this physic
發問:
A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle 2 = 44° from the neutron's initial direction. The neutron's initial speed is 6.0 * 10^5 m/s. Determine the angle at which the... 顯示更多 A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle 2 = 44° from the neutron's initial direction. The neutron's initial speed is 6.0 * 10^5 m/s. Determine the angle at which the neutron rebounds, angle 1, (1)measured from its initial direction. (2)What is the speed of the neutron after the collision? (3)What is the speed of the helium nucleus after the collision?
Since the collision is elastic, the problem satisfy the conservation of momentum and energy. Let the final speed of neutron and helium be u and v. Let the angle 2 be z (opposite rotational direction of angle 1) conservation of energy: u^2 + 4v^2 = 6x10^5 ------------------------(1) conservation of horizontal momentum u cos z + 4v cos 44° = 6x10^5 u cos z = 6x10^5 - 4v cos 44° ------------------------(2) conservation of vertical momentum u sin z = 4v sin 44° -----------------------------------(3) (2)^2 + (3)^2 u^2 = 16v^2 - 8v cos 44° x 6x10^5 + (6x10^5)^2 sub. into (1) 20v^2 - 8v cos 44° x 6x10^5 = 0 v = 172641.5521 m/s or v = 0 m/s (rejected) sub. v = 172641.5521 m/s into (1) u = 490692.9569 m/s sub. the values of u and v into (3) z = 77.85343611° Therefore, angle 2 is 77.85343611° the speed of the neutron is 490692.9569 m/s the speed of the helium nucleus is 172641.5521 m/s
其他解答:
who can help me this physic
發問:
A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle 2 = 44° from the neutron's initial direction. The neutron's initial speed is 6.0 * 10^5 m/s. Determine the angle at which the... 顯示更多 A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle 2 = 44° from the neutron's initial direction. The neutron's initial speed is 6.0 * 10^5 m/s. Determine the angle at which the neutron rebounds, angle 1, (1)measured from its initial direction. (2)What is the speed of the neutron after the collision? (3)What is the speed of the helium nucleus after the collision?
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最佳解答:Since the collision is elastic, the problem satisfy the conservation of momentum and energy. Let the final speed of neutron and helium be u and v. Let the angle 2 be z (opposite rotational direction of angle 1) conservation of energy: u^2 + 4v^2 = 6x10^5 ------------------------(1) conservation of horizontal momentum u cos z + 4v cos 44° = 6x10^5 u cos z = 6x10^5 - 4v cos 44° ------------------------(2) conservation of vertical momentum u sin z = 4v sin 44° -----------------------------------(3) (2)^2 + (3)^2 u^2 = 16v^2 - 8v cos 44° x 6x10^5 + (6x10^5)^2 sub. into (1) 20v^2 - 8v cos 44° x 6x10^5 = 0 v = 172641.5521 m/s or v = 0 m/s (rejected) sub. v = 172641.5521 m/s into (1) u = 490692.9569 m/s sub. the values of u and v into (3) z = 77.85343611° Therefore, angle 2 is 77.85343611° the speed of the neutron is 490692.9569 m/s the speed of the helium nucleus is 172641.5521 m/s
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