✔ 最佳答案
m1 = 5 kg, u1 = +2 m/s, m2 = 7.5 kg, v1 = 0 m/s, u2 = ?, v2 = ?
(a) Since the collision is elastic, kinetic energy is conserved
m1u1 +m2v1 = m1u2 + m2v2 -------------- (1)
½m1(u1^2) + ½ m2(v1^2) = ½m1(u2^2) + ½m2(v2^2) ----------- (2)
From equation (1),
(5kg)(2m/s) +(7.5kg)(0 m/s) = (5kg)u2 + (7.5kg)v2 -------------- (3)
10 + 0 = 5u2 + 7.5v2
2 = u2 + 1.5v2
u2 = 2 - 1.5v2 -------------------------------------- (4)
From equation (2)
½(5)(2^2) + ½ (7.5)(0^2) = ½(5)(u2^2) + ½(7.5)(v2^2)
20 + 0 = (5)(u2^2) + (7.5)(v2^2) --------------- (5)
Two equations (4) and (5) with 2 unknown, u2 and v2, solve for the unknown
Substitute equation (4) into equation (5)
20 = 5(2 - 1.5v2)^2 + (7.5)(v2^2)
Multiply the above equation by 2
40 = 10(2 - 1.5v2)^2 + 15(v2^2)
Then divide by 5(v2^2)
8 = 2(2 - 1.5v2)^2 + 3(v2^2)
8 = 2[4 - 6v2 + 2.25(v2^2)] + 3(v2^2)
8 = 8 - 12v2 + 4.5(v2^2) + 3(v2^2)
0 = - 12v2 +7.5(v2^2)
Multiply by 2
15(v2^2) - 24v2
3v2(5v2 – 8) = 0
5v2 – 8 = 0
v2 = 8/5 = 1.6 m/s
From equation (4)
u2 = 2 - 1.5v2 = 2 – 1.5(1.6)
u2 = - 0.4 m/s in the opposite direction
After collision, 5 kg ball moves with 0.4 m/s in opposite direction and 7.5 kg ball moves with 1.6 m/s in the same direction. (5 kg ball is bounced backward)
(b) Since the collision is inelastic, some kinetic energy is lost
Let v3 be the final velocity of both balls
5kg)(2m/s) +(7.5kg)(0 m/s) = (5kg)v3 + (7.5kg)v3
10 = 12.5 v3
v3 = 10/12.5 =
v3 = 0.8 m/s in the same direction
The final velocity of both balls is 0.8 m/s in the same direction
2011-02-23 04:07:38 補充：
In part (a), no where did the question mention that final velocity of 5 kg ball is zero. The assumption is wrong. It is also wrong that kinetic energy before collision (10 J) and after collision (6.66 J) is different. Kinetic energy must be conserved for elastic collision.
