physics F6

2009-10-12 1:59 am

a bullet of mass 20g strike a block of mass 0.78kg attached to a spring.
the bullet embeds into the block and they slide across the smooth table. the spring of force constant 2000Nm^-1 has a maximum compression of 10 cm
(a) find the maximum potential energy stored in the spring.
(b) find the speed of the bullet jaust after the impact
(c) fingd the speed of the bullet before the impact
(d) calculate the percentage loss of kinetic enegy during the impact.

更新1:

why 1/2 (M + m)v2 is the loss in ke?

回答 (1)

Prof. Physics

2009-10-12 2:30 am

✔ 最佳答案

a. Maximum potential energy stored in the spring

= 1/2 kx2

= 1/2 (2000)(0.1)2

= 10 J

b. By the law of conservation of energy

Loss of K.E. of the system = Gain in E.P.E.

1/2 (M + m)v2 = 10

1/2 (0.02 + 0.78)v2 = 10

Speed of the bullet just after the impact, v = 5 m/s

c. By the law of conservation of momentum (Since no external force acts on the system.)

mu = (M + m)v

(0.02)u = (0.78 + 0.02)(5)

Speed of the bullet just before impact, u = 200 m/s

d. Initial K.E. = 1/2 mu2 = 1/2 (0.02)(200)2 = 400 J

Final K.E. = E.P.E. stored = 10 J

Percentage loss = (400 - 10)/400 X 100% = 97.5%

參考: Physics king

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