A)A bullet of mass 50g is shoot with speed v towards a hanging wooden block.Mass of the block is 1.5kg.The bullet embeds in the block and the block swings up a maximum height of 10 cm.Estimate the speed of the bullet.
B)10 kg particle slides down from rest along rough track ABC.AB is a circular track with radius 1m and BC is horizontal with 1 m.Assume that the friction is 10N.
1)Find speed of the particle at C.
2)Find the highest point that the particle can reach.
3)Describe the subsequent motion after the particle reaches the highest point.
Mechanics
2009-12-08 5:46 pm
回答 (1)
2009-12-08 10:46 pm
✔ 最佳答案
A)maximum change P.E. of bullet and block=mgh=(1.55)(10)(10)=155J
K.E. after collision,=(1/2)mv^2
By conservation of energy,
K.E.of bullet and block after collision=maximum change P.E. of bullet and block
(1/2)mv^2=155J
v=14.14m/s
By conservation of momentem,
m1v1+m2v2=(m1+m2)v3
(0.05)v+(1.5)(0)=(0.05+1.5)(14.14)
v=438m/s
B)1)By conservation of energy,
P.E. loss=Gain in K.E.+work done against friction
mgh=(1/2)mv^2+fs
(10)(10)(1)=(5)v^2+(10)(╥/2+1)
v=3.85m/s
2)A
3)momentary ststionary
2009-12-08 14:48:22 補充：
The information of Q2 is no clear
收錄日期: 2021-04-13 16:58:16
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