Asked by Anonymous
a bullet of mass 0.01kg and travelling at a speed of 500m/s strikes a block of mass 2kg which is suspended by a string length 5m. the centre of gravity of block is found to rise a vertical distance of 0.1m .Calculate the speed of the bullet when it emerges from the block.
Answers
Answered by
Anonymous
220m/s
Answered by
bobpursley
energy of block in motion:
1/2 MV^2=Mgh
V= sqrt(2g*.1)
momentum of bullet and block after bullet leaves: MV+mv=m(500) (convervation momentum)
solve for v
2(sqrt(.2*9.8))+.01*v=(.01)(500)
v= (5-2*1.4)/.01 = you are correct. 220m/s
1/2 MV^2=Mgh
V= sqrt(2g*.1)
momentum of bullet and block after bullet leaves: MV+mv=m(500) (convervation momentum)
solve for v
2(sqrt(.2*9.8))+.01*v=(.01)(500)
v= (5-2*1.4)/.01 = you are correct. 220m/s
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