A bullet of mass 0.012 kg and horizontal speed 70 m s –1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answers
Answered by
101
initial momentum of system(bullet+block)
= 0.012×70.
final momentum of system=(0.012+0.4)×v.
(where v is the velocity of bullet- block system after the impact.)
According to law of conservation of linear momentum:
0.012×70=(0.012+0.4)×v.
=> v=2.03m/s.
further,
initial KE of system=(1/2)×0.012×70×70
=29.4J
final KE of system =(1/2)×(0.012+0.4)×2.03×2.03=0.848J
HEAT PRODUCED= LOSS IN KE OF SYSTEM.
=> heat produced= 29.4-0.848=28.552J
further,
FROM THE LAW OF CONSERVATION OF ENERGY :
0.848= (0.012+0.4)×g×h.
(where g is acceleration due to gravity and h is the height gained by the system )
=> h= 0.20m
= 0.012×70.
final momentum of system=(0.012+0.4)×v.
(where v is the velocity of bullet- block system after the impact.)
According to law of conservation of linear momentum:
0.012×70=(0.012+0.4)×v.
=> v=2.03m/s.
further,
initial KE of system=(1/2)×0.012×70×70
=29.4J
final KE of system =(1/2)×(0.012+0.4)×2.03×2.03=0.848J
HEAT PRODUCED= LOSS IN KE OF SYSTEM.
=> heat produced= 29.4-0.848=28.552J
further,
FROM THE LAW OF CONSERVATION OF ENERGY :
0.848= (0.012+0.4)×g×h.
(where g is acceleration due to gravity and h is the height gained by the system )
=> h= 0.20m
Answered by
20
Answer:
Mass of bullet =m1=0.012kg
mass of wooden block =m2=0.04kg
velocity of bullet =70m/s
p1=m1v1
=(0.012)(70)
p2=m2v2
=(0.04)(0) =0
initial momentum =(0.012)(70)+0
final momentum =(0.04+0.012)v
according to law of conservation of momentum
(0.012)(70)=0.052 v
v=(0.012)(70)/0.052
=(12)(70)/52
=210/13 m/s
this is the acquired velocity by the block
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