A bullet of mass 20g starts from rest from one end of the barrel of gun 1.5m long and weighing 40kg. The
bullet leaves at the other end with a velocity of 350m/s. Calculate (i) the time taken by the bullet to leave the
gun and (ii) the velocity of the recoil of the gun..
Answers
Answered by
13
hey dude
hope it helps
first we will find recoil velocity of gun.Vre=Mass of bullet*velocity of bullet /mass of gun
Vrecoil=0.02*350/40
recoil velocity of gun=0.175 m/s
After that we will find time taken by bullet to cross the barrel if no recoil velocity is there
v=d/t so,t=d / v
t=1.5/350 time =0.00428571 second we will take it as 0.0043 seconds
so,during this time how much must have gun moved backwards decreasing the distance the bullet has to travel d=v*t distance =0.175*0.0043 =0.0007525 meters
reduced distance =1.5-0.00075approx
=1.49935 meters the bullet has to travel so t=d/v
time=1.49925/350 time =0.0042835714
so time reduce to 0.0042835714 from 0.00428571
answer is 0.00428357
hole this answer is correct
hope it helps
first we will find recoil velocity of gun.Vre=Mass of bullet*velocity of bullet /mass of gun
Vrecoil=0.02*350/40
recoil velocity of gun=0.175 m/s
After that we will find time taken by bullet to cross the barrel if no recoil velocity is there
v=d/t so,t=d / v
t=1.5/350 time =0.00428571 second we will take it as 0.0043 seconds
so,during this time how much must have gun moved backwards decreasing the distance the bullet has to travel d=v*t distance =0.175*0.0043 =0.0007525 meters
reduced distance =1.5-0.00075approx
=1.49935 meters the bullet has to travel so t=d/v
time=1.49925/350 time =0.0042835714
so time reduce to 0.0042835714 from 0.00428571
answer is 0.00428357
hole this answer is correct
Kundank:
yes...
i can explain you
First how he can find time taken to cross barallel without finding acceleration
oooooo!
you are right!!!
here bullet is not moving with uniform velocity but with uniform acceleration
Now he applied the logic that When bullet will move forward than gun will move backward . Hence shorting the length of a barrrel
which is also wrong
As he found the recoil velocity of gun when the bullet was having velocity of 350m/s.... means that time bullet have already escaped the barrel
So the run recoils after bullet leave....hence length of barrel will remain unchained
Answered by
10
Initial momentum of the system = = 0
Final momentum of the system = mv + MV
Momentum conservation: mv + MV = 0
The velocity of the recoil of the gun.V = − (m/M) v
= − (20 x 10−3/40) x 350 m/s
= −0.175 m/s
acceleration a =( v2 −u2)/2S
= [(350 m/s)2 − 0]/(2 x 1.5 m)
= 40833 m/s2
Time taken for the bullet to leave the gun t = (v − u)/ a
=[350 m/s ]/40833 m/s2
= 0.008sec
Final momentum of the system = mv + MV
Momentum conservation: mv + MV = 0
The velocity of the recoil of the gun.V = − (m/M) v
= − (20 x 10−3/40) x 350 m/s
= −0.175 m/s
acceleration a =( v2 −u2)/2S
= [(350 m/s)2 − 0]/(2 x 1.5 m)
= 40833 m/s2
Time taken for the bullet to leave the gun t = (v − u)/ a
=[350 m/s ]/40833 m/s2
= 0.008sec
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