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24 Nov 2007 11:56:35 IST
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Book H.C. Verma
Chapter-9
Page 161
Problem 27
A gun is mounted on a railroad car. The mass of car, the gun , the shells and the operator is 50m. where m is mass of 1 shell. if thge muzzle velocity of the hell is 200m/sec. what is the recoil speed of the car after secont shot.
Plz solve this prob.
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24 Nov 2007 12:08:03 IST
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Help plzzzz.
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24 Nov 2007 12:31:07 IST
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ok
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NIT silchar electrical engineering |
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24 Nov 2007 12:32:29 IST
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it can be solved by conservation of momentum
when 1st hell is shot, mass of system (car+shell+gun) becomes 50m-m =49m
so 49mv1=200m or v1=200/49
after 2nd shot system mass becomes 48m
so similarly v2=200/48
so net speed of system after 2 shots = 200/49 +200/48 m/s
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I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
Rahul Dey
Dept. of Electronics & Electrical
Communication Engineering,
IIT Kharagpur
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24 Nov 2007 12:40:57 IST
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for 1st recoil,
V 1 = 200/49 m/s, this is the velocity of first recoil
for 2nd recoil,
49m x 0= 200 x m - 48m x V 2
V 2 = 200/48 m/s
therefore velocity of recoil= 200(1/49 + 1/48) m/s
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NIT silchar electrical engineering |
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24 Nov 2007 12:45:00 IST
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explanation:
after the first shot has been fired, the mass of the railroad car system becomes 49m (50m-m)
after the second shot, the mass of the railroad car system becomes 48m.
the two velocities are added because they are in the same direction
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NIT silchar electrical engineering |
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24 Nov 2007 13:39:43 IST
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Solution posted by Ramayani and rhd is correct but why cann't i use conservation of momentum in initial and final situation.
Secondly after first shot velocity of 49m mass should be v= 200/49 and not zero.
So if we apply conservation of momentum now then
what shoulbe equation accoreding to me it should be
49m*200/49= 48v + m*(200-200/49).
Whats wrong with above equation.
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24 Nov 2007 14:02:53 IST
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conservation of momentum in initial and final situation is what i used
vel after 1st shot is definetely 200/49
for 2nd shot we calculate vel imparted by shot =200/48 which is added to 200/49 which was its initial velocity
hope its clear now
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I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
Rahul Dey
Dept. of Electronics & Electrical
Communication Engineering,
IIT Kharagpur
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24 Nov 2007 14:13:41 IST
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But what wrong with the above equation written by me.
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24 Nov 2007 23:40:12 IST
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Vcm=0
0=49xv+mx200
V=200/49
so 200/49m/sec towards left
V=200/49
V1=200/48
so velocity of the car with respect to the earth is 200(1/48+1/49)===0.475m/sec
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