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Let velocity of the gun is V and shell is v after the shell comes out.
so, by conservation of linear momentum :
mv=kmVcos45°
or,v/V=k/√2
or,(v/V)2=k2/2
so, energy of shell : energy of bullet=1/2mv2 : 1/2 kmV2
=k2/2 : k
=k : 2
1) a shell of mass m is fired from a gun of mass km which can recoil freely on a horizontal plane , the elavation of the gun is 45° .prove that the ratio of energy of the shell to that of the gun is (2k2+2k+1)k.
[ im getting only 1/k]
2) a gun is mounted on a rail road car . the mass of the car with all of its components is 80m and mass of each shell to be fired is 5m. the muzzle velocity of the shell is 100 m/s , what is the recoil speed of the car after the second shot.
[answer : 100(1/15 + 1/16)]
3) a block of mass 4m is placed on the top of a movable inclined plane of mass 20m and base length l.when system is released from rest find the distance moved by the incline till the block reaches the ground . all surfaces are frictionless.
[answer l/16]
please gimme the complete sollutions..
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7 Answers
2121The second answer will depend on the no. of shells present in the gun after the firing the the first bullet.
Because after the firing of the first bullet the car system will gain some velocity and to calculate that momentum we need to know the mass of the system
1dude after 2 shots the mass of the system will be 80m - 10m. read the Q carefully
