21
Arnab Kundu
·2011-09-04 07:11:07
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
21
Arnab Kundu
·2011-09-04 07:33:30
The 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
1
rishabh
·2011-09-04 09:07:18
dude after 2 shots the mass of the system will be 80m - 10m. read the Q carefully
1
Debosmit Majumder
·2011-09-04 09:44:11
@rishab:for the 2nd qstn is θ given??
1
rishabh
·2011-09-04 10:56:20
θ is 0. its all horizontal.
1
Debosmit Majumder
·2011-09-04 18:05:13
i`m sorry....i meant for the 3rd qstn....
1
rishabh
·2011-09-05 05:29:29
nope. not given for the 3rd Q