Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 27)
27.
A bullet weighing 200 g is fired horizontally with a velocity of 25 m/sec from a gun carried on a carriage which together with the gun weighs 100 kg. The velocity of recoil of the gun, will be
Discussion:
4 comments Page 1 of 1.
Shruti said:
4 years ago
Initial momentum = final momentum.
Use the concept of conservation of momentum.
Use the concept of conservation of momentum.
(1)
Sujoy said:
5 years ago
MV=mv Where, Gun of mass (M)=100kg, fired velocity (v)= 25m/s, bullet mass (m)=200gm =0.2 kg.
So, V=mv/M.
V= (0.2*25)/100 =0.05m/s.
So, V=mv/M.
V= (0.2*25)/100 =0.05m/s.
(4)
Arnav said:
8 years ago
a1v1=a2v2.
100*v1=25*0.200.
v1=0.05m/sec.
100*v1=25*0.200.
v1=0.05m/sec.
(1)
Raihan said:
1 decade ago
m1u1=m2u2.
100*u1=25*0.200.
u1=.05m/sec.
100*u1=25*0.200.
u1=.05m/sec.
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