A bullet of mass 50g is fired from a gun of mass 2 kg if total kinetic energy produced is 2050 J ,the energy of the gun and bullet separately are
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Answered by
23
let velocity of bullet is v and velocity of gun is V.
from conservation of linear momentum,
initial momentum = final momentum
0 = mv + MV
where m = 50g = 0.05kg
and M = 2Kg
so, 0 = 0.05v + 2V
=> v = -40V
kinetic energy of bullet = 1/2 mv²
= 1/2 × 0.05(-40V)²
= 1/2 × 0.05 × 1600V²
= 40V²
kinetic energy of gun = 1/2 × 2 × V²
= V²
here it is clear that, kinetic energy of bullet = 40 × kinetic energy of gun
Let x is the kinetic energy of bullet
so, 40x is the kinetic energy of gun.
now, x + 40x = 2050
41x = 2050
x = 50
hence, kinetic energy of bullet = 50J
kinetic energy of gun = 2000J
from conservation of linear momentum,
initial momentum = final momentum
0 = mv + MV
where m = 50g = 0.05kg
and M = 2Kg
so, 0 = 0.05v + 2V
=> v = -40V
kinetic energy of bullet = 1/2 mv²
= 1/2 × 0.05(-40V)²
= 1/2 × 0.05 × 1600V²
= 40V²
kinetic energy of gun = 1/2 × 2 × V²
= V²
here it is clear that, kinetic energy of bullet = 40 × kinetic energy of gun
Let x is the kinetic energy of bullet
so, 40x is the kinetic energy of gun.
now, x + 40x = 2050
41x = 2050
x = 50
hence, kinetic energy of bullet = 50J
kinetic energy of gun = 2000J
Answered by
0
Answer:
2000J
Explanation:
Mass of bullet and that of gun are m
b
=0.05 kg and 2 kg respectively.
Kinetic energy of bullet K
b
=
2m
b
P
b
2
Kinetic energy of riffle K
r
=
2m
r
P
r
2
Energy of explosion E=K
b
+K
r
=
2m
b
P
b
2
+
2m
r
P
r
2
But according to conservation of law of momentum, P
b
=P
r
=P
∴ 2050=
2(0.05)
P
2
+
2(2)
P
2
⟹ P
2
=200
Thus kinetic energy of bullet K
b
=
2m
b
P
2
=
2(0.05)
200
=2000 J
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