Question

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

Answer 1

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

Answer 2

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