a machine gun bullet has a mass of 200 gram leaves the barrel at a spe

1.	a machine gun bullet has a mass of 200 gram leaves the barrel at a speed of 150 metre per second one find the kinetic energy of a bullet
Answer» Step-by-step EXPLANATION: MASS of bullet, m=100g= 1000 100 kg= 10 1 kg VELOCITY of bullet, v=100ms −1 Mass of gun, M=20kg Let recoil velocity of gun =V Before firing, the system (gun+bullet) is at rest, therefore, initial momentum of the system =0 FINAL momentum of the system = momentum of bullet + momentum of gun =mv+MV= 10 1 ×100+20V=10+20V Applying law of conservation of momentum, Final momentum = Initial momentum i.e., 10+20V=0 20V=−10 or V=− 20 10 =−0.5ms −1 . Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.

a machine gun bullet has a mass of 200 gram leaves the barrel at a speed of 150 metre per second one find the kinetic energy of a bullet

Answer»

Step-by-step EXPLANATION:

MASS of bullet, m=100g=

1000

100

kg=

VELOCITY of bullet, v=100ms

−1

Mass of gun, M=20kg

Let recoil velocity of gun =V

Before firing, the system (gun+bullet) is at rest, therefore, initial momentum of the system =0

FINAL momentum of the system = momentum of bullet + momentum of gun

=mv+MV=

×100+20V=10+20V

Applying law of conservation of momentum, Final momentum = Initial momentum

i.e., 10+20V=0

20V=−10

or V=−

=−0.5ms

−1

Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.