Question

A bullet of mass 100 kilogram is fired from the rifile of mass 200 kilogram with the speed of 50 m/s calculate the recoil velocity of gun

Answer 1

Answer:

-0.5 m/s

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.

Answer 2

Answer:-

• Mass of bullet=m1=100g=0.1kg

Fact:-

No bullet weighs 100kg

• Mass of Rifle=200kg=m2
• initial velocity=u=50m/s
• recoil velocity=v=?

We know

$\boxed{\sf ∆P=P}$

$\\ \sf\longmapsto m_1u=m_2v$

$\\ \sf\longmapsto (0.1)(50)=200v$

$\\ \sf\longmapsto 5=200V$

$\\ \sf\longmapsto v=\dfrac{5}{200}$

$\\ \sf\longmapsto v=0.02m/s$

Recoil velocity is 0.02m/s.