Question

A bullet of mass 50 g is horizontally fired with a velocity 200 m/s from the pistol of mass 3.5 kg. What is the
recoil velocity of the pistol?

Answer 1

$\huge\bold\red{\mathfrak{Answer:-}}$

Given:

Mass of bullet =50g = 0.05kg

Velocity of bullet =200m/s

Mass of pistol = 3.5 kg

To Find:

Recoil velocity

Formula:

mass of bullet ×velocity of bullet =mass of

gun×recoil velocity

let mass of bullet be m1 and velocity of bullet be v1

let mass of pistol be m2 and recoil velocity be R

$\boxed{\boxed{\mathbb{m1 × v1 = m2 × R}}}$

Solution:

m1 ×v1 = m2 × R

=> 0.05 ×200 = 3.5 × R

=> 10 =3.5R

=> 3.5R = 10

=> R = 10/3.5

=> R = 2.85 m/s

$\boxed{\boxed{\boxed{\mathfrak{R = 2.85m/s}}}}$

Answer 2

Answer:20/7

Explanation:mass of bullet=50g=0.05kg

Velocity of bullet=200m/s

mass of gun =3.5kg

Velocity of gun=?

Formula of momentum =

M(g)V(g)=M(b)V(b)

0.05×200=3.5×V(b)

10=3.5×v(b)

V(b)=10/3.5

V(b)=20/7