Question

A rifle of mass 3kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.03 s to move through this barrel, calculate the force experienced by the rifle due to its recoil.​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\purple{Answer:-}$

⏩ * For bullet : m1= 0.03 kg, u1= 0, v1= 100 m/s

* For rifle: m2= 3 kg, u2= 0, v2=?

Sol. :- By conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

= 0 + 0 = 0.03 × 100 + 3 × v2

or v2 = - 0.03 × 100/ 3

= -1 m/s

[Negative sign indicates that the rifle moves in a direction opposite to that of the bullet.]

Force exerted on the rifle,

F= m2 [v2 - u2/t]

= 3 [-1-0/0.003]

= - 1000 N . Ans.

[Negative sign indicates the force of recoil (backward kick) of the gun].

✔✔✔✔✔✔✔

Hope it helps...:-)

✨♥⭐❤✨♥⭐❤✨

Be Brainly...✌✌✌

♣ WALKER ♠

Answer 2

m1 = 3 kg

m2=0.03 kg

v2= 100 m/s

v1= 0

Inittial momentum =0 N.S

Final momentum =100*0.03

                            = 3 N.S

time = 0.03 s

force =3/0.03

        =3*100/3

       = 100 N

(Ans): 100 N