Question

From a rifle of mass of 4kg ,a bullet of mass 50g is fired with an initial velocity of 35m/s . Calculate the initial recoil velocity of the rifle.​

Answer 1

Answer:

Since momentum is conserved,

m1×v1=m2×v2

4×v=0.05×35

v= 0.4375 m/s

Answer 2

Answer:

→ -0.4375

Explanation:

Given,

    Mass of the rifle ($m_{1}$)     = 4kg.

    Mass of the bullet ($m_{2}$) = 50g.

                                                = 0.05kg.

    Final velo. of the bullet ($v_{1}$)  = 35m/s.

    Initial velo. for the both obj. ($u$)  = 0m/s.

    Final velo. of the rifle ($v_{2}$)    = ?

Now,

    By using law of conservation of momentum,

i.e. $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

But,

     ∵ $u_{1} = u_{2}$ = 0m/s.

∴ $v_{2}$ = $\frac{-v_{1}m_{2}}{m_{1}}$

            = $\frac{-(35 * 0.05)}{4}$

            = $\frac{-(1.75)}{4}$

            = -0.4375m/s.

∴ The final velocity of the rifle, $v_{2}$ = -0.4375. Its is in negative because the rifle moves to the direction opposite to the direction in which the bullet is being fired.

Please make this the brainliest. Thanks