Question

. A marksman holds a rifle of mass mR = 3.50 kg loosely, so it can recoil freely. He fires a bullet of mass mB = 10 g horizontally with a velocity relative to the ground of vBx = 300 m/s. What is the recoil velocity vRx of the rifle? Find kinetic energy of the bullet and rifle right after firing and tell me which gets more kinetic energy

Answer 1

Answer:

bullet = 450 J

rifle = 1.3 J

So, the kinetic energy of bullet is much more than the kinetic energy of gun.

Explanation:

mass of rifle, mR = 3.50 kg

mass of bullet, mB = 10 g

velocity of bullet, vBx = 300 m/s

Let VG is the recoil velocity of gun.

By using the conservation of momentum

mass of gun x recoil velocity of gun = mass of bullet x velocity of bullet

3.50 x V = 0.01 x 300

V =  0.86 m/s

kinetic energy of bullet, KB = 0.5 x mB x vB²

KB = 0.5 x 0.01 x 300 x 300 = 450 J

kinetic energy of rifle, KG = 0.5 x mG x v²

KG = 0.5 x 3.5 x 0.86 x 0.86 = 1.3 J

So, the kinetic energy of bullet is much more than the kinetic energy of rifle.

Answer 2

1-The recoil velocity of the rifle is,

$vRx= 0.86\dfrac{m}{sec}$

2-The kinetic energy of the bullet

$KE_{B} =450J$

3-The kinetic energy of a rifle

$KE_{r} =1.3J$

The kinetic energy of bullet and rifle we can say that after firing the bullet gets more kinetic energy.

Given-

The mass of rifle=3.50kg

The mass of Bullet=10g

The velocity of the bullet=300m/s

1- Recoil velocity of the rifle-

We know that within some problem domains, the amount of momentum remains constant. For this case,

$m_{g} \times vRx= m_{B}\times vBx$

$3.5\times vRx= \dfrac{10}{1000} \times 300$

$3.5\times vRx= 0.01 \times 300$

$vRx= \dfrac{0.01 \times 300}{3.5}$

$vRx= 0.86\dfrac{m}{sec}$

The recoil velocity of the rifle is,

$vRx= 0.86\dfrac{m}{sec}$

2- Kinetic energy of the bullet and rifle-

The formula for the Kinetic energy can be given as,

$KE=\dfrac{1}{2} MV^{2}$

The kinetic energy of the bullet

$KE_{B} =\dfrac{1}{2} m_{B} vBx ^{2}$

$KE_{B} =\dfrac{1}{2} \times \dfrac{10}{1000} \times 300^{2}$

$KE_{B} =450J$

The kinetic energy of a rifle

$KE_{r} =\dfrac{1}{2} m_{r} vRx ^{2}$

$KE_{r} =\dfrac{1}{2} \times 3.5\times 0.86^2$

$KE_{r} =1.3J$

On comparing the kinetic energy of bullet and rifle we can say that after firing the bullet gets more kinetic energy.

To know more about the kinetic energy, follow the link below,

https://brainly.com/question/18461965