Question

a 60g bullet fired from a 5 kg gun leave with a speed of 500 m/s. find the speed (velocity) with which the gun recoil (ferk backward) ​

Answer 1

Answer:

$\bigstar{Recoil\:velocity\:=\:-6m/s}$

Explanation:

Given:

• Mass of gun(m1) = 5kg
• Mass of bullet(m2) = 60g = 0.06kg
• Velocity of bullet(v2) = 500m/s

To Find:

• Recoil velocity of gun(v1)

Solution:

→ According to conservation of momentum, recoil velocity is given by the  equation

→ $v_1\:=\:\frac{m_2v_2}{-m_1}$

→ Substituting the given datas, we get

   $v_1\:=\:\frac{0.06\times500}{-5}$

   $\boxed{v_1\:=\:-6m/s}$

→ Recoil velocity is negative since the gun moves in the direction opposite to that of the bullet.

Notes:

• Remember to convert all the given datas to their SI units before calculating.
• Momentum is the product of mass and velocity. It's unit is kgm/s
• In an isolated system, momentum is conserved. Momentum before collission is equal to the momentum after collission.

Answer 2

Explanation:

Mass of the bullet, m = $\bold {60 g = 0•06kg.}$

Mass of the gun, M = 5 kg

Velocity of bullet, v = 500 m/s

Velocity of gun, V = ?

Momentum of the gun and bullet before firing = 0 [since all were at rest]

Momentum of the bullet after firing = mv = 30

Momentum of the gun after firing = MV = 5V

Using conservation of momentum,

Momentum before firing = tex]\bold {momentum after firing}[/tex]

=> 0 = 30 + 5V

=> V = - 6 m/s

The negative sign indicates that the velocity of the gun is directed opposite to the velocity of the bullet.