Question

a) From a rifle of mass 5kg, a bullet of mass 50g is fired with an initial velocity of 108km/h. Calculate the initial recoil velocity of the rifle. 2 b) What is the acceleration produced by a force of 5 N acting on a body of mass 20kg?

Answer 1

Explanation:

b)

F = ma

5 = 10a

5/10 = a

1/2 = a

0.5m/s² acceleration

5 point pe ek hi sum milega

Answer 2

Question:

From a rifle of mass 5kg, a bullet of mass 50g is fired with an initial velocity of 108km/h. Calculate the initial recoil velocity of the rifle.

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Mass of the rifle (m₁) = 5 kg
• Mass of the bullet (m₂) = 50 g = 0.05 kg
• Velocity of bullet (v₂) = 108 km/hr = 30 m/s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Recoil velocity of gun (v₁)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Recoil velocity is given by the equation,

 v₁ = -m₂v₂/m₁

→ Substituting the datas we get,

  v₁ = -0.05 × 30 / 5

  v₁ =  - 1.5/5

  v₁ = -0.3 m/s

$\boxed{\bold{Recoil\:Velocity=-0.3\:m/s}}$

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

Question:

What is the acceleration produced by a force of 5 N acting on a body of mass 20kg?

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Mass = 20 kg
• Force = 5 N

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Acceleration

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ By Newton's second law of motion,

  F = ma

→ Substituting the given datas we get the value of a

  5 = 20 × a

  a = 0.25 m/s²

$\boxed{\bold{Acceleration=0.25\:m/s^{2} }}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Momentum is always conserved in an isolated system. The initial momentum is always equal to the final momentum.

→ Newton's second law of motion state that the rate of change of momentum is directly proportional to the force applied and the change takes place in the direction of force.