Question

A 7.8 g bullet leaves the muzzle of a rifle with a speed of 474.5 m/s. What constant force is exerted on the bullet while it is traveling down the 0.7 m length of the barrel of the rifle? Answer in units of N.

Answer 1

G I V E N :

• Mass of the bullet (m) = 7.8 g
• Initial speed of the bullet (u) = 0 m/s
• Final speed of the bullet (v) = 474.5 m/s
• Distance (s) = 0.7 m

T O F I N D :

• Constant force exerted on bullet (F) = ?

M E T H O D :

• Write down all the known variables and their values and the one you don't know with a question mark next to it. Like this (F = ?)
• Find the kinematic equation where you have all the variables except the one you are missing.
• Re-arrange the equation to solve for the missing variable.

To answer any question to do with mass, acceleration, velocity ,etc. you can use one or more of this set of equations. They are called the 'kinematic' equations and They are really easy to re-arrange, when you need to find the 'unknown quantity', that the question is asking for. Below these equations I will show you the general problem solving strategy to solve these quickly and easily.

The Kinematic Equations :

• v = u + at
• s = ut + ½ at²
• v² - u² = 2as
• Sₙₜₕ = u + a/2 (2n - 1)

This question takes things one step further and is asking you to find the force on the bullet. To convert to a force in any situation you need to find the acceleration and multiply it by the mass.

S O L U T I O N :

We need to find the force exerted on the bullet while it is traveling down the 0.7 m length of the barrel of the rifle. Let a be the acceleration of the bullet. So, by using third equation of motion we get :

→ v² - u² = 2as

→ (474.5)² - (0)² = 2(a)(0.7)

→ 225150.25 = 1.4a

→ a = 160821.6 m/s²

Now, Let's find the force exerted :

➝ F = ma

➝ F = 7.8 × 10⁻³ × 160821.6

➝ F = 1254.40848 N

⛬ The constant force exerted on the bullet is 1254.40848 N .

Answer 2

Solution

Given ,

• mass of the bullet (m) = 7.8 m
• initial speed of the bullet (u) = 0m/s
• final speed of the bullet (v) =474.5m/s
• distance (s) = 0.7 m

To find ,

• we have to find the force exerted on the bullet

So ,

• latest find out the acceleration of the bullet by using the third equation of the motion : v^2 - u^2 = 2as

putting the value we get ;

=> (474.5)^2 - (0)^2=2(a) (0.7)

=> 225150.25 = 1.4 a

=> a = 160821.6m/s^2

Now ,

• to find the first exerted we have the formula ; F = Ma

putting the value we get ;

=> F = 7.8 × 10^-3 × 160821.6

=> F = 1254.4 N

the Constant force exerted on the bullet is 1254.4 Newton