Question

A 0.05 kg bullet travelling at 500M/s horizontally strike a thick vertical wall. It Stops after penetrating through the wall a horizontal distance of 0.25m. what is the magnitude of the average force the wall exert on the bullet?​

Answer 1

So, the magnitude of the average force the wall exert on the bullet is 25,000 N. (See the solution steps in the attached image).

Introduction

Hi ! I will help you to discuss about the relationship between impulse and momentum. However, now it's not just counting the amount of impuls and momentum. But we must look at the problem and relate it to the things we learned in previous chapters, like deceleration on straight motion. Here too, we will calculate the value of the force, so pay attention to the formulas that will be given.

Formula Used

Here, you will be given three formulas that you will use at each step.

Step I : Relationship Between Acceleration, Distance, and Initial-Final Velocity.

Although you see the word "acceleration", here it can also apply to deceleration. The deceleration always have negative value of acceleration. Pay close attention to the formula:

`\boxed{\sf{\bold{(v_t)^2 = (v_0)^2 + 2 \cdot a \cdot s}}}`

With the following condition:

• 
  `\sf{v_t}` = final velocity respect to the time (m/s)
• 
  `\sf{v_0}` = initial velocity (m/s)
• a = acceleration that happen (m/s²)
• s = shift or distance (m)

Step II : Relationship Between Acceleration, Time, and Initial-Final Velocity

`\boxed{\sf{\bold{a = (v_t -v_0)/(t)}}}`

With the following condition :

• a = acceleration that happen (m/s²)
• 
  `\sf{v_t}` = final velocity respect to the time (m/s)
• 
  `\sf{v_0}` = initial velocity (m/s)
• t = time (s)

Step III : Relationship between Force, Mass, and Initial-Final Velocity in Impulse and Momentum

Impulse is a change in the value of momentum. The impulse is influenced by the force and time factors, while the momentum is influenced by the mass and velocity factors. Consider the following formula.

`\sf{I = \Delta p}`

`\sf{F * t = m \cdot (v_t-v_0)}`

`\boxed{\sf{\bold{F = (m \cdot (v_t-v_0))/(t)}}}`

With the following condition :

• I = impulse (N.s)
• 
  `\sf{\Delta p}` = change of momentum (kg.m/s)
• 
  `\sf{v_t}` = final velocity respect to the time (m/s)
• 
  `\sf{v_0}` = initial velocity (m/s)
• F = the force (N)
• m = mass of the object (kg)
• t = the interval of the time (s)

Problem Solving

We know that:

• m = mass of the object = 0.05 kg
• 
  `\sf{v_0}` = initial velocity = 500 m/s
• 
  `\sf{v_t}` = final velocity respect to the time = 0 m/s >> see at the "it stops after ..."
• s = shift or distance = 0.25 m

What was asked :

• F = the force that exerted by the wall = ... N

Step by step :

[See at the picture that attached in this answer]

Conclusion

So, the magnitude of the average force the wall exert on the bullet is 25,000 N.