Understanding Resistive Forces for Decelerating a Bullet

[Ratiocinator]

I am a little confused over a problem I found whilst reading through an example A-level Mechanics exam paper.

The problem is on the Mechanics M2 paper, question 2.

This question asks for the magnitude of a resistive force that slowed a bullet, of mass 6 grams, from 400m/s to 250m/s.

If F = ma, then should this resistive force not be F = 0.006 * 150² = 135N, as the force to decelerate a mass is equal and opposite to that force required to accelerate it (ignoring air resistance, etc.)?

Should the momentum of this bullet not be P = mV = 0.006 * 400 = 2.4N ?

But the answer given in the paper is very different. The question and answer from the exam paper are:


Q. A bullet of mass 6 grams passes horizontally through a fixed, vertical board. After the bullet has traveled 2cm through the board its speed is reduced from 400m/s to 250m/s. The board exerts a constant force on the bullet.

Find, to 3 significant figures, the magnitude of this resistive force.


A. F * 0.002 = 0.5 * 0.006 (400² - 250²)

F = 14600N


The exam paper can be found here:

http://www.edexcel.org.uk/VirtualContent/83441/GCE_Mech_M1_M5_Specimen_Paper_mkscheme.pdf

 

[cristo]

I am a little confused over a problem I found whilst reading through an example A-level Mechanics exam paper.

The problem is on the Mechanics M2 paper, question 2.

This question asks for the magnitude of a resistive force that slowed a bullet, of mass 6 grams, from 400m/s to 250m/s.

If F = ma, then should this resistive force not be F = 0.006 * 150² = 135N, as the force to decelerate a mass is equal and opposite to that force required to accelerate it (ignoring air resistance, etc.)?

Yes, f=ma holds, but you have used a velocity not an acceleration.

Should the momentum of this bullet not be P = mV = 0.006 * 400 = 2.4N ?

This is the correct initial momentum of the bullet (modulo units-- momentum is measured in kgm/s)

But the answer given in the paper is very different. The question and answer from the exam paper are:


Q. A bullet of mass 6 grams passes horizontally through a fixed, vertical board. After the bullet has traveled 2cm through the board its speed is reduced from 400m/s to 250m/s. The board exerts a constant force on the bullet.

Find, to 3 significant figures, the magnitude of this resistive force.


A. F * 0.002 = 0.5 * 0.006 (400² - 250²) (*)

F = 14600N


The exam paper can be found here:

http://www.edexcel.org.uk/VirtualContent/83441/GCE_Mech_M1_M5_Specimen_Paper_mkscheme.pdf

To answer this question, one needs to use the fact that the work done by the block on the bullet is equal to the kinetic energy lost by the bullet on traveling through the block. The left hand side of (*) is the work done (which is equal to the resistive force multiplied by the distance the bullet travels through the block). The right hand side is equal to the change in kinetic energy of the bullet. Equating the two enables one to find the resistive force.

 

[Ratiocinator]

Yes, that was a silly mistake I made regarding acceleration and velocity!

Many thanks for your help. It is much clearer now.

 

[cristo]

Yes, that was a silly mistake I made regarding acceleration and velocity!

We all make them!

Many thanks for your help. It is much clearer now.

You're very welcome!

 

[Ratiocinator]

It would seem my brain fog is yet to clear completely.

I tried to calculate the resistive force met by the bullet by working out the deceleration rate.

First I transposed the basic linear motion equation for displacement to make time the subject:

t = S / ((u + V) / 2)

t = 0.002 / ((400 + 250) / 2)

t = 6.1538s

I am not sure where I went wrong with this, as 6 seconds seems like an excessively long time for a speeding bullet to pass through such small distance. But, continuing:

Now I have a figure for time taken I can calculate acceleration:

a = V / t

a = 150 / 6.1538

a = 24.3750m/s/s

And returning to Newton’s equation:

F = ma

F = 0.006 * 24.3750

F = 0.1462500N

As the proper answer to the exam question is 14625N, my answer is just one tenth of what is should be.

Please could you tell me where I went wrong?

 

[hage567]

What you've done is used the average velocity of the bullet as it travels that distance. That doesn't make sense in this problem.

You can use kinematics to get the deceleration, and you don't need to worry about finding time. You can find the deceleration given the initial and final velocities and the distance the bullet traveled in that time. Do you know which kinematic equation might fit what you know?

 

[Ratiocinator]

Thanks for your reply.

I have found a more suitable equation from my notes. Transposing for acceleration gives:

a = (V² - u²) / S * 0.5

a = (400² - 250²) / 0.002 * 0.5

a = 24375000m/s/s

Back to Newton’s equation:

F = ma

F = 0.006 * 24375000

F = 146250N

So this way seems more accurate! Now I have to sort out in my head the fundamental differences between the methods I used to prevent a reoccurrence of my error!

 

[PhanthomJay]

You can use either method, your problem is in the math.
2 cm is 0.02m, not 0.002m. And you've slipped on decimal points more than once.

 

[Ratiocinator]

Of course! I tend to make silly mistakes like that. I have even lost marks in tests due to making similar stupid mistakes (using unsquared numbers in trigonometry, mistaking + for * in matrices, etc.). One of my maths tutors recommended I be tested to see if I have any 'condition' that causes me to make such errors. I will soon be seeing an educational psychologist to be tested for dyscalculia, and similar things.

Thanks for pointing that out! It is odd how I arrived at the correct figure on the last attempt, though? I will go through this problem once again.