Speeding bullet, average force by post on bullet

serendipityfox

Homework Statement
bullet initial speed 600m/s, mass 15g, penetrates 10cm, average force exerted by post on bullet
Homework Equations
f= delta p/t, d=vi+vf/2 *t
used a) change in momentum / time, time from from kinematics equation d = vi+vf/2 *t ... 0.1 = 300*t
=3.33*10^-4
then 600/0.00333 = 1.8*10 ^5
how to proceed?

Last edited:
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Delta2

Homework Helper
Gold Member
What you calculated as 1.8x10^5 is the average acceleration (or to be more accurate deceleration) of the bullet. How can you find the average force if you know the average acceleration? Newton's 2nd law of course! :D

serendipityfox

thankyou delta2
i bungled the second equation... should be 9 / 0.00033 of course

a) 9.0*10^3
b) 5.3*10^4
c) 2.4*10^5
d) 1.6*10^6

Delta2

Homework Helper
Gold Member
I get the same answer as you, don't know whats going on, maybe recheck the data given.

PeroK

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2018 Award
thankyou delta2
i bungled the second equation... should be 9 / 0.00033 of course

a) 9.0*10^3
b) 5.3*10^4
c) 2.4*10^5
d) 1.6*10^6
If you assume a constant force, hence constant acceleration, then the work done is $Fd$ and this must equal the initial KE of the bullet. So:

$F = \frac{mv^2}{2d}$

This agrees with the formula derived from the kinematics of constant acceleration.

Assuming SI units (units are important but missing from your work and your answers), then this should be $F = 27kN$.

serendipityfox

... average force exerted by post on bullet, will be same as the bullet on post?

serendipityfox

@Delta2 - thankyou for the assistance
@PeroK - sorry will endevour to be more precise

PeroK

Homework Helper
Gold Member
2018 Award
... average force exerted by post on bullet, will be same as the bullet on post?
That's Newton's third law.

serendipityfox

so still no viable answer according to the exam sheet... well this discourse has been helpful anyway, thanks

haruspex

Homework Helper
Gold Member
2018 Award
so still no viable answer according to the exam sheet... well this discourse has been helpful anyway, thanks
As @PeroK notes, the question embodies a blunder: you have to assume the force is constant.
Average force (a vector) is defined as change in momentum (a vector) divided by elapsed time (a scalar). Dividing work done by distance produces a scalar, and this does not in general match the magnitude of the average force. It is an interesting exercise to calculate the difference in the case of a spring.

serendipityfox

physics question section on an MCAT practice paper

"Speeding bullet, average force by post on bullet"

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