Forces, Friction and James Bond

[SilentBronco]

Homework Statement

James bond is standing on a large frozen lake and fires his handgun horizontally. This exerts 1400N of force on the bullet over a distance of 10cm before leaving the barrel at 200m/s. What is the coefficient of kinetic friction between bond and the ice if he slid for 24.2 s before coming to rest after the bullet is fired.

Homework Equations

F(net)=m*a
a=(delta)v/(delta)t

The Attempt at a Solution

ΣFx(bond)=ƒk-F=mass(bond)*a(bond)
F=recoil force
ƒk = kinetic friction = (mu)k*n
n=m(bond)*g
g=9.8 m/s2

No idea how to move forward with this problem. I am unable to cancel a variable and continually work with multi variable equations. Please help.

 

[haruspex]

You are not using the information regarding the bullet.
You are given the force on it and the distance over which that acts. What does that allow you to find? You also know the exit speed, so what else can you determine?

 

[Chestermiller]

SilentBronco: Let me guess. You are learning about kinetic energy and work in your present course unit. Am I correct?

Chet

 

[SilentBronco]

Chestermiller: not quite. My class is focusing on kinematics and how forces cause acceleration, currently.Work is a later chapter.

Haruspex:
From Newtons third law, I know that the force the bullet generates on bond is equal and opposite the force the bond applies to the bullet.

From knowing the exit speed, the initial speed, and the barrel length I can calculate the acceleration of the bullet. With acceleartion of the bullet I can then find the mass of the bullet. I don't see how this information is relevant to bond. Any information about the distance over which the force was applied is not the same for bond, because of the mass difference between bond and the bullet.
The same force on both bond and the bullet will produce very different accelerations.

Still, if the acceleration of bond = (Net force)/(mass of bond)
and
net force on bond = (Friction kinetic) - (Force of the bullet on bond)

I will always have two variables the friction constant (mu kinetic) and the acceleration of bond.

Moreover, what does the stopping distance on ice tell me, and where is that information useful?
I can setup the function
(acceleration of bond)=(change in velocity of bond)/24.2
Change in velocity of bond = Vfinal - V initial
Vfinal = 0
Change in velocity of bond = -Vinitial

I don't know v initial, because I don't know Fnet.
What am I missing.
My logic is running in circles.

 

[Chestermiller]

You're actually off to a very good start. You were not quite right about the force balance on Bond (you got the sign wrong), but not too bad:
$$m\frac{dv}{dt}=F_{bullet}-F_{friction}$$

Now, how long (timewise) do you think the 1400 N force that the bullet applied to Bond lasted (compared to the 24.2 seconds he was sliding)? Pretty short time right? If that's the case, you can split this problem into two parts: (a) when the bullet is speeding Bond up and (b) when the ice is slowing Bond down.
Let's focus on part (a). Do you think that the effect of the ice friction is significant during this time? If not, you can neglect ice friction during this short time interval. Have you learned about the law of conservation of momentum? According to this law, the impulse of a force on a body is equal to the change in momentum of the body. How can you use this law to determine Bond's velocity immediately after firing the gun, but before he really starts sliding on the ice yet. Let's see if you can use this information to solve part (a).

Chet

 

[SilentBronco]

Hmmm, in my free body diagram, I assumed, the bullet was moving to the right (while looking down at the piece of paper). Therefore force on bond would be opposite, and to the left. I had setup my coordinate system so x was positive to the right, and y positive upward. Why are the directions opposite? Friction, opposes motion. If bond were sliding backwards from the recoil of the bullet why would the force he experiences be positive and friction negative?

I can use kinematics to determine the acceleration of the bullet:
a(bullet) = (V^2)/(2*(change in x))
a(bullet) = (200^2)/(2*(.1))
a(bullet) = 2x10^5 m/s

m(bullet)=1400/2x10^5
m(bullet)=.007 kg

The time that force lasted on the bullet
t = sqrt((2*(displacement x))/a)
t = sqrt((2*(.1))/2x10^5)
t = 1X10^(-3) s

Transitioning to bond:
a(bond) = (F)/m
a(bond) = (1400)/100 = 14 m/s^2
If the acceleration occurred with negligible friction from the ice

So then the known velocities of bond are
V0=0
V1=?
V2=0

Solving for V1, the acceleration of bond must have occurred in the same time as the acceleration of the bullet
V1=a(bond)t(contact between bullet and bond)
V1=14*1x10^-3
V1 = 1.4X10^-2

Assuming constant deacceleration between bond and the ice after the initial recoil force
deltaV = (V1-V2)/2
deltaV = 7x10-3

a(bond) = deltaV/deltaT
a(bond) = 7x10^-3/24.4
a(bond) = -3.125x10^-4

uk=(m(bond)*a(bond)+F/)(m(bond)*g)
uk=(m(bond)*a(bond)+1400/)(100*9.8)
uk=1.42

That is too large a coefficient of friction for ice. Idk where I went wrong.
Also I wrote the prompt with a typo, the time of the slide is 24.4s

 

[Chestermiller]

Hmmm, in my free body diagram, I assumed, the bullet was moving to the right (while looking down at the piece of paper). Therefore force on bond would be opposite, and to the left. I had setup my coordinate system so x was positive to the right, and y positive upward. Why are the directions opposite? Friction, opposes motion. If bond were sliding backwards from the recoil of the bullet why would the force he experiences be positive and friction negative?

I can use kinematics to determine the acceleration of the bullet:
a(bullet) = (V^2)/(2*(change in x))
a(bullet) = (200^2)/(2*(.1))
a(bullet) = 2x10^5 m/s

m(bullet)=1400/2x10^5
m(bullet)=.007 kg

The time that force lasted on the bullet
t = sqrt((2*(displacement x))/a)
t = sqrt((2*(.1))/2x10^5)
t = 1X10^(-3) s

Transitioning to bond:
a(bond) = (F)/m
a(bond) = (1400)/100 = 14 m/s^2
If the acceleration occurred with negligible friction from the ice

So then the known velocities of bond are
V0=0
V1=?
V2=0

Solving for V1, the acceleration of bond must have occurred in the same time as the acceleration of the bullet
V1=a(bond)t(contact between bullet and bond)
V1=14*1x10^-3
V1 = 1.4X10^-2

You did very nicely so far. Everything up to here is correct.

Assuming constant deacceleration between bond and the ice after the initial recoil force
deltaV = (V1-V2)/2
deltaV = 7x10-3

This is not correct. Where did the 2 come from? ΔV= 14x10^-3-0=14x10^-3
So

a(bond) = deltaV/deltaT
a(bond) = 14x10^-3/24.4
a(bond) = -6.25x10^-4

uk=(m(bond)*a(bond)+F/)(m(bond)*g)
uk=(m(bond)*a(bond)+1400/)(100*9.8)
uk=1.42

You went wrong here when you included the bullet force as applying during the time interval that Bond was sliding on the ice. It shouldn't be in there.

Overall, very nicely done.

Chet

 

[SilentBronco]

So the force of the bullet is only applied for 1 mili second, afterwhich friction is the only force opposing the recoil motion.

Solving with your advice:
uk = 6.37 x 10^-5

Seems almost immeasurable. Thank you for helping. I did not think to separate the problem into multiple steps.

One final question: Why can friction force be ignored during the initial calculation of bonds acceleration?

 

[Chestermiller]

So the force of the bullet is only applied for 1 mili second, afterwhich friction is the only force opposing the recoil motion.

Solving with your advice:
uk = 6.37 x 10^-5

Seems almost immeasurable. Thank you for helping. I did not think to separate the problem into multiple steps.

One final question: Why can friction force be ignored during the initial calculation of bonds acceleration?

It was a judgement call, based on experience. I just felt that the frictional force was going to be a lot smaller than the 1200 N force of the bullet. Bond's weight was only 1000 N, so, even if the coefficient of friction of the ice was 0.1, that would only amount to a 100 N force. So I decided it could tentatively be neglected. If it later turned out to be important, we could go back and re-solve the problem with the friction force included for the 0.001 sec.

Chet