How Fast Does Robin Hood's Arrow Propel the Sheriff's Hat?

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The discussion revolves around a physics problem involving projectile motion and energy conservation. The scenario features Robin Hood shooting an arrow that strikes the Sheriff's hat, prompting questions about the speed of the hat after impact and its subsequent motion as it falls to the ground.

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Approaches and Questions Raised

  • Participants explore the calculation of the spring constant of the bow and the energy of the arrow. There are discussions about using kinetic energy to determine the arrow's speed before impact and applying conservation of momentum to find the final velocity of the hat and arrow together. Questions arise regarding the approach to solving the projectile motion aspect of the problem.

Discussion Status

Participants have made progress in calculating the spring constant and the energy of the arrow. There is a clear direction towards finding the speed of the arrow and the subsequent velocity of the hat with the arrow lodged in it. However, there is ongoing exploration regarding the best method to determine the time it takes for the hat to hit the ground after being struck.

Contextual Notes

Participants are considering the assumptions of neglecting air resistance and treating the collision as inelastic. There is also a focus on the vertical motion of the hat after the collision, with some uncertainty about the appropriate equations to use for the projectile motion calculations.

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Robin Hood is walking through Sherwood Forest when he comes upon the Sheriff of Nottingham. The Sheriff tells Robin Hood he is going to arrest him for numerous misdeeds. Robin pulls out his bow, loads an arrow, pulls the bow back 0.70 m, and shoots an arrow into the Sheriff's hat, knocking the hat with the arrow lodged in it off his head. The hat's mass is 0.5 kg. It takes 4N to pull Robin's bow back 0.10 m, and the arrow has a mass of 0.20 kg.
a) What is the speed of the hat with the arrow in it just after it is struck? Assume the arrow travels horizontally between Robin and the Sheriff and neglect air resistance.
b) If the Sheriff is 2.0 m high, how far behind him will his hat with the arrwo in it hit the ground?

For (a) I used F=kx and found it to be 40 N/m
 
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jenha14 said:
For (a) I used F=kx and found it to be 40 N/m
OK, you found the value of k, the spring constant of the bow. Keep going. How much energy does the arrow have after it leaves the bow?
 
I found the energy to be:

W = 1/2(k)(x)^2
 
jenha14 said:
I found the energy to be:

W = 1/2(k)(x)^2
Good. Keep going. What's the arrow's speed just before it hits the hat? Then figure out the speed of "hat + arrow" after the the arrow hits the hat.

Hint: Treat the arrow sticking into the hat as an inelastic collision. What's conserved?
 
I found that the speed of the arrow before the collision was:
KE = 1/2mv^2
9.8 J = 1/2(0.2 kg)(v)^2
v = 98 m/s

And used the equation:
m1v1i + m2v2i = (m1 + m2)vf
And found that the final velocity was 2.8 m/s

How do you start part (B)
 
Part (B) is a projectile motion problem. You just found the initial velocity, which is horizontal. How long does it take before it hits the ground?
 
Would I use F=ma to find the acceleration and then plug that into v=v0 + at to find the time?
 
jenha14 said:
Would I use F=ma to find the acceleration and then plug that into v=v0 + at to find the time?
You'd be better off using a different formula, since you don't have the final velocity in the vertical direction. You have the distance, so find a kinematic formula with distance and time.

You shouldn't have to use F=ma to find the acceleration (though you could, of course). Since this is just a projectile, you should know the acceleration.
 

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