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

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Robin Hood confronts the Sheriff of Nottingham and shoots an arrow that strikes the Sheriff's hat, which has a mass of 0.5 kg. The bow's spring constant is calculated to be 40 N/m, allowing for the determination of the arrow's speed just before impact, found to be 98 m/s. After the inelastic collision, the combined speed of the hat and arrow is calculated to be 2.8 m/s. For the projectile motion part, the time it takes for the hat to hit the ground is derived from kinematic equations, focusing on the vertical distance and acceleration due to gravity. The discussion emphasizes the application of physics principles in solving the problem.
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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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