How Long Does It Take a Bullet to Stop Inside a Tree?

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SUMMARY

The discussion focuses on calculating the time it takes for a bullet to stop inside a tree after penetrating to a depth of 4.07 cm. A bullet with a mass of 4.90 g and an initial speed of 609 m/s experiences a constant frictional force, resulting in a calculated stopping time of 1.33e-4 seconds. The calculations utilize kinetic energy, work-energy theorem, and Newton's second law to derive the force and acceleration, confirming the time elapsed through two different methods. Both methods yield consistent results, demonstrating the effectiveness of applying physics principles in real-world scenarios.

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  • Understanding of kinetic energy and its formula: KE = 1/2 * m * v^2
  • Familiarity with the work-energy theorem and its application
  • Knowledge of Newton's second law: F = m * a
  • Basic grasp of average velocity calculations under constant acceleration
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  • Study the work-energy theorem in-depth to understand its applications in various physics problems
  • Learn about the principles of constant acceleration and how to apply them in real-world scenarios
  • Explore advanced topics in dynamics, including frictional forces and their effects on motion
  • Investigate the relationship between force, mass, and acceleration through practical experiments
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Physics students, educators, and anyone interested in the dynamics of motion and energy transfer in real-world applications, particularly in understanding the effects of forces on projectiles.

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A bullet with a mass of 4.90 g and a speed of 609 m/s penetrates a tree to a depth of 4.07 cm. Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped.

I thought I could use the equation time = distance divided by velocity, but that's not right. Apparently the answer is 1.33e-4 and I can't figure out how to get it!
 
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You CAN'T use "time = distance divided by velocity" because there is no constant velocity.

My first thought was to use "energy". At first the bullet has kinetic energy (1/2)m v2= (1/2)(0.0049)(609)2= 909 Joules (I converted mass from grams to kg). At the end it has no kinetic energy. The "constant friction force" must have done work equal to that, and since the force IS constant, that is "force times distance". You can use that to find the force: F*0.0407= 909 so
F= 909/0.0407= 22326 Newtons. Once you know the force, you can find the acceleration (f= ma): 22326= 0.0049a so a= 22326/.0049= 4556278 m/s2, and then determine the time for the velocity to reduce to 0: 4556278 t= 609 so t= 609/4556278= 0.00134= 1.34 x 10-4 seconds.


Another way, MUCH simpler (I thought of it AFTER I did the above!) is to use "average velocity". As long as the acceleration is constant, which it is here because the friction force is constant, the average velocity is just "(initial velocity+ final velocity)/2" which, here, is (609+ 0)/2= 304.5 m/s.
At that average velocity, the bullet will require 0.0407/304.5 =
0.000134= 1.34 x 10-4 seconds to penetrate 0.0407 m.
(And we didn't need to use the mass!)
 


To calculate the time elapsed, we need to use the equation for kinetic energy: KE = 1/2 * m * v^2. We know the mass of the bullet (4.90 g) and its initial velocity (609 m/s), so we can plug these values into the equation.

KE = 1/2 * 0.00490 kg * (609 m/s)^2
KE = 1/2 * 0.00490 kg * 370881 m^2/s^2
KE = 907.018 J

Now, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the frictional force is equal to the change in the bullet's kinetic energy as it slows down and stops.

W = F * d
907.018 J = F * 0.0407 m
F = 22277.3 N

Now, we can use Newton's second law, which states that force is equal to mass times acceleration (F = m * a). In this case, the acceleration is due to the frictional force acting on the bullet.

22277.3 N = 0.00490 kg * a
a = 4547143 m/s^2

Finally, we can use the equation for acceleration (a = Δv/Δt) to calculate the time elapsed (Δt).

4547143 m/s^2 = (0 - 609 m/s)/Δt
Δt = -609 m/s / 4547143 m/s^2
Δt = 1.33e-4 seconds

Therefore, the time elapsed between the moment the bullet entered the tree and the moment it stopped is 1.33e-4 seconds.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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