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Physics mechanics problems, please help

  1. Jul 26, 2013 #1
    I have attempted the questions but have not managed to get the required answer because my knowledge of mechanics is very weak i have a problem visualising the scenario, could you please explain the steps behind each question to get to the final answer, thank you very much

    I managed to find the answer to number 2) but do not understand what they were asking for when they said "without giving it any velocity components". I realised that usually i would use the horizontal component to find the time which i would then put in the y=1/2gt^2 formula to find the height, but here the height was already given as 100m so i just rearranged the equation to find the time. Once i found the time i put it in the x=vt formula to find the answer, x=vt usually would give you the range, so when they are saying no velocity components i am sure there is a link somewhere. Because the flyer has a horizontal speed? therefore does no velocity components mean that i find my answer in the same direction via x=vt which is the horixontal component of velocity.

    Here are the problems.


    1) A gun fires a shell which hits a target on the same level as the gun and is 1300m away. The muzzle velocity is 120 m/s. What angle of elevation to the horizontal was the barrel at? answer is (31 degrees)

    2) A powered hang glider is moving at 15 m/s parallel to and 100m above flat ground. The flyer wishes to hit Dr. Mustoe, who is sitting on a park bench, with a bag of flour. He intends to release the bag without giving it any velocity components relative to the hang glider. How far away from the bench should he release the bag? Answer is (67.8m)

    3) A 600kg car is moving on a level road at 30 m/s. How large a retarding force is required to stop it in a distance of 70m and what is the required coefficient of friction between tyre and road for this to be possible? Answer is (-3.86kN, 0.66)

    4) A 20kg box sits on an incline of 300. If the coefficient of sliding friction between the box and incline is 0.3 what is the acceleration of the box down the incline? Answer is (2.35 m/s2)


    Thank you very much
     
  2. jcsd
  3. Jul 26, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    For 1), you know the horizontal displacement AND the horizontal velocity. You can use those to find time.

    Consider the parabolic nature of the bullet due to gravity. You know it reaches its peak height at (time/2).

    Using that fact, you can find the vertical velocity.

    Wait, now you have vertical AND horizontal velocity. Consider tan(theta).....
     
  4. Jul 26, 2013 #3
    Problem 1

    With a weak understanding of mechanics, knowing how to solve these 4 problems is not likely to help very much in the long run. However, it is a start. Make sure that you read the textbook BEFORE class and then reread the example problems from the textbook AFTER class. Without this type of a significant commitment, you will likely find your class unmanageable.

    That being said, we are here to help.

    1) A gun fires a shell which hits a target on the same level as the gun and is 1300m away. The muzzle velocity is 120 m/s. What angle of elevation to the horizontal was the barrel at? answer is (31 degrees)
    So we have motion in two direction: x and y.

    To decompose the velocity vector, we use [itex]v_{0x} = v_{0} cos (\theta) = 120(m/s) cos(\theta)[/itex] and [itex]v_{0y} = 120 sin (\theta)[/itex]. [itex]\theta[/itex] is the elevation angle we are trying to find.

    In the x direction, there is no acceleration. In the y-direction, there is acceleration due to gravity. ALL projectiles travel in the x direction for the same amount of time they travel in the y direction, so we can use that fact later.

    In the x-direction, the bullet travels at a speed given by [itex]v_x = \frac{x}{t}[/itex], and we know that x is 1300m. We also know that v is constant because there is no acceleration. So we use [itex]v_{0x} = v_{0} cos (\theta)[/itex] to get [itex] 120*cos (\theta)=\frac{1300}{t}[/itex]. There are two unknowns here, [itex]\theta[/itex] and t. We will need a second equation.

    We can get another equation from the y direction. There is a very common "trick" to find [itex]v_y[/itex]. We can use the fact the at the top of the path, the object has traveled for half of the total time it will travel for and also, at the top of the path of any projectile, the vertical speed is 0m/s because it has stopped moving upward and is about to start moving down.

    [itex] v_{fy} =v_{0y} +a_y t [/itex]. [itex] v_{fy}[/itex] is 0m/s, a is -9.8m/s2.

    So using [itex]v_{0y} = v_0 sin (\theta)[/itex] and [itex] v_{fy} =v_{0y} +a_y t [/itex]. [itex] v_{fy}[/itex] we can see that [itex] 0 = 120 * (\theta) -9.8 t [/itex]

    Now we have two equations and two unknown variables, so the rest of the problem is algebra. Solve the y-equation to get t by itself, and plug it into the other equation. You will end up with a tangent to solve, like the first reply mentioned.

    I hope that helped. I might have time later to look at some of your other problems. You should start by going back and reading through the examples in the chapter and see if you can relate those example problems to these problems.

    Dr Peter Vaughan
    BASIS Peoria Physics
     
  5. Jul 27, 2013 #4
    Hi, thanks for replying guys. So the key in the question 1 is to realise that as there is an angle in the flight path we know that it has a velocity component in the y direction? and in-order to find the velcity in the y direction we know that its final velocity will he zero half way through the flight path because at its highest point the v in the y direction is always zero.

    So we have equation 1 which is 120∗cos(θ)=1300/t and equation 2 which is 0=120∗(θ)−9.8t

    I rearranged equation 2 to find t which was, t=120*sin(θ)/9.8, t=12.24sin(θ). I then enter this in equation 1 to find the angle.

    So from equation 1, 120∗cos(θ)=1300/t and this becomes 120∗cos(θ)=1300/12.24sin(θ) when equation 2 is entered.

    Then 120∗cos(θ)=106.2*sin(θ), so sin(θ)/cos(θ)=1.13, tan(θ)=1.13, and the inverse of tan gives 48.5 degrees.

    where have i gone wrong as the answer is supposed to be 31 degrees, thanks.
     
  6. Jul 27, 2013 #5
    This is correct.

    Earlier, you said "half way through" - where is this used in the equations?
     
  7. Jul 27, 2013 #6

    gneill

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    Staff: Mentor

    hbk69, you should create a separate thread for each separate question and follow the provided posting template.

    Regarding the first question, you know that the vertical and horizontal motions of the projectile are independent, so that you can write an equation for each. Hint: replace 't' in the horizontal equation of motion with an expression for t obtained from the vertical motion equation (consider the moment of impact, that is, when the projectile reaches its final location).
     
  8. Jul 27, 2013 #7
    I think that is where the problem lies, i don't know how to reflect it in the equation, i tried dividing the t by 2 in the equation but i still got 29.5 degrees which is close to the 31 degrees but still wrong
     
  9. Jul 27, 2013 #8
    Okay, i will create separate threads for each question. Can mods please edit my first post, by deleting questions 2,3 and 4. Thanks. This thread will now be just for the first question.

    I had done exactly that as shown in my previous post but i think i made a mistake with the time
     
  10. Jul 27, 2013 #9
    Which equation is "the equation"?
     
  11. Jul 27, 2013 #10
    the equation in the x direction is 120∗cos(θ)=1300/t and equation in the y direction is 0=120∗(θ)−9.8t, i rearranged the equation in the y direction to find t and then put that value for t into equation in the x direction, but my calculation as i have shown in the previous post was wrong.
     
  12. Jul 27, 2013 #11
    In one of these equations, you have to make use of the fact that the time to the apex of the trajectory is half the time of the entire flight.

    Then you need to be careful with algebra, the derivation you showed earlier has a mistake that leads to a completely wrong result.
     
  13. Jul 27, 2013 #12
    this is the equation which i need to show that t is half the time, 0=120∗(θ)−9.8t but how can i show that? and what is the mistake in the derivation? thanks
     
  14. Jul 27, 2013 #13
    If "t" is the duration of the entire flight, then this equation must have "t/2" instead of "t".

    At some stage you had 120∗cos(θ)=1300/12.24sin(θ), which really should have been 120∗cos(θ)=1300/(12.24sin(θ)), that is, ## 120 \cos \theta = \frac {1300}{12.24 \sin \theta} ##, but you interpreted that as ## 120 \cos \theta = \frac {1300 \sin \theta}{12.24} ##. Then you divided the equation by cos(θ), and ended up with tan(θ), which is completely wrong.
     
  15. Jul 27, 2013 #14
    so i could multiply each side by 2 to get the correct value for t?

    also what would be the correct way of rearranging the equation to get tan(θ)?
     
  16. Jul 27, 2013 #15
    Multiplying each side by 2 does not change anything at all. Use "t/2" instead of "t", but just in that one equation.

    The whole point is that you should not be getting tan θ - that is an error.
     
  17. Jul 27, 2013 #16
    but shouldn't sinθ/cosθ give me tan θ? and that way i can find the angle?
     
  18. Jul 27, 2013 #17
    sin θ / cos θ is indeed tan θ. But you got sin θ / cos θ by a mistake. Reread my message.
     
  19. Jul 27, 2013 #18
    In this problem you will not end up with a tangent (sorry for saying that you would earlier, I hadn't solved the problem and I assumed it would come out that way). When you look through what voko said and are careful, you will end up getting the product of sin*cos. Then you can use the trig identity that [itex]sin(2\theta)=2sin(\theta)cos(\theta)[/itex].

    There are a few things to keep in mind here. When you get down into the long calculation part in physics, it's important to never skips any steps. It may save you one minute on any particular problem, but you give all that time back when you make a mistake. The other thing to remember is that this algebra part of the problem is absolutely the least important part in doing physics. Figuring out how you would get to the answer is far more important that not making a math error in actually arriving at the answer.

    So make sure that instead of writing "t" in the equation for the y-direction, you write "t/2" since we are only going for half of the total time in the y direction. Then be careful with your multiplication and division, and if you don't end up with [itex]sin(\theta)cos(\theta)[/itex], double check you work, and if you still can't get it, come back and we'll help.

    Dr Peter Vaughan
    BASIS Peoria Physics
     
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