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Linear momentum and impulse problem

  1. Apr 19, 2008 #1
    [SOLVED] Linear momentum and impulse problem

    1. The problem statement, all variables and given/known data

    A 3.0-kg steel ball strikes a massive wall with a speed of 10 m/s at an angle of 60 degrees with the surface. It bounces off with the same speed and angle (see Figure 9.30). If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

    [Description of accompanying diagram (Figure 9.30) ...]

    There are x and y axes, with the y-axis representing the wall. A ball is shown coming in at an angle of 210 degrees relative to the positive x-axis (or 60 degrees to the left of the negative y-axis), and leaving at an angle of 150 degrees relative to the positive x-axis (or 60 degrees to the left of the positive y-axis).

    2. Relevant equations

    [tex]\overrightarrow{I} = \int (from t_i to t_f) \overrightarrow{F}dt = \Delta\overrightarrow{p} = \overline{F} \Delta t[/tex]

    Note: [tex]\overrightarrow{I} = impulse; \overrightarrow{F} = force; \Delta\overrightarrow{p} = change in momentum; \overline{F} = time-averaged force; and \Delta t = change in time [/tex]

    3. The attempt at a solution

    The answer in the back of the book says 260 N to the left of the diagram.

    If I reverse engineer the problem, I find that 260 N of force corresponds to 52 N of impulse. Then ...

    [tex]\overline{F} = \frac{\overrightarrow{I}}{\Delta t} = \frac{52 N}{0.20 s} = 260 N[/tex]

    But I can't for the life of me figure out where the impulse comes from. Since the ball enters and exits at the same angle and speed, it looks like the impulse should be zero!

    Please help. Thank you.
     
    Last edited: Apr 19, 2008
  2. jcsd
  3. Apr 19, 2008 #2

    Nabeshin

    User Avatar
    Science Advisor

    Well, momentum is a vector quantity. Therefore, direction is important in figuring out whether an impulse has acted on an object.
     
  4. Apr 19, 2008 #3
    This should read ...

    If I reverse engineer the problem, I find that 260 N of force corresponds to
    [tex]52 N \cdot s[/tex]
    of impulse. Then ...

    [tex]\overline{F} = \frac{\overrightarrow{I}}{\Delta t} = \frac{52 N \cdot s}{0.20 s} = 260 N[/tex]
     
    Last edited: Apr 19, 2008
  5. Apr 20, 2008 #4
    draw the motion of the ball, what i mean is to draw the speed of the ball as in x and y vector components

    so if the ball is striking the surface at 60 degree angle, then it has sin and cosine components of speed,
     
  6. Apr 20, 2008 #5
    I tried doing this with the momentum instead, and it worked! Thank you! Here's my solution ...

    Calculation of momentum of ball before it hits the wall ...

    [tex]\sin \Theta = \frac{opp}{hyp} = \sin\Theta = \frac{p_{y_b}}{p_{before}}[/tex]
    [tex]\Rightarrow p_{y_b} = p_{before}\sin\Theta[/tex]

    Similarly, ...

    [tex]p_{x_b} = p_{before}\cos\Theta[/tex]

    Calculation of momentum of ball after it hits the wall ...

    You'll need to draw your own diagram, but the calculations come to:

    [tex]p_{y_a} = p_{after} \sin\Theta[/tex]
    [tex]p_{x_a} = -p_{after} \cos\Theta[/tex]

    Since
    [tex]p_{before} = mv_i[/tex]
    and
    [tex]p_{after} = mv_f[/tex]
    ; and also because
    [tex]\Theta = 30\circ[/tex]
    (You'll have to draw your own diagrams to see that theta is not 60 degrees) ...

    We obtain the following:

    [tex]p_{y_b} = mv_i \sin 30 \circ[/tex]
    [tex]p_{x_b} = mv_i \cos 30 \circ[/tex]
    [tex]p_{y_a} = mv_f \sin 30 \circ[/tex]
    [tex]p_{x_a} = -mv_f \cos 30 \circ[/tex]

    We then can calculate the impulse ...

    [tex]I_y = p_{y_a} - p_{y_b} = mv_f \sin 30 \circ - (mv_i \sin 30 \circ)[/tex]
    . . . [tex] = ((3 kg) * (10 m/s) * \sin 30 \circ) - ((3 kg) * (10 m/s) * \sin 30 \circ) = 0[/tex]
    [tex]I_x = p_{x_a} - p_{x_b} = -mv_f \cos 30 \circ - (mv_i \cos 30 \circ)[/tex]
    . . . [tex] = (-(3 kg) * (10 m/s) * \cos 30 \circ) - ((3 kg) * (10 m/s) * \cos 30 \circ) = -52 N \cdot s[/tex]

    One can then solve for the average force by using my calculation given in my most recent post!
     
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