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Linear momentum and impulse problem

[SOLVED] Linear momentum and impulse problem

1. Homework Statement

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. Homework 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.
 
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Answers and Replies

Nabeshin
Science Advisor
2,204
16
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!
Well, momentum is a vector quantity. Therefore, direction is important in figuring out whether an impulse has acted on an object.
 
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]
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]
 
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36
0
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,
 
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,
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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