Help with this physics problem (acceleration in two dimensions).

[JetsetPyro]

A hockey puck rebounds from a board as shown in Figure 16.

Figure 16 looks like this.

http://img174.imageshack.us/img174/2571/fig16vr6.png

The puck is in contact with the board for 2.5ms. Determine the average acceleration of the puck over the interval.

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So I know you have to use vector components. The answer says the avg. acceleration is 7.3x10^3m/s^2 [7.5 N of W] (degrees).

How do I solve it? =/

I have made an attempt.

I found that the vector components for v1x and v2x to add up to 43.57, and the components of v1y and v1x to add up to (-1.87). However, I'm not getting the correct average acceleration value in the end.

 

[torresmido]

For the second vector the vector in the ight of the picture), what angle did you use?
The angle should be 180°-22° = 158°

 

[JetsetPyro]

I just tried it using 158 as the angle and I didn't get the correct answer given again.

I'm calculating the vector components using sin(theta)(Ay), sin(theta)By, cos(theta)Ax, and cos(theta)Bx of course.

I can't seem to find out what I'm doing wrong.

 

[rohanprabhu]

First you need to fix a coordinate system within which you'll work. Let the 'N' be our y-axis and 'E' our z-axis.

Now try to find the vector of the two velocities. For the first one, it is simply

\overrightarrow{v_i} = 26(\cos{(22)}\hat{i} - \sin{(22)}\hat{j})

it is important to note here that i have taken component on the 'y' axis to be negative. This is because, the direction of the 'y' component is downwards. Here is the diagram to resolve this vector:

http://img142.imageshack.us/img142/4125/resolvingonefc5.png

Now.. find the vector v_f yourself. And then, remember that acceleration is given by:

\overrightarrow{a} = \frac{\overrightarrow{v_f} - \overrightarrow{v_i}}{\Delta t}

 

[JetsetPyro]

|vf| = 21(sin(22)+cos(22))
|vf| = 27.337

|a| = (|vf|-|vi|)/(/\t)
|a| = (27.337-14.37)/0.0025s
|a|= 5186.8

Hmmm, did I do something wrong?

 

[rohanprabhu]

|vf| = 21(sin(22)+cos(22))
|vf| = 27.337

well.. for one.. you did many things wrong. First of all, you did not assign unit vectors to the \overrightarrow{v_f}. Also, there is a difference between \overrightarrow{v_f} and |v_f|. Also, |v_f| cannot be anything other than 21.

When finding the acceleration, you need to subtract vectors first and then take modulus. I suggest u refer your textbook and understand vectors.

 

[JetsetPyro]

I couldn't assign unit vectors to vf (with direction) as I don't know how to show that on this forum. =/

Also, we haven't been taught "modulus" or is that modules? I don't know what they are.

 

[ablemerle1]

impulse

 

[rohanprabhu]

I couldn't assign unit vectors to vf (with direction) as I don't know how to show that on this forum. =/

Also, we haven't been taught "modulus" or is that modules? I don't know what they are.

To learn how to write in LaTeX, check this: https://www.physicsforums.com/showthread.php?t=8997

also.. u can show unit vectors using 'i' or 'j' like: vf = 21(cos(22)i + sin(22)j). Modulus of a vector is the 'magnitude' of the vector. If there is a vector:

<br /> \overrightarrow{l} = a\hat{i} + b\hat{j}<br />

then,

<br /> |\overrightarrow{l}| = \sqrt{a^2 + b^2}<br />

the modulus gives the length of the vector. For a velocity vector, the modulus is the magnitude of the vector, in the case of v_f, it is 21. For a vector in the form:

<br /> \overrightarrow{l} = p(\cos{(\theta)}\hat{i} + \sin{(\theta)}\hat{j})<br />

the magnitude will always be 'p'. Also, even if u do it using impulse, it eventually will end up the same away. You'll assume a mass.. you'll find the change in momentum [which is nothing but the change in velocity multiplied by mass].. and then you'll divide it by time to find force and then reduce the 'm' to give u acceleration. Taking an extra 'm' will serve no purpose. So don't do it using impulse or collisions since only the acceleration is asked.