<p>Do not tell me the answer, because I want to try it myself. But do tell me the formula(s) and what I need to do first.</p>
<li><p>A tennis ball of mass m = 0.060 kg and speed v = 25 m/s strikes a wall at a 45 degree angle and rebounds with the same speed at 45 degrees. What is the impulse given the wall?</p></li>
<li><p>A 115-kg fullback is running at 4.0 m/s to the east and is stopped in 0.75 s by a head-on-tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.</p></li>
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<ol>
<li><p>what is the definition of impulse? how does it relate to momentum? hint: impulse = change in momentum. (look at the momentum in the x and y directions seperately)</p></li>
<li><p>look at 17.</p></li>
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<h1>17 is asking for the impulse, basically it's asking for the "change in momentum". But is this "change in momentum" between before the ball hit the wall, and after it has hit it, or what?</h1>
<p>Seiously, I do not know what to do.</p>
<p>As Bobobabob said, look at the velocity & momentum of the ball in x & y directions separately; see what it was before hitting the wall, and then after hitting the wall. The difference between before/after is the "change in momentum".</p>
<p>The problem says that the velocity is the same after it hits the wall. Obviously the mass is also the same. So isn't the momentum of before and after the same?..since the "change in momentum" is the product of the mass and the velocity.</p>
<p>Careful. The ball has the same speed (= magnitude of its velocity vector, if I remember my definitions), but <em>not</em> the same velocity.</p>
<p>Try this. If you had to show the initial velocity of the ball (remember, it's travelling at a 45 degree angle to the wall, or 'north-east') as its components in the x- and y- directions (or 'east' and 'north'), what would be the size of these components? HINT: If the ball had x- and y- components each of size 1m/sec, its net velocity would be sqrt(2) m/sec .</p>
<p>Now do the same thing, except now for the ball's velocity in the 'north-west' direction after striking the wall. What's the delta in the y-components? What's the delta in the x-components?</p>
<p>Ok. For the magnitude of the x and y components before the ball hits, is it 17.68 for both of them?</p>
<p>How would I find the x and y components after the ball hits, without knowing the magnitude of of the vector?</p>
<p>'...rebounds..with the same speed..'</p>
<p>So the magnitude after the ball hits is also 25. Then the components of x- is 17.68, and y- is -17.68 again?</p>
<p>Yes, depending on how you defined your x- and y- axes. For the sake of convenience, let's associate them with directions. Is your y-axis pointing 'north', the x-axis pointing 'east', and the wall parallel to the y-axis? Or do you have a different (not necessarily wrong) set of assumptions?</p>
<p>Note that one of your velocity components switched sign; what's the before/after difference in velocity along this direction?</p>
<p>This is how I drew my axes. I put the wall on the y- axis. I put the point where the ball hits the wall at the origin. So the vector after the ball hits is pointing at 135 degrees, and the vector before the ball hits is pointing at 225 degrees.</p>
<p>The difference is 0?</p>
<p>If that's the way you have your axes, then your y-component should be unchanged; the x-component should change from +17.68 to -17.68 .</p>
<p>I already found the components of each velocity vector. If i subtract, I will get 0?</p>
<p>'..difference is 0?'</p>
<p>Nope. Delta x-velocity = x<em>velocity</em>after - x<em>velocity</em>before
= -17.68 - (+17.68)
= -35.36 m/sec</p>
<p>Since this has a minus sign, the ball loses (35.36)(0.060) units of momentum in the positive x direction; the wall gains this, in the positive x direction, or 'east'.</p>
<p>Do I do the same with the y- components?</p>
<p>Oh, oh...look at your post#12. The 'before' vector should be at 45 degrees, not 225.</p>
<p>Oh. The y- difference is 0.</p>
<p>As for the y-components, they stay unchanged, so the difference will be zero.</p>
<p>....but if you really want to be sure :), you can compute 17.68 - 17.68 = 0...</p>