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A ball with an initial velocity of 10 m/s moves at an angle $60^\circ$ above the $+x$ direction. The ball hits a vertical wall and bounces off so that it is moving $60^\circ$ above the $-x$ direction with the same speed. What is the impulse delivered by the wall?
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Final Answer
$-10m$, where $m$ is the mass of the ball in kilograms.
Solution Video

OpenStax College Physics Solution, Chapter 8, Problem 20 (Problems & Exercises) (5:53)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A ball is hitting a wall and approaches at an angle of 60 degrees with respect to the horizontal and then bounces off with an angle of 60 degrees with respect to the horizontal by going the other way and I'm gonna answer this question in sort of a long answer. You could notice that since the final speed and the initial speed we are told is the same that the change in velocity is going to be entirely horizontal to the left and you can just find the horizontal component of this and then double it but I'm gonna answer this question more generically which will account for what if these speeds are different... what if these angles are different, that will be accounted for in this method of solution. Okay! So our job is to find the impulse so that is the change in momentum which is the mass times the final velocity minus the mass times the initial velocity and then we can factor out the mass because the mass isn't changing and that is <i>m</i> times <i>Δv</i> then <i>Δv</i> being the change in velocity. So <i>Δv</i> is the final velocity minus the initial velocity and I like to write subtractions for vectors as addition of a negative because when you are adding vectors, you are putting the tail of the second on the head of the first and the second vector is gonna be negative <i>v</i> so that will be this initial velocity switched around in the opposite direction so it will be going this way... this is negative <i>v</i>. So we have <i>v prime</i>—the final velocity— 60 degrees with respect to horizontal and then at the head of that vector, we put the tail of negative <i>v</i> and that goes here also 60 degrees with respect to horizontal and then this resultant here connecting the tail of the first to the head of the last is our change in velocity: it's <i>v prime</i> plus negative <i>v</i>. Okay so we need to figure out what this is then. So the x-component of this resultant is going to be the sum of the x-components of these vectors that we are adding together. So that's... now we have 'right' to be defined as the positive direction so since this x-component for <i>v prime</i> is to the left and that is this little portion here in this right triangle here, this is <i>v prime x</i> is gonna be <i>v prime</i> times <i>cosine</i> of 60 because this is the adjacent leg of this right triangle. So that's negative because it's pointing to the left, <i>v prime</i> times <i>cos</i> 60, and then add to that a negative number because the x-component for minus <i>v</i> is also to the left so we have minus <i>v</i> times <i>cos</i> 60 and so that's this component here for this triangle. Okay and then we can factor out the negative <i>cos</i> 60 and multiply that by the sum of the magnitudes of these two vectors here. And then for <i>Δv y</i>, we have <i>v prime</i> times <i>sin</i> 60— that's <i>v prime y-component</i>; that's going upwards so it's positive— and then minus because this is downwards this is the minus <i>v y-component</i> so minus <i>v</i> times <i>sin</i> 60 again and we factor out the <i>sin</i> 60 and we get that multiplied by <i>v prime</i> minus <i>v</i>. Now <i>v prime</i> equals <i>v</i>—we are told the speed after impact is equal to the speed before impact— and so we can replace <i>v prime</i> with <i>v</i> and for <i>Δv x-component</i>, we have negative 2<i>v</i> when you add the <i>v</i> plus the <i>v</i> negative 2<i>v cos</i> 60 and for <i>Δv y</i>, we have these are the same so they make zero and so the y-component is zero. And so the magnitude of this resultant <i>Δv</i> then is the square root of the sum of the squares of the x- and y-components but the y-component being zero gives us <i>Δv x</i>, in the end, is the magnitude of <i>Δv</i>. So <i>Δv</i> then is negative 2<i>v cos</i> 60 since that's what <i>Δv x-component</i> is. And then we plug that into our impulse formula for <i>Δv</i> and we have then the impulse is <i>m</i> times bracket negative 2<i>v cos</i> 60 and this solution is a long one but the reason it's useful is because it shows you what to do in a situation where the speeds are not necessarily the same and it emphasizes the fact that this <i>ΔP</i> is a vector— this is not a scalar quantity, this is a vector quantity— and so you have to take the difference in momenta as vectors. Okay! Often you don't think really about the fact that this is a vector because often your questions are along a straight line but here's an example where it isn't and so you have to really treat the vector properties of momenta in this case. Okay! So we have negative 2<i>m</i> times 10 meters per second times <i>cos</i> 60 and that works out to negative 10 meters per second times whatever <i>m</i> is—the mass—and it has to be expressed in kilograms in order for this formula to work.