OpenStax College Physics, Chapter 8, Problem 20 (Problems & Exercises)

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 m times Δv then Δv being the change in velocity. So Δv 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 v so that will be this initial velocity switched around in the opposite direction so it will be going this way... this is negative v. So we have v prime—the final velocity— 60 degrees with respect to horizontal and then at the head of that vector, we put the tail of negative v 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 v prime plus negative v. 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 v prime is to the left and that is this little portion here in this right triangle here, this is v prime x is gonna be v prime times cosine of 60 because this is the adjacent leg of this right triangle. So that's negative because it's pointing to the left, v prime times cos 60, and then add to that a negative number because the x-component for minus v is also to the left so we have minus v times cos 60 and so that's this component here for this triangle. Okay and then we can factor out the negative cos 60 and multiply that by the sum of the magnitudes of these two vectors here. And then for Δv y, we have v prime times sin 60— that's v prime y-component; that's going upwards so it's positive— and then minus because this is downwards this is the minus v y-component so minus v times sin 60 again and we factor out the sin 60 and we get that multiplied by v prime minus v. Now v prime equals v—we are told the speed after impact is equal to the speed before impact— and so we can replace v prime with v and for Δv x-component, we have negative 2v when you add the v plus the v negative 2v cos 60 and for Δv y, we have these are the same so they make zero and so the y-component is zero. And so the magnitude of this resultant Δv 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 Δv x, in the end, is the magnitude of Δv. So Δv then is negative 2v cos 60 since that's what Δv x-component is. And then we plug that into our impulse formula for Δv and we have then the impulse is m times bracket negative 2v cos 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 ΔP 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 2m times 10 meters per second times cos 60 and that works out to negative 10 meters per second times whatever m is—the mass—and it has to be expressed in kilograms in order for this formula to work.