(II) A mass mₐ = 2.0 kg, moving with velocity vₐ ...

Question

(II) A mass mₐ = 2.0 kg, moving with velocity vₐ (→ above v) = (4.0 î + 5.0 ĵ ― 2.0 k̂) m/s, collides with mass m₈ = 3.0 kg, which is initially at rest. Immediately after the collision, mass mₐ is observed traveling at velocity v'ₐ (→ above v') = (― 2.0 î + 3.0 k̂) m/s. Find the velocity of mass m₈ after the collision. Assume no outside force acts on the two masses during the collision.

Accepted Answer

Hey, everyone in this problem, a ball with mass M one equals 1.5 kg is initially at rest. Another ball with mass M two equals 2.5 kg. And velocity V two is equal to three I hat minus four J hat plus two K hat meters per second collides with it. Post collision. The second ball moves with the velocity V two prime of two I hat minus three J hat minus two K hat meters per second. And we're asked to calculate the velocity of the first ball after the collision. We're told to consider that there are no external forces that affect the system during this collision. We have four answer choices all in meters per second. Option A I hat minus J hat plus four K hat. Option B 0.6 I hat mine is 0.6 J hat plus 2.4 K hat. Option C 1.7 I hat minus 1.7 J hat plus 6.7 K hat. And option D 2.5 I hat minus 2.5 J hat plus 7.5 K hat. So let's start by writing out what we have, we have that the mass M one is 1.5 kg. OK. Initially, it's at rest. And so its initial velocity which will write as V one not, is just going to be the zero vector. OK? Zero I hat plus zero J hat plus zero K hat and the final velocity of V one F. Well, this is what we're trying to find. OK. So that's the velocity after the collision. Now, first ball two, it has a mass M two of 2.5 kg. We know it has an initial velocity before the collision. V two knot of three, I had minus four J hat plus two K hat meters per second. And it has a final velocity after the collision, which we're gonna call V two F instead of using the prime notation that they've used in there in the text. And so it's gonna be two I hat minus three J hat minus two K hat meters per second. This is all of the information we were given in the problem and that we're gonna need to use and we have a collision where we have no external forces. And so we can think about the conservation of moment and I, and because we have no external forces and we know conservation um the momentum is conserved, sorry, we have that the initial momentum P not, is gonna be equal to the final momentum. PF Now, in both cases before and after the collision, the momentum is gonna be made up of two components. We have the momentum of ball one and we have the momentum of ball two. So PN one plus PN two is equal to P F one. What P F two recall that momentum is given by mass multiplied by velocity. So we can use that in all four of these terms, we get M one V one not plus M two. Me do not. It's gonna be equal to M one B one F plus M two V two. We can substitute in our values. And we're gonna start with the one not we know that that's zero. So that term is actually just gonna go to zero. We have no momentum from um ball one initially and then we can fill in the rest of the information. And again, we're looking for this V one F value. So substituting in everything we've written above, we get mass to 2.5 kg multiplied by V two N. We're gonna put big square brackets. We have three I hat minus four J hat plus two K hat meters per second. That's gonna be equal to M 11.5 kg multiplied by V one F. That value we're looking for om 22.5 kg multiplied by its velocity after the collision. Two I hat minus three J hat minus two K hat meters per second. All right. So let's keep going. We're gonna give ourselves some more space. And what we wanna do here is simplify, we wanna solve for VF isolate for VF. So we're gonna move this 2.5 kg multiplied by the final speed of vault two. We're gonna move that to the left hand side. OK? And then we're just gonna swap sides. So what we get is 1.5 kg multiplied by V one F is going to be equal to OK? Both of the vectors that we have V two knot and V two F have this 2.5 kg being multiplied out in front of it. So we can factor that out. We have 2.5 kg. And what we're left with is three I hut minus four J hat plus two K hat meters per second minus two. I had minus three J hat minus two K hat meters per second. Ok. Let's go ahead and simplify those vectors inside of the brackets. So we get 1.5 kg multiplied by V one F is equals to 2.5 kg multiplied by one. Well, when we're subtracting vectors, we're gonna do it component wise. And if we subtract the I hat components three minus two. Well, that just gives us I hat and we have minus four J hat minus negative three J hat and that's minus four plus three. It gives us minus one. So we get minus J hat. We have two K hat minus negative two K hat. So two K hat plus two K hat, we get plus four K hat and this is still in meters per second. And so now I've got that simplified down to a single vector. We're gonna go ahead and divide both sides by 1.5 kg. And again, trying to isolate V one F. So now we have V one F is gonna be 2.5 kg divided by 1.5 kg multiplied by this vector. We have I hat minus J hat plus four K hat meters per second. And the last thing we have to do is bring this constant 2.5 kg divided by 1.5 kg into the brackets and apply it to each term of our vector. When we do that, we are gonna get that this final velocity V one F is going to be 1.6666 repeated I ha minus 1.6666 repeated J hat was 6.6666 repeated K hat. And again, still in meters per second. And that is the final velocity of ball one OK? Or the velocity of ball one after the collision that we were looking for. Now, we're gonna go up to our answer choices and we need to round and we're gonna round to one decimal place. And when we do that, we can see that we're gonna get 1.7 I hat minus 1.7. J hat plus 6.7 K hat meters per second. And that is going to correspond with answer choice. C thanks everyone for watching. I hope this video help see you in the next one.

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