A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. b. What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?

Video transcript

Hey, everyone. Welcome back in this problem. An experiment consists of releasing a metallic sphere from rest from the top of a vertical tube of length L and at the same instant, a pop gun located at the bottom of the tube fires a ping pong ball at a speed of five m per second. We're asked to determine the greatest length the tube can have in order for the collision to take place before the ping pong ball hits the bottom of the tube. We're given four answer choices. A is 0.2 m. B 1.2 m C 2.8 m and D 5.10 m. Now let's go ahead and draw a diagram of what's going on in this situation. So we have a tube and it's vertical and it has a link no little. Now, we have a metallic sphere at the top of the tube and it's going to be released from rest. We're gonna say that V and we're gonna call it S for the sphere V, not S is equal to zero m per second. And we have a ping pong ball at the bottom that is going to get fired upwards at five m per second. So the initial speed would be not P for the ping pong ball is five m per second. All right. And sometime later you can imagine they're both gonna travel and they're gonna end up colliding and we're not sure what that collision looks like, but they are going to collide. Now we're gonna call the distance that the beer travels is going to be delta Y S in the distance the ping pong ball travels is going to be delta Y P. And we can see from this diagram that the sum of delta Y S and delta Y P is going to be L. Now let's take up to be the positive wider direction for our problem. And I think we have everything we need in our diagram. So we can get started on writing out this information. Now, we want to rate the variables we have for our sphere and for the ping pong ball. So for the sphere, we've already written that the initial speed Vena S is 0m/s. We don't know anything about the speed when they collide, we know the acceleration will be the acceleration due to gravity because we've chosen up to be a positive direction. This acceleration is going to be negative 9.8 m per second squared delta Y S we've drawn this in our diagram and we have the time T S you're doing the same for the ping pong ball. The initial speed V not P is going to be five m per second. We don't have information about the final speed. The acceleration is the same acceleration negative 9.8 m per second squared. OK. The acceleration due to gravity we have delta Y P and we have S T E P. OK. So if we want to solve for any of these variables, yeah, we can't do that just yet. For each of the sphere in the ping pong ball, we only have two known values. Hm We can write two equations. So we can write an equation for the sphere and we can write an equation for the ping bong ball. So we can relate the two through some of these variables. So we may have to substitute one of the equations into the other. So remember that I mentioned when we drew delta Y S and delta Y P on our diagram that the sum of the two is going to equal to L. Now for delta Y S, we're starting at the top and moving down. So this is going to be a negative distance and it's gonna move a distance of L minus delta Y P. OK. So delta Y S is equal to negative L minus delta Y P. And the time, well, when they collide it's at the same time in the same place. And so the time T S is just going to be equal to the time T P. So now we have two equations that we can write with two unknowns. So we can solve for the values we want. Now we don't have information about the final speed. So let's leave that out of our equation. We're gonna choose the kinematic equation that does not include V F. Let me give that delta Y S is equal to V R S multiplied by T S plus one half multiplied by A S multiplied by T S squared. Substituting in what we know delta Y S is going to be equal to negative in brackets little L minus delta Y P, this is equal to the initial speed is zero. So this first term on the right hand side goes away, we have one half multiplied by negative 9. m per second squared multiplied by T P squared because T S and T P are equivalent. Now we can write the same equation for the ping pong ball. So we have that delta Y P is equal to V, not P multiplied by T S OS, not T S T P plus one F multiplied by A P multiplied by T P squared. Substituting in the values we know we have delta Y P is equal to five m/s, multiplied by T P plus one half multiplied by negative 9.8 m per second squared multiplied by T P squared. Now we're gonna label these equations. So the equation we wrote for the ping pong ball is going to be equation two. The equation we wrote for this sphere is going to be equation one and we are substitute one of these equations into the other, right. And you know what, before we label this first equation, let's simplify it a little bit further to make our life easier. So we currently have the equation negative and in brackets little L minus delta Y P is equal to one half multiplied by negative 9.8 m per second squared, multiplied by T P squared. We're gonna move this negative L term to the right hand side. And we have delta Y P OK, because we have negative multiplied by a negative. So we have positive delta Y P Is equal to one half multiplied by negative 9.8. This is going to give us negative 4. meters per second squared, Multiplied by TP 2ared minus oops plus L OK. We had a negative on the left hand side. So we have to add it to move it to the right hand side. Now let's call this equation one. And we can see that equation one in delta Y P equals something. Equation two and delta Y P equals something. So let's substitute equation one into Equation two. What we're essentially doing is setting them equal to each other because they're both equal to the same delta YP. And so we get negative 4.9 m per second squared, multiplied by T P squared plus L is equal to five m per second. Multiplied by T P. We have one half multiplied by 9.8 m per second squared. So we get negative 4.9 meters per second squared multiplied by T P squared. All right. Now you'll notice that we have this minus 4.9 m per second multiplied meters per second squared, multiplied by T P squared term on both sides. OK. So we can move one to the other side and those are going to cancel out. So we are left with T P Is equal to L over five m/s. Now remember that T P is the time when the collision is going to happen. OK. So we know what time the collision will happen. Now let's go ahead and figure out where the collision will happen. OK? So we wanna find delta Y P, the distance that ping pong ball has traveled in order to set a condition on it so that it the collision happens before it hits the bottom of the tube. So we have our condition on the time. Let's move over And substitute this back into our equation number one. Now, equation one was delta Y P Is equal to negative 4.9 meters per second squared, Multiplied by TP two. Now we're gonna replace T P with L over five m per second. OK? We found that using the substitution method minus whoops plus L. Now, in order for this collision to occur before the ball hits the ground, we want delta Y P to be greater than zero. So we want delta YP to be greater than zero, which tells us that negative 4.9 meters per second squared multiplied by L squared over 25 m squared per second squared plus Should be greater than zero. All right. So now we have an inequality to solve. We've worked through the physics. We have to solve this inequality. We have an L in both terms. So let's factor that out. We get L multiplied by negative 4. m per second squared, divided by 25 m squared per second squared multiplied by L 41 Is greater than zero. Now, what do we know about L that can help? Well, L is the length of the tube. So L has to be greater than zero. OK. That means that this first term which is just L is greater than zero. So in order for L multiplied by this second term to be bigger than zero, Then the second term also needs to be greater than zero. OK. If it's negative, we have a positive times a negative which will give us a value less than zero. So we need -4. meters per second squared divided by m squared per second squared Plus one greater than zero. This tells us that 4. meters per second squared, Divided by meter squared per second squared. L has to be less than one. OK? Because we moved that term to the right hand side. So now it's on the side with the less then and we can divide by 4.9 m per second squared, divided by 25 m squared per second squared. And this gives us that L must be less than 25 m squared per second squared, divided by 4.9 m per second squared. OK. And if we work this out on our calculator, this tells us that L must be less than 5.1 m. All right. So that was a long problem. Let's just walk through what we've done. We have two balls, one being dropped from the top, one being fired from the bottom of this two. And we wanna make sure they collide before the ping pong ball hit the ground. So we used our kinematic equations, we had to substitute one equation into the other in order to find the time that they would collide. Once we knew what time the balls could collide, we could go back and look at the displacement of the ping pong ball at the time of collision. And we know we need that displacement to be greater than zero in order for the balls to collide before it hits the ground. And so we find that the length of the tube L must be less than 5.1 m. So we go back up to our answer choices that tells us that the greatest length the tube can have is 5.1 m which corresponds with answer choice. D Thanks everyone for watching. I hope this video helped see you in the next one.