Closest Approach and Constrained Motion

Transcript

0:00Introduction

Hello everyone! First, one small puzzle from the sea. Ship A is due north of ship B, okay? Ship A sails east, and ship B sails north, straight towards the spot where ship A started. Will they crash? And if not, how close will they come, and when? Hmm, tricky, right? Today, one smart idea answers all of it. And look, closest approach and strings over pulleys both come in JEE again and again. So, let's begin.

0:32Sit on A: B moves in a straight line

So, both bodies move; that is what makes it hard. Remember the trick from earlier in this chapter: work in the frame of one body, say body A: now it is at rest, and body B moves with v B A vector. The arrow on top means vector, and B slash A means B relative to A, okay? It equals v B vector minus v A vector, constant because both velocities are constant. Now tell me. Sitting on A, you see B move with one constant velocity. So what path does it follow? Yes, a straight line! B starts at r zero vector, its position relative to A; the small zero below means at time zero. From there, it moves along one straight line. Clear? Now picture it. Body A sits still, and B slides past along this line. Where is B closest to A? At the foot of the perpendicular, exactly! So the least distance is the perpendicular distance from A to this line. Two moving bodies became one fixed point and one line. Simple, isn't it?

1:39How close, and when?

Now, how close? d min, the small min means minimum, equals mod of r zero vector cross v B A vector, by mod of v B A vector. The straight bars mean mod, the magnitude, and the cross is the cross product. This is exactly that perpendicular distance. Write it down. And when? t equals minus, r zero vector dot v B A vector, by mod v B A squared; the dot is the dot product. Look, don't drop this minus sign. While B comes closer, this dot product is negative, so the minus makes the time positive. But if this dot product is positive, B is already moving away, so the closest distance is right at the start: mod r zero vector, not this d min. Got it? Alright, a topper's shortcut, since most JEE questions are in a plane. Write r zero vector as x zero comma y zero, and v B A vector as v x comma v y. The double arrow means so: the cross product is just x zero v y minus y zero v x, inside mod. That's it!

2:43Do they collide?

Now the other big JEE question: do they collide? Sitting on A, when will B actually hit it? When B heads straight at A, that's right! r zero vector points from A to B, so the velocity must point straight back towards A, okay? So r zero vector is exactly opposite to v B A vector. Then the cross product is the zero vector, the zero with an arrow, so d min is zero. They collide. Not exactly opposite? Then no collision, however close it looks. With me?

3:15Strings: only the part along the string counts

Alright, now the second idea: strings. An inextensible string, one that cannot stretch, has the same speed at every point along it, okay? Now, a block on a table. Its string goes over a pulley, at height h above the block's line. You pull the free end at a steady speed u. Now, theta is the angle the string makes with the horizontal. The block slides along the table, so its speed is the part of u along the table: u cos theta, right? No, no, careful! That is the classic JEE trap. The block moves across the string, so only its component along the string equals the string's speed. Along the string, the block's velocity v has the component v cos theta. So v cos theta equals u. Not u cos theta! Divide by cos theta: v equals u by cos theta. Now think. As the block nears the pulley, the string gets steeper. What happens to v? It speeds up, yes! Nearer the pulley, theta grows, cos theta falls, so u by cos theta grows. Steady pull, but the block keeps speeding up. Interesting, right?

4:23Example 1: least separation of two moving particles

Now, Example one. Particle A starts at the origin with velocity three i cap metres per second; i cap, the i with a hat, is the unit vector along x, and m s to the power minus one is metres per second. Particle B starts at zero comma eight metres, with velocity minus four j cap, the unit vector along y. Find their least separation. Pause the video, and try it first. Done? Okay, let's check. Sit on particle A, at the origin. So what are r zero vector and v B A vector? Yes! r zero vector is just B's position: eight j cap metres. And v B A vector is v B minus v A: minus four j cap minus three i cap, that is, minus three i cap minus four j cap. Its magnitude: the square root of three squared plus four squared. Our old friend, three four five, remember? Five metres per second. Now our shortcut: x zero is zero, y zero is eight, v x is minus three, v y is minus four. Zero into minus four, minus eight into minus three; the dots mean multiply. That gives twenty four. Twenty four by five: d min is four point eight metres. Now the time. The dot product: zero into minus three, plus eight into minus four. Minus thirty two. Negative, see? B is coming closer. So t is minus, minus thirty two, by five squared. Thirty two by twenty five: one point two eight seconds. Done!

5:57Example 2: two ships, least distance and when

Now, remember our two ships? Here they are, with numbers. Ship A is ten kilometres north of ship B, okay? Ship A sails east at thirty kilometres per hour, that is km h to the power minus one, and ship B sails north at forty. Find their least separation, and when it occurs. Pause and try. Hint: sit on ship B this time. Done? Okay, let's check. Sit on ship B, at the origin, x towards east, y towards north. Ship A starts ten kilometres north: r zero vector is zero comma ten kilometres. Its velocity relative to ship B is v A B vector; A slash B means A relative to B, okay? Thirty comma zero, minus zero comma forty: thirty comma minus forty kilometres per hour, east and also southwards. Now, back to our puzzle. Ship B sails straight at the spot where ship A started. So, do they collide? No, they don't! r zero vector points north, so for a crash, v A B vector must point due south, straight back at ship B, okay? But thirty comma minus forty also goes east. Not opposite, so no collision. Clear? Then how close? The relative speed: the square root of thirty squared plus forty squared, fifty kilometres per hour. Shortcut again: zero into minus forty, minus ten into thirty, inside mod, gives three hundred. Three hundred by fifty: d min is six kilometres. And when? On top, the dot product: zero into thirty, plus ten into minus forty. That is minus four hundred. With the minus sign in front, t is four hundred by fifty squared, that is four hundred by two thousand five hundred, in hours. Zero point one six hours, or nine point six minutes. And there it is, puzzle solved! The ships never crash; they come within six kilometres.

7:56Quick revision: closest approach and string constraint

So, quick revision. Sit on one body: the other moves in a straight line. The least distance is the perpendicular from the body to that line. And for strings, only the component along the string equals its speed. Now go and try the practice problems on this page. You can do it. All the best!