Oblique Impact: Walls, Floors and Smooth Spheres

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Shine light on a mirror at a slant, and it comes off at exactly the same angle, right? Now throw a ball at a slant onto a smooth floor. Does the ball also bounce off at the same angle, just like the light? Most students say yes. By the end of this lesson, you will know the exact angle, and whether that answer is right. This is oblique impact, collisions at a slant, and it comes up often in JEE, so let's begin.

0:31Line of impact and common tangent

So, look at the figure. Two smooth spheres, capital A and capital B, touching at one point. Sphere A comes in at a slant, with velocity u, the black arrow. The orange dashed line through both centres is called the line of impact. And the thin black dashed line, where the spheres touch, is the common tangent. On the right, the figure already lists the two rules we are going to build. With me? Now think. At contact, the spheres push on each other for a very short time. The caption under the figure says this push, the impulse, acts along the line of centres. But why along the line of centres, and not along the tangent? Because the spheres are smooth, yes, exactly! Smooth surfaces can only push straight out, along the common normal, and for two spheres that normal is the line joining their centres. So the whole impulse acts along this one line. Clear? Now, these spheres are smooth, so there is no friction between them. So tell me, along the common tangent, what happens to each sphere's velocity component? Nothing at all! Correct! No force acts along the tangent, so for smooth bodies, each velocity component along the common tangent stays unchanged. But remember, this needs smooth bodies. Okay? And along the line of impact, we work exactly as in one D, that is, one dimension. Total momentum is conserved, and e, the coefficient of restitution, connects the speeds of separation and approach. So one two dimensional problem becomes a one dimensional problem, plus a line where nothing changes. Right?

2:11Ball on a smooth fixed surface

Now, the simplest oblique impact: a ball on a smooth fixed surface, the hatched line in the figure. The ball comes in with speed u, the dark blue arrow, and leaves with speed v, the orange arrow. The dashed line is the normal to the surface. Theta, a Greek letter, is the angle of the incoming ball, measured from the normal. And theta with a small dash is theta prime, the angle going out. On the right are the results, and we will now build them one by one. Got it? Now, the surface is smooth and fixed. The figure says the part along the surface stays the same. But why? What is missing along the surface? Friction, yes! No friction means no force along the surface. So the parallel component does not change. v sine theta prime equals u sine theta. Okay? Now the normal component. It is reversed, and multiplied by e, just like a ball dropped straight down. So, v cosine theta prime equals e times u cosine theta. Clear? Now divide the first line by the second. The v and the u cancel, and we get: tan theta prime equals tan theta over e. Write this one down, with both angles measured from the normal. With me? So, back to our puzzle. Does the ball come off at the same angle it came in, like light from a mirror? Not for a real ball! For a real ball, e is less than one, so tan theta prime is bigger than tan theta. So theta prime is greater than theta: the ball comes off closer to the surface, flatter. Only a perfectly elastic ball, with e equal to one, would bounce like light. So most students got the puzzle wrong. Right?

3:55A projectile bouncing on smooth level ground

Now, let's put this bounce inside projectile motion. The ball lands on smooth level ground, bounces, and flies again. The horizontal velocity is along the ground: the parallel component. So it stays unchanged at every bounce. Okay? The vertical velocity is the normal component. So at each bounce, it is reversed and multiplied by e. Clear? Now think. If the vertical speed becomes e times, what happens to the time of the next flight? It also becomes e times. Very good! The flight time depends directly on the vertical speed. And the horizontal speed has not changed, so the range of each flight is also e times the previous one. Got it? So, the total range. Capital R for the first flight, then e R, then e squared R; the small two is a power, and the three dots mean it goes on forever. This is a geometric series, and since e is less than one, it adds up to capital R over one minus e. Okay?

4:57Example 1: ball hits a smooth floor at 30 degrees, e = 0.5

Now, let's try one. Here, m s to the minus one means metres per second, and the small circle means degrees. So, a ball hits a smooth floor at ten metres per second, at thirty degrees to the vertical, with e equal to zero point five. Find its speed and direction just after the bounce. Pause the video and try it. Done? Okay, let's check. It is the same figure, with the floor as the surface. The floor is horizontal, so the vertical is the normal, and theta is thirty degrees. Normal component: ten cosine thirty, about eight point six six; the wavy equals sign means about. At the bounce it is reversed and multiplied by e, that is, halved. The arrow means it becomes four point three three. Okay? Parallel component: ten sine thirty, which is five. Smooth floor, so it stays five. Right? Now, the speed after. Five squared is twenty five, and four point three three squared is eighteen point seven five. So v is the square root of forty three point seven five: about six point six metres per second. With me? Now the direction. Before you calculate: will the ball leave at more than thirty degrees from the vertical, or less? More, yes! Because e is less than one. Tan theta prime equals five over four point three three, so theta prime is about forty nine degrees to the vertical. Flatter, closer to the floor, just as we said. Perfect!

6:26Example 2: total range of a projectile bouncing on ground

Now a harder one, JEE level. Read it with me. So, a ball is projected at twenty metres per second, at forty five degrees, over smooth level ground, with e equal to zero point five. Find the range of the first flight, of the second, and the total distance it travels horizontally before it stops bouncing. Take g, the acceleration due to gravity, as ten. Pause the video and try it. Done? Okay, let's check. The first flight is an ordinary projectile. R one, with a small one written low, equals u squared sine two theta, over g. Twenty squared is four hundred, and two theta is ninety degrees, whose sine is one. So, four hundred times one, over ten: forty metres. Clear? Now the second flight. Some students say: e is half, so the speed is halved, and the range goes as speed squared, so it becomes one quarter, ten metres. Is that right? No, no, careful! Only the vertical part is halved. The horizontal part stays the same. So the flight time halves, and so does the range. R two equals e times R one: zero point five times forty, that is twenty metres. Got it? And the total: forty, plus twenty, plus ten, and so on. That is R one over one minus e: forty over one minus zero point five, so forty over zero point five. Eighty metres. See how quick that was?

7:55Quick revision: line of impact, bounce angle, total range

So, quick revision. The impulse acts along the line of impact, and for smooth bodies, the tangent parts do not change. On a smooth surface, the normal part is multiplied by e, so tan theta prime equals tan theta over e. And for a bouncing projectile, each range is e times the last, with total capital R over one minus e. Now go and try the practice problems on this page. You can do them. All the best!