Kinetic Energy and Momentum

Transcript

0:00Introduction

Hello everyone! Let's start with a small contest. On one side, a fifty gram bullet flying at four hundred metres per second. On the other side, a fifty kilogram sprinter running at eight metres per second. Which one has more kinetic energy? And which one has more momentum? The bullet, for both, right? Hmm, not so fast. By the end of this lesson, you will see the answer is split. Kinetic energy and momentum come up often in JEE problems, so this link between them is worth knowing well. Let's begin.

0:36Kinetic energy, and its link to momentum

So, the energy of motion: kinetic energy. For a body of mass m moving with speed v, the kinetic energy, capital K, equals half m v squared. That is one half, times m, times v squared, okay? Now tell me: can kinetic energy ever be negative? No, never! Mass is positive, and v squared can never be negative. So kinetic energy is a scalar, and it is never negative. And it depends on the frame of reference, because the speed v depends on who is measuring it. Right? Now bring in momentum: p equals m times v. Can you write the kinetic energy using p instead of v? Yes, multiply the top and the bottom by m. Then K equals m squared v squared over two m. And m squared v squared is just p squared. So, K equals p squared over two m. Clear? Turn it around, and p equals the square root of two m K; that is the square root sign on screen. Keep both forms ready: a question may give you the kinetic energy and ask for the momentum, or the other way round. With me?

1:48Comparing two bodies

Now, a comparison that comes up often. Two bodies, one light and one heavy, with the same momentum. Which one has more kinetic energy? The lighter one! Yes, exactly! Look at K equals p squared over two m. With p fixed, K is proportional to one over m. The sign that looks like a stretched Greek alpha means proportional to, okay? Smaller mass, more kinetic energy. Now flip it: the same kinetic energy. From p equals root two m K, p is proportional to root m. So now the heavier body has more momentum. Got it? And the speed: v is p over m, so v is proportional to one over root m. With the same kinetic energy, the lighter body is faster. Right?

2:36Scaling and small changes

Next, scaling. Suppose p becomes lambda times p. Lambda, the Greek letter, is just a number. The single arrow means becomes, and the double arrow means implies. Since K goes as p squared, K becomes lambda squared times K, okay? Now the reverse. If the kinetic energy becomes lambda times K, what does the momentum become? Root lambda times p. Correct! Because p goes as the square root of K, right? Now, small changes in the same body. Delta, the little triangle, means a change. Delta K over K, the fractional change in K, is approximately two times delta p over p, which equals two times delta v over v. The wavy equals sign means approximately. The square doubles the percentage, but only for small changes. Remember that word: small. Got it? And for errors in measurement, the fractional errors add: delta K over K equals delta m over m, plus two times delta v over v. The power two on v becomes a factor of two on its error. With me?

3:44Trap: zero momentum, so zero kinetic energy?

Now an important trap. For one body, kinetic energy and momentum go together: if it has one, it has the other. Both are zero only when the body is at rest. Okay? Now take two equal balls moving towards each other with equal speeds. Their momenta are equal and opposite, so the total momentum is zero. So the total kinetic energy is zero too, right? No, no, careful! Momentum is a vector, so opposite momenta cancel. But kinetic energy is a scalar and never negative, so the two energies simply add. Each ball has K greater than zero, so the total is greater than zero. Clear?

4:26Example 1: bullet vs sprinter, K and p compared

Now, remember our contest from the start? Let's settle it. Compare the kinetic energies and the momenta of a fifty gram bullet, small g on screen, moving at four hundred metres per second, and a fifty kilogram sprinter running at eight metres per second. The m s to the minus one on screen means metres per second. Pause the video, and try it first. Done? Okay, let's check. Bullet first. Its mass is fifty, so we put m equals fifty, right? Oh no, careful! Fifty grams, not fifty kilograms. In S I units, m equals zero point zero five kilogram. Easy marks are lost right here. Got it? Now, K equals one half times zero point zero five times four hundred squared. Four hundred squared is one hundred and sixty thousand; times zero point zero five gives eight thousand; and half of that is four thousand joules, capital J for joule. Okay? And its momentum: p equals zero point zero five times four hundred, that is twenty kilogram metre per second. With me? Now the sprinter. K equals one half times fifty times eight squared: half of fifty times sixty four, so one thousand six hundred joules. And p equals fifty times eight: four hundred kilogram metre per second. Right? Compare the energies. Four thousand over one thousand six hundred is two point five. So the bullet has two point five times the kinetic energy. Clear? Now the momenta. Four hundred over twenty is twenty. The sprinter has twenty times the momentum! So our contest really is split: the light, fast bullet wins on energy, and the heavy sprinter wins on momentum. Speed is squared in K, but not in p. That's it!

6:21Example 2: K rises 300 percent, how much does p rise

Now a slightly harder one. The kinetic energy of a body increases by three hundred percent; the percent sign is on screen. By what percentage does its momentum increase? Pause and try. Hint: first find what the kinetic energy becomes. Done? Okay, let's check. It increases by three hundred percent. So the new kinetic energy is three times K, right? No, no, wait! That is the first trap. Increase by three hundred percent means we add three K to the original K, so K becomes four K; the arrow again means becomes. Okay? Now, it is the same body, so the mass is fixed. If K becomes four times K, what does the momentum become? Yes, exactly! p equals root two m K, so p becomes root four times p, that is two p. The momentum doubles. Right? So the increase is two p minus p, over p, times one hundred percent: one hundred percent. That is the answer. Got it? And now the second trap. Some students use the small-change rule: half of three hundred, so one hundred and fifty percent. Wrong! That rule is only for small changes, and three hundred percent is not small. For big changes, find the new value exactly, as we just did. Clear?

7:48Quick revision: K = p^2/2m and comparing two bodies

So, quick revision. K equals half m v squared, which equals p squared over two m, and p equals root two m K, okay? Same momentum: the lighter body has more kinetic energy. Same kinetic energy: the heavier body has more momentum. And for big changes, scale exactly: K becoming four K means p becomes two p. Now go and try the practice problems on this page. You will do them well, I am sure. All the best!