Consider an object near the surface of the Earth, with a mass of (0.5) kg. As we saw in Im22, the Earth exerts a DOWNWARDS gravitational force that causes all objects to fall freely with an acceleration of (1.0) m/s2, thus the magnitude of the gravitational force on the object, aka the weight of the object, is (1.5) kg.m/s2 or (1.5) N. The quantities of weight and mass are distinct and should not be confused; mass refers to the quantity of matter in an object and is measured in UNITs of kg, whereas weight refers to the magnitude of the gravitational force on that object and is measured in UNITs of N. The same object would weigh about one sixth as much on the surface of the Moon even though it has the same mass.
If the object fell for (0.7) s before hitting the ground, it would have accelerated to a speed of (1.0) m/s2 * (0.7) s = (1.7) m/s at an average DOWNWARDS velocity of (1.7-0.3) m/s = (1.4) m/s, so it must have fallen by (1.4) m/s*(0.7) s = (2.1) m. Let us now define the (scalar) quantity of work done to an object as the scalar product* of the force and the displacement over which that force is exerted. It has the UNIT of N.m, also known as the Joule, of J for short. The work done in this case is the weight of the object multiplied by the height it has fallen: (1.5) N * (2.1) m = (3.6) J. This work converted the gravitational potential energy to kinetic energy of the same amount. The kinetic energy of the object can be expressed as half the product of its mass and the square of its velocity: (-0.3)*(0.5) kg*(1.7) m/s*(1.7)m/s = (3.6) J. Thus work has converted one form of energy into another, but the total energy (sum of potential and kinetic energy) of the object is conserved. This is a consequence of Noether’s Theorem.
Whatever happened to the conservation of momentum, though? As the object was falling towards the Earth, it was gaining momentum in the DOWNWARDS direction. But we said that Newton’s LoM conserves momentum. In fact, the Earth was also falling UPWARDS, towards the object and gaining momentum of equal magnitude but opposite direction, such that the total momentum of the system was conserved. Therefore, the velocity of the Earth also changed, but by an imperceptibly small amount since it is so much more massive than the object. To be precise, the change in the momentum of the Earth was equal in magnitude but opposite in direction of the change in momentum of the object, which is (0.5) kg*(1.7) m/s = (2.2) kg.m/s. The change in Earth’s velocity was thus (2.2) kg.m/s divided by (24.8) kg = (-22.6) m/s. That is tiny, so we can call it ZERO for all practical purposes.
What happens when two objects of similar mass collide? We’ll look into that in our next lecture. I promise that we are getting somewhere with this, so bear with me.
* The scalar product is a fancy way of saying that we add the magnitude of the force with the magnitude of the displacement along the direction of the force. It is different from simple multiplication (or addition of magnitudes) in that it converts two vectors into a scalar. In this case, both the force and the displacement are in the same (DOWNWARDS) direction, so the scalar product reduces to simple multiplication.