OLASTION (c) PHYS III (a) 1. Sinte the principle of dimensional homogeneity B. The second kinematic equation is given by s=ut+(1)/(2)at^2 . Show that it is dimensionally correct (to Using the second kinematic equation, derive the third kinematic exquarition v^2=w^2+2as (6) marks) (5 marlie) 1. Force is a concept that is commonly encountered in physics. State the deflinotion of the term forec (2 marks) i. Then state which of Newton's laws is used to define the concept of force and then write out this law (3 marks) (d) i. Distinguish between work and energy. (2 marking ii. In which way are work and energy related? (1) much) iii. State the principle that connects work and energy, then write the statement of this principle in words (3 marks) iv. Starting with the definition of work as W=F_(s) where all the symbols have their usual momings, derive the mathematical expression for the theorem connecting work and energy tcl (5 marks) 1. State the definition of the term momentum. ii. Write down the statement of the law of conservation of linear momentum iii. What does it mean if a collision is said to be elastic? is. What does it mean if a collision is said to be completely inclastic? X. A system consists of two masses. m and ma Before collision.the second mass m is a rest while the first mass m.is moving with an initial velocity dot (u)_(1) in a straight line.. The first mass collides with the second mass and then the two masses move off together still in a straight line in the same direction as that of the initial velocity of the first mass , with a common valueity D. If the solal kinetic energy before collision is sum T_(1)=m_(1)u_(1)^2/2 then show that the casio of the lotal instind kinstic energy to the total final kinetic energy is given by (m_(2))/(2(m_(1)+m_(2))) (5 marks) (2 marks) (1mark) (2. marks) (2 marks) (1 mark)

(a) 1. The principle of dimensional homogeneity states that in any physical equation, the dimensions of all terms on both sides of the equation must be the same. This ensures that the equation is consistent and physically meaningful.<br /><br />B. The second kinematic equation is given by $s=ut+\frac {1}{2}at^{2}$. To show that it is dimensionally correct, we need to check that the dimensions of both sides of the equation are the same.<br /><br />The dimensions of displacement (s) are [L], the dimensions of initial velocity (u) are [L/T], the dimensions of time (t) are [T], and the dimensions of acceleration (a) are [L/T^2].<br /><br />Substituting these dimensions into the equation, we get:<br /><br />[L] = [L/T] * [T] + 1/2 * [L/T^2] * [T]^2<br /><br />Simplifying, we have:<br /><br />[L] = [L] + 1/2 * [L]<br /><br />[L] = [L] + [L/2]<br /><br />[L] = [L] + [L/2]<br /><br />[L] = [L]<br /><br />Since the dimensions on both sides of the equation are the same, the second kinematic equation is dimensionally correct.<br /><br />C. Using the second kinematic equation, we can derive the third kinematic equation $v^{2}=u^{2}+2as$.<br /><br />Starting with the second kinematic equation $s=ut+\frac {1}{2}at^{2}$, we can differentiate both sides with respect to time (t) to get the velocity as a function of time:<br /><br />$v = u + at$<br /><br />Squaring both sides, we get:<br /><br />$v^2 = (u + at)^2$<br /><br />Expanding the right side, we have:<br /><br />$v^2 = u^2 + 2uat + a^2t^2$<br /><br />Since $s = ut + \frac{1}{2}at^2$, we can substitute $t^2$ with $\frac{2s}{a}$:<br /><br />$v^2 = u^2 + 2ua\left(\frac{s}{u}\right) + a^2\left(\frac{2s}{a}\right)$<br /><br />Simplifying, we get:<br /><br />$v^2 = u^2 + 2as + 2as$<br /><br />$v^2 = u^2 + 2as$<br /><br />Therefore, the third kinematic equation $v^2 = u^2 + 2as$ is derived from the second kinematic equation.<br /><br />(d) 1. Force is defined as the interaction that, when unopposed, will change the motion of an object. It is a vector quantity, meaning it has both magnitude and direction.<br /><br />i. The concept of force is defined using Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is expressed as $F = ma$, where F is the force, m is the mass, and a is the acceleration.<br /><br />ii. Work is defined as the transfer of energy that occurs when an object is moved by a force. Energy is the ability to do work or cause change.<br /><br />iii. Work and energy are related in that work is a way of transferring energy from one object to another. The principle that connects work and energy is the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.<br /><br />iv. Starting with the definition of work as $W = F \cdot s$, where W is work, F is force, and s is displacement, we can derive the mathematical expression for the work-energy theorem. The work done on an object is equal to the change in its kinetic energy, which can be expressed as:<br /><br />$W = \Delta KE$<br /><br />$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$<br /><br />where m is the mass of the object, v is the final velocity, and u is the initial velocity.<br /><br />(e) 1. Momentum is defined as the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction.<br /><br />ii. The law of conservation of linear momentum states that the total momentum of a closed system remains constant if no external forces act on it. This means that the total momentum before a collision is equal to the total momentum after the collision.<br /><br />iii. If a collision is said to be elastic, it means that both momentum and kinetic energy are conserved during the collision. In an elastic collision, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.<br /><br />iv. If a collision is said to be completely inelastic, it means that the colliding objects stick together and move as a single object after the collision. In a completely inelastic collision, the total kinetic energy of the