{"id":6248,"date":"2020-10-13T15:27:06","date_gmt":"2020-10-13T09:57:06","guid":{"rendered":"http:\/\/astan.lk\/al_virtualclassroom\/?p=6248"},"modified":"2020-10-13T15:27:27","modified_gmt":"2020-10-13T09:57:27","slug":"forces-and-motion","status":"publish","type":"post","link":"https:\/\/astan.lk\/al_virtualclassroom\/forces-and-motion\/","title":{"rendered":"Forces and motion"},"content":{"rendered":"

Freefall<\/strong><\/span><\/p>\n

A free fall is a downward motion without any initial force or velocity. Our earth has the inherent property of attracting items towards it. Hence a free fall is a natural phenomenon on earth for any object at any height without any support.<\/p>\n

<\/div>\n
\n
<\/div>\n
\n

Free Fall Speed<\/strong><\/p>\n<\/div>\n

An object in motion is governed by two parameters, the velocity and the acceleration. The same concept also applies to free fall. It is too obvious to conclude that an object under free fall has \u2018only one\u2019<\/b> direction, invariably, vertical downwards. Hence in cases of free fall the velocity can be called in its scalar terms as \u2018Free Fall Speed\u2019<\/b>.<\/div>\n

Now we observe that in a free fall, the speed is not uniform. You might have observed that two stones dropped from different heights hit the ground with different velocities. The higher the height, greater is the speed of an object when it reaches the ground. It means that there is some acceleration is imparted on objects under free fall and the same is defined as ‘free fall acceleration’.<\/p>\n

\n

Free Fall Acceleration<\/strong><\/p>\n<\/div>\n

\n
As per Newton\u2019s second law a force is needed to change the state of rest of any object and hence the law applies to free fall of objects. The action of \u2018free fall\u2019 takes place, as we mentioned already, due to an attraction force or gravitation force. When this force is exerted on the object, it is subjected to an acceleration called \u2018Free Fall Acceleration\u2019.\u00a0<\/span><\/div>\n
<\/div>\n
This free fall acceleration is constant but naturally the free fall speed is a variable quantity, varies with the time or distance of motion.<\/span><\/div>\n
<\/div>\n
In cases of free fall, the distance is better referred as height as the direction is invariably vertical. The free fall acceleration is well known as \u2018acceleration due to gravity\u2019. For all practical purposes it is considered as constant at all places of earth and is denoted by the letter \u2018g\u2019.<\/div>\n
<\/div>\n<\/div>\n
A free fall occurs for an object only due to free fall acceleration and hence its parameters of motion are independent of its mass, when you ignore the air resistance. Normally, the air resistance is very low except in cases of objects with large areas and hence it is a safe bet to ignore that. But at the same time one must realize that the force acquired by an object under free fall is certainly proportional to the mass of the object. In fact such force of the object is nothing but the Net Weight of the object and also its momentum as it hits the ground (mass times the final velocity) depends on the mass of the object.<\/div>\n
<\/div>\n
<\/div>\n
<\/div>\n
Let us derive the three important free fall equations considering the free fall without air resistance. Let us assume \u2018v\u2019 is the final velocity, \u2018h\u2019 is the height and \u2018t\u2019 is the time related to a free fall. Obviously the initial velocity for a free fall is 0 and the acceleration in this case is \u2018g\u2019.<\/div>\n
<\/div>\n
The final velocity is generally defined as the sum of the initial velocity and the product of acceleration and time.<\/div>\n
<\/div>\n
Therefore, for a free fall,<\/div>\n
v = 0 + gt\u00a0<\/b><\/div>\n
v = gt ………(1)<\/b><\/div>\n
<\/div>\n
The second free fall equation is derived as follows :<\/div>\n
<\/div>\n
The distance covered by an object in any motion is the product of average velocity and time.<\/div>\n
<\/div>\n
Therefore, for a free fall,<\/div>\n
\n
\n
h =<\/b> \u00bd(v + 0)(t) =<\/b> \u00bd\u00a0vt =<\/b> \u00bd\u00a0gt \u00d7<\/span><\/span><\/span><\/span>\u00a0t\u00a0<\/b><\/div>\n<\/div>\n
h =<\/b> \u00bdgt2<\/sup> ……….(2)<\/b><\/div>\n<\/div>\n
<\/div>\n
<\/div>\n
Let us eliminate \u2018t\u2019 from these two equations to arrive at the third free fall equation,<\/div>\n
<\/div>\n
v2 = 2gh ……….(3)<\/b><\/div>\n
<\/div>\n<\/div>\n

Projectile motion<\/span><\/strong><\/p>\n

Projectile motion is an example of curved motion with constant acceleration. It is two dimensional motion of a particle thrown obliquely into the air.<\/b><\/span><\/span><\/div>\n

 <\/p>\n

\n

Consider the motion and path followed by the ball when it moves in the curved path. We will make <\/b>two assumptions <\/b>here:<\/p>\n

a) First assumption is that the free fall acceleration (g) remains constant and does not change its value during the motion of the ball.<\/p>\n

b) Resistance offered by the ball is negligible.<\/p>\n<\/div>\n

If we consider the motion and the assumptions stated above, we will find that :
\n<\/span><\/p>\n

    \n
  1. The path of the projectile (ball here) is always a parabola<\/i><\/b>. <\/i><\/li>\n
  2. The path followed by the projectile is termed as the “trajectory of the projectile<\/i>“.<\/b><\/li>\n
  3. Projectile feels only one force while in motion, which is the force of gravity<\/b><\/i>.<\/li>\n<\/ol>\n

    \"Projectile\"<\/p>\n

    \n

    Projectile Motion Equation<\/h3>\n<\/div>\n

    Projectile motion is a two dimensional concept and it follows the two dimensional kinematics. A projectile has both the horizontal and the vertical components of motion.<\/p>\n

    Projectile motion can be stated as the:
    \n
    \n<\/span><\/p>\n

    y = 1<\/span>2<\/span><\/span><\/span><\/span>12<\/span><\/span><\/span> (at2<\/sup>) + v0<\/sub> t + y0<\/sub><\/b><\/div>\n

    Where,
    \ny = height
    \nt = time
    \na = acceleration of the projectile because of gravity
    \nV0<\/sub> = Initial velocity of the projectile
    \nY0<\/sub> = Initial height of the projectile<\/p>\n

    Horizontal Component of the Velocity :<\/b> Whenever the projectile is thrown or follows the trajectory, the horizontal component of the velocity does not changes and the displacements covered by the horizontal components of the velocity are uniform. In other words, final horizontal velocity component is equal to the initial velocity component.<\/div>\n
    Now the point here to note is that when the projectile follows the trajectory, gravity force does not affect or does not make any change in the horizontal velocity component of the velocity.<\/div>\n

     <\/p>\n

    Vertical Component of the Velocity : <\/b>Vertical component of the velocity does not remain constant during the projectile motion. Gravity force acts on it and changes the vertical component of the velocity of the projectile. The displacements covered by the vertical component of the velocity are not uniform.<\/div>\n
    \n

    For the vertical component of the velocity during the projectile motion, change in both the magnitude and direction takes place. If the projectile is moving in the upward direction, then the vertical component of the velocity is in the upward direction and decrease in its magnitude takes place.<\/p>\n<\/div>\n

    On the other hand, when the projectile moves in the downward direction, the direction of the vertical components of the velocity is in the downward direction and increase in the magnitude takes place.<\/div>\n
    \n

    Range Equation for Projectile Motion<\/h3>\n<\/div>\n\n\n\n\n\n\n\n\n\n
    Equations involving Vertical Motion<\/strong><\/td>\nEquations involving Horizontal Motion<\/strong><\/td>\nExplanation of Symbols used<\/strong><\/td>\n<\/tr>\n
    <\/td>\n<\/td>\n<\/td>\n<\/tr>\n
    V(iy)<\/sub> = Vi<\/sub> sin\u03b8<\/span><\/span><\/span><\/span><\/span><\/td>\nV(ix)<\/sub> = Vi <\/sub>cos\u03b8<\/span><\/span><\/span><\/span><\/span><\/td>\n\n
      \n
    • Vi <\/sub>=<\/span> Magnitude of Initial Velocity<\/li>\n
    • V(iy)<\/sub> =<\/span> Y component of Initial Velocity<\/li>\n
    • V(ix)<\/sub> =<\/span> X component of Initial Velocity<\/li>\n<\/ul>\n<\/td>\n<\/tr>\n
    V(fy)<\/sub> = V(iy)<\/sub> + ay <\/sub>t<\/span><\/td>\nV(fx)<\/sub> = V(ix)<\/sub><\/span><\/td>\n\n
      \n
    • Vfy<\/sub><\/span> = Y component of Velocity at time ‘t’<\/li>\n
    • Vfx <\/sub><\/span>= X component of Velocity at time ‘t’ (note that the X component remains constant)<\/li>\n
    • ay <\/sub><\/span>= acceleration in vertical direction, which in the case of projectile motion would be -9.8 m\/s2<\/sup><\/li>\n<\/ul>\n<\/td>\n<\/tr>\n
    Yf<\/sub> – Yi<\/sub> = V(iy) <\/sub>t + \u00bday<\/sub>t2<\/sup><\/span><\/td>\nXf<\/sub> – Xi<\/sub> = V(ix) <\/sub>t<\/span><\/td>\n\n
      \n
    • Yi<\/sub><\/span> = Initial Y co-ordinate of Projectile<\/li>\n
    • Yf<\/sub><\/span> = Y co-ordinate of Projectile at time ‘t’<\/li>\n
    • Xi<\/sub><\/span> = Initial X co-ordinate of Projectile<\/li>\n
    • Xf<\/sub><\/span> = X co-ordinate of Projectile at time ‘t’<\/li>\n<\/ul>\n<\/td>\n<\/tr>\n

    \n(Vfy<\/sub>)2<\/sup> = (V(iy)<\/sub>)2<\/sup> + 2ay<\/sub>(Yf<\/sub>– Yi<\/sub>)<\/span><\/span><\/td>\n    		<\/td>\nSymbols already described above<\/td>\n<\/tr>\n

    \nYf<\/sub> – Yi<\/sub>=t.<\/span>\u00bd(<\/span>V<\/span>(<\/span>i<\/span>y<\/span>)<\/span><\/span><\/span><\/span>+<\/span>V<\/span>(<\/span>f<\/span>y<\/span>)<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n    		<\/td>\nSymbols already described above<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    <\/h3>\n

    Maximum Projectile Range : Expression<\/b><\/p>\n

    Now, lets look at the expression for projectile range using the above formula, Let the projectile start at (0, Yi) co-ordinates with a speed of Vi = v, and angle \u03b8<\/span><\/span><\/span><\/span>\u00a0with the horizontal surface. After some time t, it strikes the ground at a distance of Xf. The value of Xf gives the range of the projectile<\/p>\n

    The figure given below aids the visualization of the motion :<\/p>\n

    \"Projectile
    \n<\/b><\/p>\n

    In this figure, the range of the projectile is given by the formula,<\/p>\n

    d = X<\/span>f<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= <\/strong>(<\/span>(<\/span>V<\/span>c<\/span>o<\/span>s<\/span>\u03b8<\/span>)\/<\/span><\/span>g<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span> (<\/span>V<\/span>s<\/span>i<\/span>n<\/span>\u03b8<\/span>+\u221a{<\/span>(<\/span>V<\/span>s<\/span>i<\/span>n<\/span>\u03b8<\/span>)\u00b2<\/span><\/span>+2<\/span>g<\/span>Y<\/span>i<\/span><\/span><\/span><\/span>)})<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong><\/p>\n

    Using the above equation one can make a graph of `theta` versus `d` for different `theta`, and see where the value of `d` maximizes. This will be the value of maximum projectile range. Moreover, this equation reduces to a very simple form when the projectile starts form ground level, that is when Y<\/span>i<\/span><\/span><\/span><\/span><\/span><\/span>Yi<\/span><\/span> = 0.<\/p>\n

    The equation then becomes :
    \n<\/b><\/p>\n

    d =<\/b> X<\/span>f<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= (<\/span>2<\/span>V\u00b2<\/span><\/span>c<\/span>o<\/span>s<\/span>\u03b8<\/span>s<\/span>i<\/span>n<\/span>\u03b8<\/span>)\/<\/strong><\/span><\/span>g<\/span><\/span><\/span><\/span><\/span><\/span>
    \n<\/b><\/div>\n

     <\/p>\n

    d =<\/b> X<\/span>f<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= <\/b>(<\/span>V\u00b2<\/span><\/span>s<\/span>i<\/span>n<\/span>(<\/span>2<\/span>\u03b8<\/span>)<\/span>)\/<\/strong><\/span><\/span>g<\/span><\/span><\/span><\/span><\/span><\/b><\/span><\/div>\n

    Using the above equation we can very easily find the expression for maximum projectile range in this simple situation. We know that the maximum value of sin 2\u03b8<\/span><\/span><\/span><\/span>\u00a0is 1.<\/p>\n

    Therefore, the maximum range of the projectile is
    \nd = <\/b>X<\/span>f<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= V\u00b2<\/span>\/<\/span><\/span><\/span><\/span>g<\/span><\/span><\/span><\/span><\/span><\/span>
    \n<\/b><\/p>\n


    \n<\/b>Also, the value of 2\u03b8\u00a0<\/span><\/span><\/span><\/span>for which sin 2\u03b8<\/span><\/span> = 1 is 90\u00b0<\/span><\/span><\/span><\/span><\/span>. Therefore, the value of \u03b8<\/span><\/span><\/span><\/span>\u00a0= 90\/2 = 45\u00b0<\/span><\/span><\/span><\/span><\/span><\/p>\n

    \n

    Horizontal Projectile Motion<\/h3>\n<\/div>\n
    \n

    This is a type of Projectile motion in which projectile does not follow path in the upward direction or it does not have upward trajectory and the initial velocity of the projectile is also zero. This type of projectile motion is called horizontal projectile motion. This motion generally occurs when the projectile is shot straight without forming any angle with the horizontal surface and the projectile falls downward until it hits the ground.<\/p>\n<\/div>\n

    Exemplary Horizontal Projectile motion is shown in the figure below.<\/p>\n

    As shown in the figure below, the initial component of the vertical components of the velocity is zero. Horizontal velocity component of the projectile remains constant as the gravity does not affect it. Direction of the vertical component of the velocity is in downward direction during the trajectory. The magnitude of the vertical component of the velocity increases as the projectile moves downward, the force of gravity acts on it, results in acceleration of the projectile.<\/div>\n
    \n
    \"Horizantal<\/h5>\n

    The figure above illustrates a body thrown horizontally from a point O with a velocity \"\" The point O is at a certain height above the ground. Let x and y be the horizontal and vertical distances covered by the projectile, respectively, in time t. Therefore, at time t, the projectile is at p.<\/p>\n

    In order to calculate x, let us consider the horizontal motion, which is uniform motion. This is because the only force acting on the projectile is the force of gravity. This force acts vertically downwards and hence the horizontal component in zero. Therefore, the equations of motion of the projectile for the horizontal direction is just the equation of uniform motion in a straight line.<\/p>\n

    \u2234<\/span><\/span><\/span><\/span>\u00a0x = vt —————— (i)<\/p>\n

    In order to calculate y, the vertical motion of the projectile is considered. Since the vertical motion is controlled by the force of gravity, it is an accelerated motion. The initial velocity, vy<\/sub> (0), in the vertically downward direction is zero. Since the Y-axis in the figure above is taken downwards, the downward direction is regarded as the positive direction. So, the acceleration of the projectile is + g.<\/p>\n

    \u2234\u00a0<\/span><\/span><\/span><\/span>from the equation
    \ny(t) = Vy<\/sub>(0)t + \u00bd\u00a0ay<\/sub> t2<\/sup>
    \nWe have y(t) = \u00bd\u00a0gt2<\/sup> ————-(2)<\/p>\n

    Here vy<\/sub> (0) is taken as zero because both distance and time are being measured from the origin O.<\/p>\n

    From equation (1)
    \nt = x\/<\/span><\/span><\/span><\/span>v<\/span><\/span><\/span><\/p>\n

    Substituting for t from the above equation in equation (2) we have,<\/p>\n

    y(t) = \u00bdg(<\/span>x\/<\/span>v<\/span><\/span>)\u00b2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= (\u00a0g\/<\/span>2<\/span>v\u00b2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0)x2<\/sup><\/p>\n

    \u2234<\/span><\/span><\/span><\/span>\u00a0y = kx2<\/sup> Where k = g\/<\/span>2<\/span>v\u00b2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>……………..(3)<\/p>\n

    is a constant for a projectile projected upwards with a definite velocity v and at a place with a definite value of ‘g’.<\/p>\n

    Equation (3) is a second-degree equation in x, a first-degree equation in y and is the equation of a parabola. Therefore, a body thrown horizontally from a certain height above the ground follows a parabolic trajectory till it hits the ground.<\/p>\n

    Resultant Velocity of a Horizontal Projectile:
    \n<\/b><\/p>\n

    In this section, let us calculate the resultant velocity of the projectile V<\/span>\u20d7\u00a0<\/span><\/span><\/span><\/span><\/span><\/span>V\u2192<\/span><\/span>, at any point p on the trajectory, in an interval of time t. Vx<\/sub>and Vy<\/sub> are the horizontal and vertical components of V<\/span>\u20d7\u00a0<\/span><\/span><\/span><\/span><\/span><\/span>V\u2192<\/span><\/span> as illustrated in the figure below.<\/p>\n

    \"Velocity<\/h5>\n

    Since, the horizontal motion of the projectile is uniform, V<\/span>\u20d7\u00a0<\/span><\/span><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= V<\/span>\u20d7\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    However, the motion in the vertical direction is an acceleration one.<\/p>\n

    \u2234<\/span><\/span><\/span>\u2234<\/span><\/span> Vy<\/sub>(t) = Vy<\/sub> (0) + ay<\/sub> t<\/p>\n

    Since O is considered to be the origin, Vy<\/sub> (0) = 0<\/p>\n

    \u2234<\/span><\/span><\/span>\u2234<\/span><\/span> Vv<\/sub> (t) = gt
    \n\u2234<\/span><\/span><\/span>\u2234<\/span><\/span> The magnitude of the resultant velocity V<\/span>\u20d7\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0is given by,<\/p>\n

    |V<\/span>\u20d7 \u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span>| = V =\u221a (V\u00b2<\/span>x<\/span><\/span><\/span><\/span>+<\/span>V\u00b2<\/span>y)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \u2234<\/span><\/span> V = \u221a(V\u00b2<\/span><\/span>+<\/span>g\u00b2<\/span><\/span>t\u00b2<\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    The direction is given by tan\u03b2<\/span><\/span><\/span><\/span>\u00a0= V<\/span>y\/<\/span><\/span><\/span><\/span>V<\/span>x<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>= g<\/span>t\/<\/span><\/span>V<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \u2234<\/span><\/span><\/span><\/span>\u03b2<\/span><\/span> = tan-1 <\/sup>(gt\/V)<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    \n<\/div>\n

    Motion -The change of position of a body with time is called motion.<\/p>\n

    It was Galileo who first realized that motion of a body is independent of its mass, it is the change of state of motion or state of rest.<\/p>\n

    Inertia<\/strong><\/span><\/p>\n

    A body is said to possess a property called inertia \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 The inertia of a body is measured by its reluctance to change its state of motion or state of rest. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Examples<\/p>\n