2 Force and Momentum Notes B.docx

**Mass** [(*m*)]{.math.inline} is a measure of the amount of matter in a body. - It is a **scalar** quantity with an SI unit of **kilogram (kg)**.\ The kilogram is the only SI unit that has a prefix, kilo. - **Momentum** ([*p*]{.math.inline}) is the product of mass and velocity. - I...

**Mass** [(*m*)]{.math.inline} is a measure of the amount of matter in a body. - It is a **scalar** quantity with an SI unit of **kilogram (kg)**.\ The kilogram is the only SI unit that has a prefix, kilo. - **Momentum** ([*p*]{.math.inline}) is the product of mass and velocity. - It is a **vector** quantity with an SI unit of\ **kilogram metre per second (kg m s^-1^)**. - momentum = mass × velocity - The total momentum before an interaction\ is equal to the total momentum after,\ within a closed system. Sample Problem -------------- Two identical hockey pucks, moving in opposite directions, collide head-on. The red puck is moving to the right at 5 m s^-1^, and the green puck is moving to the left at 2 m s^-1^. The red puck is brought to rest by the collision. \(i) What is the velocity of the green puck after the collision? \(ii) What is the change in momentum of each puck? Sample Answer ------------- Using the principle of conservation of momentum, making sure all values are in their standard unit and assign a value on each direction (i.e. right is the '+' direction and left is the '--' direction): \(i) [*m*~1~*u*~1~ + *m*~2~*u*~2~]{.math.inline} [ = *m*~1~*v*~1~ + *m*~2~*v*~2~]{.math.inline} \(ii) Change in momentum of the red puck is [*m*~1~*v*~1~ − *m*~1~*u*~1~ = (*m*)(0) − (*m*)(5) =  − 5*m*]{.math.inline} - A **force** ([*F*]{.math.inline}) is that which can cause acceleration. - It is a **vector** quantity with an SI unit of **newton (N)**. - 1 newton is the amount of force that will cause a body of mass 1 kilogram to accelerate by 1 metre per second squared. - 1 N = 1 kg m s^-2^ - Sir Isaac Newton described the concept of force in his three laws of motion. **Law 1:\ **A body remains at rest or continues moving at constant velocity unless acted upon by a resultant external force. **Law 2:\ ** The rate of change of momentum is proportional to the applied force and occurs in the direction of the force. **Law 3:\ **For every action, there is an equal and opposite reaction, where action and reaction are on different bodies. **Law 1:\ **Seat belts are used to stop passengers in cars from continuing at constant velocity when the car decelerates rapidly (or crashes!). **Law 2:\ ** When a football is kicked a force is applied on the ball by the players foot; the greater the force the greater the acceleration. **Law 3:\ **In space there is no external matter for a rocket to 'push off' in order to accelerate, however when force is applied on the fuel within the rocket to expel it, an equal and opposite force is applied to the rocket causing it to accelerate. The force the rocket exerts on the gas backwards is equal to the forces the gas exerts on the rocket forward. These forces cause the gas to accelerate backwards and the rocket to accelerate forwards. - Based Newton's 2^nd^ Law we have: - [*F*]{.math.inline} [\$\\propto \\frac{\\text{change\\ in\\ momentum}}{\\text{time\\ take}}\$]{.math.inline} - [*F*]{.math.inline} [\$\\propto \\frac{final\\ momentum - initial\\ momentum}{\\text{time\\ take}}\$]{.math.inline} - [*F*]{.math.inline} [\$\\propto \\frac{mv - mu}{t}\$]{.math.inline} - [*F*]{.math.inline} [\$\\propto m\\left( \\frac{v - u}{t} \\right)\$]{.math.inline} - [*F*]{.math.inline} [ ∝ *ma*]{.math.inline} subbing '[*a*]{.math.inline}' for '[\$\\left( \\frac{v - u}{t} \\right)\$]{.math.inline}' - [*F*]{.math.inline} [ = kma]{.math.inline} to go from 'proportional to...' to 'equal to...'\ we must multiply by a constant. - [*F*]{.math.inline} [ = ma]{.math.inline} the unit 'newton' is defined in such a way\ as to make [*k*]{.math.inline} = 1. - The force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. - It is an example of an **Inverse Square Law**,\ because [\$F \\propto \\frac{1}{d\^{2}}\$]{.math.inline}, this means that as the distance\ between the centres doubles, the force is\ one quarter. - The force is always attractive. - The Earth\'s atmosphere is a layer of gases held in place by gravity. For a planet to sustain an atmosphere, it must have sufficient mass to generate enough gravitational force. The Moon\'s gravity is too weak to sustain an atmosphere due to its lower mass. - **Weight** [(*W*)]{.math.inline} is a force equal to the product of\ mass and acceleration due to the gravity**.** - As weight is a force, it is a vector quantity\ and is measured in newtons. - As a body moves through a fluid like air, it encounters\ **air resistance**, a frictional force opposing its motion which increases with velocity. When the upward air resistance equals the downward gravitational force, the body reaches **terminal velocity** and falls at a constant speed. - **Centre of gravity** is the point at which the whole weight of a body appears to act. - Acceleration due to gravity [(*g*)]{.math.inline} is proportional to the mass of the planet and inversely proportional distance squared between the centre of gravity of the planet and the accelerating object. - The value of [*g*]{.math.inline}, acceleration due to gravity,\ varies at different locations on the surface\ on the earth because locations are at different\ distances from the Earth's centre since the\ Earth is not perfectly spherical and there\ are mountains and valleys. - In the Leaving Cert exam questions,\ which ask about acceleration due to gravity\ at a given altitude above the Earth's surface\ are quite common. This formula is useful when\ answering these questions. - If three of the four values are known,\ you can use this formula to find the fourth. - **Friction** [(*F*~*f*~)]{.math.inline} is a force that opposes the motion of a body. - Friction between two surfaces may cause a moving object to slow down. For example, a book sliding along a table, or a hockey puck on the ice rink. Clearly, the nature of the two surfaces in contact, as well as the weight of the sliding body, determines the magnitude of the frictional force. This type of friction can also allow people or cars to accelerate (See image below). Friction allows tires to grip the road or the souls of your shoes to grip the path. Sample Problem -------------- A sled of mass 60 kg is being pulled along the snow with a tractive (pulling) force of 500 N.\ The resistance to its motion is 80 N. \(i) What is the resultant force on the sled? \(ii) What is its acceleration? Sample Answer ------------- \(i) Resultant force [=]{.math.inline} Tractive force -- Friction \(ii) Using the formula for force, making sure all values are in their standard unit: - An object at rest on or moving over a surface experiences a reaction force perpendicular to that surface. This force, due to the object\'s weight, is equal in magnitude and opposite in direction according to Newton\'s third law. This is called the \"normal reaction\" because \"normal\" means perpendicular. It can also be referred to simply as the \"reaction.\" - Normal reactions are usually given the letters [*N*]{.math.inline} or [*R*]{.math.inline}. Sample Problem -------------- A 20 kg object is at rest on a rough surface inclined at an angle of 30 ° to the horizontal. \(i) Draw a diagram to show the forces acting on the object. \(ii) Resolve the weight into components perpendicular and parallel to the surface. \(iii) What is the magnitude of the reaction acting on the object? Sample Answer ---------------------------------- (i) \(ii) [*W*~∥~]{.math.inline} [\$= \\left( 20g \\right)\\sin{30{\^\\circ}} = \\left( 20g \\right)\\left( \\frac{1}{2} \\right)\$]{.math.inline} \(iii) The reaction is equal in magnitude to the perpendicular component of the weight.\ [\$N = 10\\sqrt{3}g\\mathrm{\\text{\\ N}}\$]{.math.inline} Apparatus - Ticker tape, ticker timer, trolley, sloped runway, metre stick, pulley, scale pan & Newton weights. Procedure - Set up apparatus as shown in diagram, with all the Newton weights on the trolley. Attach one end of ticker tape to trolley, place other end through ticker timer. Attach string to trolley, pass it over the pulley, and attach the other end to the empty scale pan. - Slope the runway so that the trolley will move at constant velocity when given an initial push. - Move one of the weights from the trolley to the scale pan. Note the mass, measured in grams, written on the weight. Multiply the mass by 0.098 to get the weight, measured in newtons. This is the force [(*F*)]{.math.inline} which causes the trolley to accelerate. - Release the trolley and use the ticker timer to calculate the acceleration [(*a*)]{.math.inline} as described in Mandatory Experiment 1 (B). - Move weights from the trolley to the pan and repeat process, so as to get six or seven readings. Tabulation of data -------------------------------------------------------- \ [*F* / *N*]{.math.display}\ -------------------------------------- -- -- -- -- -- -- \ [*a* / m *s*^ − 2^]{.math.display}\ -------------------------------------------------------- Graph - Plot a graph of force [(*F*)]{.math.inline}, measured in newtons, against acceleration [(*a*)]{.math.inline}, measured in metres per second squared, with acceleration on the [*y*]{.math.inline}-axis. Conclusion - The straight line through the origin shows that force is proportional to acceleration. Use Graph - The mass of the system can be calculated from the graph. If the acceleration is on the [*y*]{.math.inline}-axis and force on the [*x*]{.math.inline}-axis then the mass of the system is equal to the reciprocal of the slope of the line of best fit, [\$m = \\frac{1}{\\text{slope}}\$]{.math.inline}. Explanation - Based on Newton's second law: [*F* = *ma*]{.math.inline} - Rearrange to put [*F*]{.math.inline} on the [*x*]{.math.inline}-axis and [*a*]{.math.inline} on the [*y*]{.math.inline}-axis, we get [\$a = \\frac{1}{m}F\$]{.math.inline} - Using the equation of the line [*y* = *mx* + *c*]{.math.inline}; given that [*c* = 0]{.math.inline} because the line passes through the origin, it is clear that mass of the body is equal to the reciprocal of the slope. Precautions - Moving the weights from the scale pan to the trolley ensures the mass of the system remains constant throughout experiment. - Use a low friction pulley, oil the pulley or oil the wheels of the trolley to reduce friction. - Ensure the runway is correctly sloped so that the trolley moves at constant velocity when the empty scale pan is attached to it; at this point the frictional force is equal and opposite to the gravitational force. - Ensure that the string pulling the trolley is parallel to the runway, so that all the force acts in the same direction as the acceleration. - Using more masses on the scale pan or a lighter trolley increases the acceleration; this will reduce the percentage error. Alternative Methods - This experiment can be performed using a rider and air track rather than a trolley and runway, the acceleration can be found using light gates or motion detectors, data logger & calculator rather than ticker timer and the force can be provided by a Newton balance rather than pulley and weights.  Apparatus - Ticker tape, ticker timer, two trolleys, sloped runway, metre stick, Velcro (pin & cork or magnets), electronic balance. Procedure - Find the mass of both trolleys using an electronic balance. - Set up apparatus as shown in diagram. Attach one end of ticker tape to trolley 1, place other end through ticker timer. Place trolley 2 at rest on the runway. - Slope the runway so that the trolleys will move at constant velocity when given an initial push. - Push trolley 1 and allow it to collide with and stick to trolley 2. - Use the ticker timer to calculate the velocity of trolley 1 before collision [(*u*)]{.math.inline}, as described in Mandatory Experiment 1 (A). Initial velocity of trolley 2 was zero before collision. - Use the ticker timer to calculate the velocity of the trolleys after the collision [(*v*)]{.math.inline}. Both trolleys have the same velocity after collision because they are stuck together. Calculations - Apply the formula [*m*~1~*u* = (*m*~1~+*m*~2~)*v*]{.math.inline}.\ This is equivalent to [*m*~1~*u*~1~ + *m*~2~*u*~2~ = *m*~1~*v*~1~ + *m*~2~*v*~2~]{.math.inline}. *\ * Tabulation of data -------------------------------------------------------------------------------------------------------------- \ [*m*~1~ / *kg*]{.math.display}\ ----------------------------------------------- -- -------------------------------------------------------- -- \ [*m*~2~ / *kg*]{.math.display}\ \ [*s*~1~ / *m*]{.math.display}\ \ \ [*t*~1~ / *s*]{.math.display}\ [*u* / m *s*^ − 1^]{.math.display}\ \ [*s*~1~ / *m*]{.math.display}\ \ \ [*t*~1~ / *s*]{.math.display}\ [*v* / m *s*^ − 1^]{.math.display}\ \ \ [*m*~1~*u* / kg m *s*^ − 1^]{.math.display}\ [(*m*~1~+*m*~2~)*v* / kg m *s*^ − 1^]{.math.display}\ -------------------------------------------------------------------------------------------------------------- Conclusion - The total momentum before the collision is equal to the total momentum after the collision. Precautions - Oil the wheels of the trolley to reduce friction. - Ensure the runway is correctly sloped so that the trolleys move at constant velocity. At this point the frictional force is equal and opposite to the gravitational force. - Ensure the runway is free of dirt, so that the trolley moves at constant velocity. - Giving trolley 1 a greater initial velocity will reduce the percentage error. - Trolley 1 must stick to trolley 2. Alternative Methods - This experiment can be performed using a rider and air track rather than a trolley and runway, the velocities can be found using light gates or motion detectors, data logger & calculator rather than ticker timer. Apparatus - '[*g*]{.math.inline} by free fall apparatus'\ (electromagnet, distance scale, trapdoor),\ electric circuit, switch, sphere, timer. Procedure - Set up apparatus as shown in diagram,\ with the switch in position A. - Measure the distance [(*s*)]{.math.inline} from the bottom\ of the sphere to trapdoor, (30 cm), using the distance scale on the apparatus. - Move switch to position B to release ball and start timer. - When the sphere hits the trapdoor (opening it), it stops the timer. - Repeat at this height a number of times, recording the **shortest time** [(*t*)]{.math.inline}. - Increase the height (by 10 cm) and repeat process, so as to get six or seven readings. Calculations - Square the values of time to get [*t*^2^]{.math.inline}~.~ Tabulation of data ------------------------------------------------------ \ [*s* / *m*]{.math.display}\ ------------------------------------ -- -- -- -- -- -- \ [*t* / s]{.math.display}\ \ [*t*^2^ / *s*^2^]{.math.display}\ ------------------------------------------------------ Graph - Plot a graph of distance [(*s*)]{.math.inline}, measured in metres, against time squared [(*t*^2^)]{.math.inline}, measured in seconds squared, with distance on the *y*-axis. Use Graph - Acceleration due to gravity can be calculated from the graph. If the distance is on the\ [*y*]{.math.inline}-axis and time squared on the [*x*]{.math.inline}-axis then acceleration due to gravity is twice the slope of the line of best fit, [(*g*=2×slope)]{.math.inline}. Explanation - Based on the second equation of motion; given that the initial velocity [(*u*)]{.math.inline} equals zero and acceleration [(*a*)]{.math.inline} equals acceleration due to gravity [(*g*)]{.math.inline} we get: - [\$s = ut + \\frac{1}{2}at\^{2} \\Rightarrow s = \\frac{g}{2}t\^{2}\$]{.math.inline} - Using the equation of the line [*y* = *mx* + *c*]{.math.inline}; given that [*c* = 0]{.math.inline} because the line passes through the origin, it is clear that acceleration due to gravity is twice the slope. Precautions - Measure the distance [(*s*)]{.math.inline} from bottom of sphere. - Take care to avoid parallax error when measuring distance. - For each value of *s* take several values for *t* and use the minimum time. Any experimental errors will cause an over estimation of time, therefore the minimum time found is the most accurate one. - Ensure no external force (e.g. draughts, etc.) acts on sphere (during release). - Use a small smooth dense (heavy) sphere as this reduces air resistance. - Use a metal sphere as it conducts electricity, so can be magnetised, allowing the electromagnetic/release system to work. - Adjust 'sensitivity' of trap door; adjust 'sensitivity' of electromagnet (using paper between sphere and core). - Use large values for *s*; this will reduce the percentage error. - Use a more accurate timer (i.e. one that measures in milliseconds) or a more accurate distance scale. Note - The experiment can be repeated with spheres of different masses to show that acceleration is independent of the mass of the falling object.