Objective: This equation can be re-written as 1. To investigate the relationship between the mass, velocity and momentum of an object. (me = - (m) MA (4) 2. To investigate the law of conservation of momentum for two objects. Now if we repeat the experiment by varying the mass of cart A (but keep the mass of cart B constant) then plot a Background: graph of ME against ma we expect a straight graph and the slope of the graph can be used to find a value for the The momentum of an object is defined by the equation mass of the cart B, mg. Momentum = mass x velocity (1) To assess how close this value is to the actual value of the mass of the car me, we can calculate the percentage difference using It is a vector quantity, meaning that it has a direction. If one direction is considered positive, then the opposite direction is negative. Although momentum is defined by a simple equation, practically it is a very useful concept in engineering Percentage error = [actual value - experimental value/ -X 100 because it helps us to understand what happens in collisions and explosions. It helps us to carry calculations to actual value find out who has done a violation in a road accident, leading to conviction. Safety: 1. Please follow instructions at all times. 2. Some items (e.g. the air track and the metal stands) are quite heavy and can cause damage to persons or items if fallen and hit accidentally. Handle these with care to avoid this happening When two objects collide their momentum collision is the same as their momentum after collision, and this statement is known as the law of conservation of momentum. In symbols we can write this as This lab involves the use of electrical equipment. Follow usual safety guidelines about handling electrical equipment. In particular, the mains in high voltage which can be fatal and so do not touch any exposed part MAVia + MBViB = MAYfa + mgVFB (2) of the circuitry and do not leave wires/cables hanging around which be a trip hazard. where MA = mass of car (trolley) A my = mass of car (trolley) B Apparatus: Via = initial velocity of car A ViB = initial velocity of car B 1. Horizontal friction compensated surface or dynamic track; VIA = final velocity of car A VFB = final velocity of car B 2. Two carts or trolleys; 3. Mechanism for the trolleys to be joined to start with and then to be released from rest; 4. Light gate(s) with data logging system Now if both carts have zero momentum at the before, for example in the event of an explosion, the equation (2) becomes 0 = MAVIA + mgVIB (3)Procedures: Observations and Data: (5 marks] 1. Measure the mass of cart A (with flag) and cart B (with flag) and record in the observation and data Section 1: section. Show the calculations below for the first trial. 2. Level the track. Place a cart on the track. If the cart rolls one way or the other, adjust the track to raise or Measure the mass of the cart A mA: 0.260kg lower one end. Measure the mass of the cart B me: 0.260kg 3. Set up the apparatus as shown in figure 1. Initial velocity of the cart A, VIA: 0 photogates Initial velocity of the cart B, VIB :0 Final velocity of the cart A_VIA: -0.5602m/s release mechanism Final velocity of the cart B, VFB: 0.5780m/s stopper flag flag Total momentum before release: 0 cart A cart B Total momentum after release: mAVEA + mgVIB =0.004628 track Section 2: laboratory bench Record your measurements in table below. 4. Use the MODE button on the data logger to get on the screen SPEED. Mass of cart A, Final velocity of cart A, UfA Final velocity of cart B, VFB WEB 5. Then use the SELECT button to get on the screen ONE GATE. MA VIA kg m/s m/s This means that you will be measuring the velocity of the cart as it passes through the light gate. One grate will measure the speed of the cart A and the other will measure the speed of the cart B. 0.260 0.5780 0.5602 -0.9692 6. Press the START/STOP button on the data logger. 7. At this stage call the teacher/technician to check that you are on the right track. 0.310 -0.4878 0.5222 -1.0705 8. Use the release mechanism to activate the movement of the carts. 0.360 0.4292 0.5495 1.2802 Read the readings from the data loggers for the final velocities of cart A and cart B in the observation and data section. 0.410 0.3724 0.5865 1.5749 9. Now vary the mass of the cart A by adding eg.50 grams at a time (do not change the mass of the cart B) and 0.460 0.3396 0.5970 -1.7579 repeat step 8. 10. Record mass of cart A, speed of cart A and speed of cart B in the table. 0.510 -0.3195 0.5988 -1.8741Graph Plotting: Analysis, evaluation, and conclusion: Using Microsoft Excel or MATLAB, draw a graph of ." (-axis) against ma (x-axis). Label your axes 1. Considering the data analysis in section 1, does your data suggestion that momentum is conserved or not? with the correct quantities and correct units. Justify your answer by using numbers. [3 marks] Draw the line of best fit for your data. In Excel this is called trendline. Use Excel or MATLAB to find the equation of the best fit line. Equation: y = -3.9322x + 0.0928 mark] o Record the gradient (slope) of your line. Gradient (slope) of the line: -3.9322 [1 mark] 2. Considering the trend followed by your data points explain if your data is consistent with the theory suggested by equation (4) on page 3 of this lab. [3 marks] Copy your graph from Excel/MATLAB and paste here. [8 marks] CONSERVATION OF MOMENTUM 0.1 0.2 03 0.4 0.5 0.6 3. Using the slope (gradient) of your graph find the mass me of the cart B. [2 marks] [Hint: Equation (4)] -1 fB/v fA y =-3.9322x + 0.0928 -2 4. Calculate the percentage error of your value for the mass of the car my compared with the actual mass of the car. [Note : Actual mass of the cart B is 0.260 kg] [2 marks] -2.5 Mass of cart ma3. In a vehicle safety test, a 1580 kg truck traveling at 60 km/h collides with concrete barrier and comes to a complete stop in 0.120 s. What is the magnitude of the change in the momentum of the truck? [3 marks] 4. A railroad diesel engine weighs 4 times as much as a flatcar. If the engine coasts at 5 km/h into a flatcar that is initially at rest, how fast do the two coast after they couple. [3 marks]5. In this experiment the data logger measures the time taken for the flag to pass through the light gate and then 2. Figure shows a head-on collision between two blocks on a smooth surface. it calculates the instantaneous velocity of the cart as it passes through the light gate. Explain why a stopwatch cannot be used to measure the time required for this experiment. [2 marks] JUST BEFORE COLLISION JUST AFTER COLLISION 3.4 m/'s 2.0 m/s ym's 2.5 kg 1.6 kg 2.5 kg 1.6 kg Before the collision, block A of mass 2.5 kg is moving to the right with a speed of 3.4 m s- and block B of mass 1.6 kg is moving to the left with a speed of 2.0 m s-1. The blocks coalesce (sticks together) at impact and move with a common speed of vm s-. 6. Write a brief conclusion for your experiment. [3 marks] a) Calculate the speed v. [2 mark] b) Calculate the total kinetic energy before the collision and the total kinetic energy after the collision. Laboratory Assignment [3 marks] 1. A trolley A of mass 1.0 kg is moving at 2.0 ms-]. It collides with a stationary trolley B of mass 2.0 kg. Trolley B moves off at 1.2 ms"!. Use the principle of conservation of momentum to calculate the speed of the trolley A after the collision and state of its direction.] [3 marks] c) Is this collision elastic or inelastic? Explain marks]