M5 - Theory and Procedure2

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Experiment M5 Purdue University - Physics 22000 M5-1  Purdue University 1996 - 2021 Lab 5: Impulse and Momentum (M5) Objectives • Understand the definition of momentum and its vector nature as it applies to one- dimensional collisions. • Examine the concept of impulse to explain how forces act over time. Theory By the late seventeenth century, several leading scientists, including Isaac Newton, had developed the concept of momentum to describe both elastic (in which momentum and kinetic energy are conserved) and inelastic collisions (in which only momentum is conserved). A perfectly inelastic collision between two objects is a collision in which the two objects stick together after the collision, so that their velocities are the same. In everyday life, inelastic (but not perfectly) collisions are the most common. Momentum is a vector and is defined by the product of mass and velocity (which is also a vector). Alternatively, it is equivalent to three algebraic equations for x , y and z components): p mv    or p x  mv x p y  mv y p z  mv z ì í ï î ï ü ý ï þ ï (1) Because momentum is a vector, it has both magnitude and direction. In one dimension, its sign indicates the direction of a vector. Momentum change is the difference between the final momentum and the initial momentum. Since mass is always positive, momentum has the same sign (i.e., the same direction) as velocity vector. There is a relationship between the forces experienced by an object and its momentum change. It can be shown mathematically from Newton's laws and experimentally from our own observations, that, in one dimension, the change in momentum of an object is equal to a quantity called impulse . Impulse depends on both the magnitude of the applied force at each instant of time, and the time interval over which this force acts. Impulse is defined as the change in momentum of an object caused by the force  F acting on the object during time interval  t : (2) Impulse is equal to the area under the force vs. time graph. Large force acting over a short time and small force acting over a long time can have the same impulse. For a constant value

M5-2 Experiment M5 Purdue University - Physics 22000 of the force, it is simply a product of the force and the time interval  t when the force is acting on the object. The same impulse (i.e., the same change in momentum) could be achieved with a small force acting for a long period of time as with the large force acting for a short period. An example of the first case (small force, large ∆t) would be any soft collision, whereas a large force and small ∆t is typical for hard collisions. The designers of new cars try to include crumping zones into the car body. In other words, they try to "soften" collisions. The safer design tries to make collisions last longer to reduce the maximum force and therefore to protect passengers. Real objects are not point-like objects, but they have nonzero dimensions. To analyze their motion, we need to use the concept of center of mass. The center of mass is defined as the single point that would move in response to external forces in the same way as the whole object if the total mass was concentrated in this point. For regular, symmetrical objects, it is going to be the center point of the object. For example, the center of mass of a homogeneous cube is located at the center of the cube. Set-up: We will use a plunger to create an elastic collision in which the carts are repelled. An inelastic collision will occur when we use the Velcro pads attached to both cars. After the inelastic collision, the carts will stick together. Procedure: Activity 1: Inelastic Collision Login using your Purdue career account. Download lab files for experiment M5 from Brightspace for Physics 220. First, make sure that the track is properly leveled. Double-click on the “M5 Activity 1” icon. In this activity, you are going to use two carts: the dynamics cart (with the plunger locked into the hidden position so that it is flush with the end of the cart) and the collision cart. Make sure that the black force sensor is not attached to the collision cart. To lock the plunger, push it into the cart and then shift up until it locks. Using the scale, determine the mass of your dynamics cart (the cart equipped with the plunger) and the mass of your collision cart (the cart equipped with Velcro). Record these values on your data sheet. Move the collision cart to the starting point, located in the middle of the track. Make sure that its sail is in the direction of the right motion sensor. Position the dynamics cart so that it is flush with the left bumper and so that its sail is pointing in the direction of the left motion sensor.

Experiment M5 Purdue University - Physics 22000 M5-3 Collision Cart Dynamics Cart Start recording data and after you hear the ultrasonic motion sensors working, push the dynamics cart towards the collision cart. The two carts should collide and continue moving together in the same direction. Use the “Add Coordinates Tool” icon to read five typical velocity values before and after the collision and calculate the momentum of the system before and after the collision. Do not include points that were recorded when the dynamics cart was in contact with your hand or when the two carts encountered the far bumper. Remember, before the collision, the collision cart should be at rest. If both carts have the same mass, then according to the conservation of momentum principle : p  mV before + 0  mV after + mV after  2* mV after V before  2* V after or V after  1 2 V before (3) Print a copy of this graph by selecting Print from the File menu. Find the selected data points on the printout and clearly mark these points with a pen (for example, circle them). Write this graph, "Velocity vs. Time for an Inelastic Collision." Remember that we have not eliminated friction. Therefore, the momentum after the collision would not be exactly equal to the initial momentum. Activity 2: Simulated Explosion Select “ Open Experiment ” from the “ File ” menu. Do not save any changes. Next, select: “M5 Activity 2” and open it. Two sets of velocity vs. time axes should appear on your screen. In this activity, you are going to again use the plunger cart and the collision cart. Record the mass of each cart on your data sheet.

M5-4 Experiment M5 Purdue University - Physics 22000 Align the left end of the collision cart with the middle of the track. Make sure that its aluminum sail is in the direction of the right motion sensor. Pick up the plunger cart and lock the plunger into place by pushing it towards the cart so that it is flush with the bumper. Position the right bumper of the plunger cart 1 cm to the left of the non-plunger cart. Make sure the sail is closest to the left motion sensor and its plunger is facing towards the collision cart. The plunger is released by pushing down on the trigger, which is the small knob above the plunger. Begin data collection by clicking on the “ Record ” button. Next, release the plunger by gently tapping it with a ruler. The two carts should move away from each other. Let us analyze the momentum of both carts. Initially, both carts were not moving, so the momentum before the explosion was equal to zero.  p  0 ( ) If both carts have the same mass, then according to the conservation of momentum principle :  p  0 Þ  p dynamic +  p collision  mV dynamics + mV collision ( )  m V dynamics + V collision ( )  0 V dynamics + V collision ( )  0 Þ V collision  - V dynamics (4) Using the “Smart Tool” icon, read five typical values for the velocity of both carts before and after the trigger was released. Record these values on your data sheet. Calculate the averages of these velocity values. Using the average velocity values, calculate the momentum of each cart. Do not print this graph. Remember: velocity is a vector quantity! Activity 3: Elastic Collision Select Open Activity from the File menu. Do not save any changes. Next, select: “M5 Activity 3” and open it. Two sets of velocity vs. time axes will appear on your screen. For perfectly elastic collisions, both the momentum and the mechanical energy are conserved . Therefore, we have two equations that both must be obeyed. For carts with the equal mass, we have: Conservation of momentum: 𝑚𝑉 ௗ௬௡௔௠௜௖௦ ௖௔௥௧ ௕௘௙௢௥௘ + 0 = 𝑚𝑉 ௗ௬௡௔௠௜௖௦ ௖௔௥௧ ௔௙௧௘௥ + 𝑚𝑉 ௖௢௟௟௜௦௜௢௡ ௖௔௥௧ ௔௙௧௘௥ (5) Conservation of energy: 1 2 𝑚𝑉 ௗ௬௡௔௠௜௖௦ ௖௔௥௧ ௕௘௙௢௥௘ ଶ + 0 = 1 2 𝑚𝑉 ௗ௬௡௔௠௜௖௦ ௖௔௥௧ ௔௙௧௘௥ ଶ + 1 2 𝑚𝑉 ௖௢௟௟௜௦௜௢௡ ௖௔௥௧ ௔௙௧௘௥ ଶ

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