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Activity 04: Collision Name: ___________________________ Date: ____________________________ Physics with Video Analysis 15 - 1 Cart 1 Cart 2 Cart 1 Cart 2 Collisions Part 1 When analyzing collisions between objects, physicists define the total Momentum P of the system as the sum of all the individual momenta. In particular, when no net outside force is defined, the total momentum P of the system is conserved. In this assignment you will have an opportunity to verify this law in two systems: (a) one consisting of two carts of equal mass, and (b) and two with carts of unequal mass, as they undergo collisions. 1. Preliminary Questions Note: You will receive full credit for each prediction made in this preliminary section whether or not it matches conclusions you reach in the next section. As part of the learning process it is important to compare your predictions with your results. Do not change your predictions! (a) Imagine that we have two identical carts on a track. Cart one comes in at a velocity v and collide elastically with cart 2. What do you expect is the velocity of cart 1 after the collision? What happens to the velocity of Cart 2, again after the collision? Explain mass of both cart 1 & cart 2 = 524 g =0.524 kg positive x direction has the positive velocity. Conservation of momentum: (0.524 kg)(v 1 m/s) + (0.524 kg)(0 m/s) = (0.524 kg)(0 m/s)+(0.524 kg)(v 2 m/s) v 2 = v 1 , in the original direction of cart 1. v 1 =v 2 =0.5 m in 8 frames (0.8 s), that is 0.625 m/s After the collision, cart 1 stops to rest and cart 2 moves forward at the same velocity as cart 1. (b) Now consider a system consisting of Cart 1 that has twice as much mass as Cart 2. Cart one comes in at a velocity v and collide with cart 2. What do you expect is the momentum of cart 1 after the collision? What happens to the kinetic energy of Cart 2, again after the collision? Explain

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  • Activity 04: Collision Name: ___________________________ Date: ____________________________

    Physics with Video Analysis 15 - 1

    Cart 1 Cart 2

    Cart 1 Cart 2

    Collisions Part 1

    When analyzing collisions between objects, physicists define the total Momentum P of the system as the sum of all the individual momenta. In particular, when no net outside force is defined, the total momentum P of the system is conserved. In this assignment you will have an opportunity to verify this law in two systems: (a) one consisting of two carts of equal mass, and (b) and two with carts of unequal mass, as they undergo collisions.

    1. Preliminary Questions

    Note: You will receive full credit for each prediction made in this preliminary section whether or not it matches conclusions you reach in the next section. As part of the learning process it is important to

    compare your predictions with your results. Do not change your predictions!

    (a) Imagine that we have two identical carts on a track. Cart one comes in at a velocity v and collide elastically with cart 2. What do you expect is the velocity of cart 1 after the collision? What happens to the velocity of Cart 2, again after the collision? Explain

    mass of both cart 1 & cart 2 = 524 g =0.524 kg

    positive x direction has the positive velocity.

    Conservation of momentum: (0.524 kg)(v1m/s) + (0.524 kg)(0 m/s) = (0.524 kg)(0 m/s)+(0.524 kg)(v2 m/s) v2 = v1, in the original direction of cart 1.

    v1=v2=0.5 m in 8 frames (0.8 s), that is 0.625 m/s

    After the collision, cart 1 stops to rest and cart 2 moves forward at the same velocity as cart 1.

    (b) Now consider a system consisting of Cart 1 that has twice as much mass as Cart 2. Cart one comes in at a velocity v and collide with cart 2. What do you expect is the momentum of cart 1 after the collision? What happens to the kinetic energy of Cart 2, again after the collision? Explain

  • 15 - 2 Physics with Video Analysis

    mass of cart 1 =1048 g=1.048 kg, mass of cart 2 = 524 g=0.524 kg

    Final velocity of both the carts is same.

    Applying conservation of momentum: (1.048 kg)(v1 m/s) + (0.524 kg)(0 m/s) = (1.048+0.524)(vf m/s) vf = 0.6667 v1, in the same direction of cart 1.

    v1=0.50 m in 10 frames (1 s), that is 0.50 m/s

    vf = 0.6667 *0.5 m/s=0.3333 m/s

    After the collision, cart 1 sticks with cart 2, and the combined mass moves at a diminished velocity of 0.3333 m/s.

  • Physics with Video Analysis 15 - 3

    2. Activity-Based Questions

    1D Collision for Equal Mass Carts: Before answering the questions below, open the Logger Pro experiment file . It has the movie inserted. The movie, recorded at 10 frames/second, shows two identical carts moving on a track. The movie has been scaled to meters.

    (a) Click the Add Point icon ( ) once and then click in each frame of the movie on the yellow dot at the center of Cart 1 (the cart on the left) to create X:Y series data. Notice the x vs. t graph that appears. Sketch the data points in the figure below and label them as Cart 1.

    Rewind the movie ( ) and use the Set Active Point tool ( ) to select the X2:Y2 series. Click

    Add Point one time and click on the center of Cart 2 (the one on the right) in each frame. Sketch its graph in the figure that follows. Label it as Cart 2.

    (b) Are the position vs. time of the two carts consistent with what you answered before? Account for discrepancies.

    Yes, they are as per my expectations. No discrepancies found, apparently.

    (c) Compute the momentum of the system (Momentum = mv) before the collision and after the collision. To do so easily, click on the left hand side of the graph, where the labels are. Select more. Select X velocity and X2 velocity. These are the velocities of cart 1 and cart 2 respectively. Once you have the velocities it is easy to compute the momenta, before and after the collision. To do so, select 4

    Cart 1

    Cart 2

  • 15 - 4 Physics with Video Analysis

    points of the velocity graph of cart 1 just before and just after the collision and find their average (using the Stat button in logger pro). Multiply by the mass of the cart and this will give you the momentum of one cart 1, before and after the collision. Repeat the same for cart 2. Add the individual momenta (before and after the collision). What can you say about these two numbers?

    Before the collision:

    Mean value of v1 = 0.6635 m/s, momentum of cart 1=0.524 kg*0.6677 m/s=0.3477 kg.m/s

    Mean value of v2 = 0.0 m/s, momentum of cart 2=0.524 kg*0.0 m/s=0.0 kg.m/s

    After the collision:

    Mean value of v1 = 0.0 m/s, momentum of cart 1=0.524 kg*0.0 m/s=0.0 kg.m/s

    Mean value of v2 = 0.6432 m/s, momentum of cart 2=0.524 kg*0.0 m/s=0.33704 kg.m/s

    Total momentum before collision = (0.3477+0.0) kg.m/s =0.3477 kg.m/s

    Total momentum after collision = (0.0+0.33704) kg.m/s =0.33704 kg.m/s

    There is a net loss in momentum after the collision. The energy lost due to sound, heat and frictional loss

    of velocity may be the reason behind such loss of momentum.

  • Physics with Video Analysis 15 - 5

    Center of Mass Motion in a 1D Collision for Unequal Mass Carts: Before answering the questions below, open the Logger Pro experiment file . The movie has already been inserted. The movie, recorded at 10 frames/second, shows Cart 1 on the left with iron block attached to double its mass. The movie is scaled in meters.

    (d) Click the Add Point tool ( ) once and then click in each frame of the movie on the yellow dot at the center of Cart 1 (the cart on the left) to create X:Y series data. Notice the x vs. t graph that appears. Sketch the data points in the figure below and label them as Cart 1.

    Rewind the movie ( ) and use the Set Active Point tool ( ) to select the X2:Y2 series. Click

    Add Point one time and click on the center of Cart 2 (the one on the right) in each frame. Also sketch the data points in the figure below. Label it as Cart 2.

    (e) Are the position vs. time of the two carts consistent with what you answered before? Account for discrepancies.

    Yes, exactly.

    (f) Compute the momentum of the system (Momentum = mv) before the collision and after the collision. To do so easily, click on the left hand side of the graph, where the labels are. Select more. Select X velocity and X2 velocity. These are the velocities of cart 1 and cart 2 respectively. Once you have the velocities it is easy to compute the momenta, before and after the collision. To do so, select 4 points of the velocity graph of cart 1 just before and just after the collision and find their average (using the Stat button in logger pro). Multiply by the mass of the cart and this will give you the momentum of one cart 1, before and after the collision. Repeat the same for cart 2. Add the individual momenta (before and after the collision). What can you say about these two numbers?

    Cart 1

    Cart 2

  • 15 - 6 Physics with Video Analysis

    Before the collision:

    Mean value of v1 = 0.5348 m/s, momentum of cart 1 =1.048 kg*0.5348 m/s=0.5605 kg.m/s

    Mean value of v2 = 0.0 m/s, momentum of cart 2 =0.524 kg*0.0 m/s=0.0 kg.m/s

    After the collision:

    Mean value of v1 = 0.3544 m/s, momentum of cart 1=1.048 kg*0.3544 m/s=0.3714 kg.m/s

    Mean value of v2 = 0.3473 m/s, momentum of cart 2=0.524 kg*0.3473 m/s=0.1820 kg.m/s

    Total momentum before collision = (0.5605+0. 0) kg.m/s =0.5605 kg.m/s

    Total momentum after collision = (0.3714+0.1820) kg.m/s =0.5534 kg.m/s

    Again, there is a net loss in momentum after the collision. The energy lost due to sound, heat and

    frictional loss of velocity may be the reason for such loss of momentum.

    3. Reflections on Your Findings

    (a) You can compute the percent variation of momentum with the equation

    %100

    )(2

    1% x

    PP

    PPVariation

    afterbefore

    afterbefore

    If the total momentum P is conserved exctly, this number should be zero. Any number leass than 10% is a good result. Show what you obtain in the case of equal masses:

    %Variation equal masses =(0.3477-0.33704)/(1/2*(0.3477+0.33704))*100=3.1% % Variation non-equal masses= (0.5605-0.5534)/(1/2*(0.5605+0.5534))*100=1.3%

    If the total momentum of a system is a constant, then we say that the total momentum is conserved. A consequence of Newtons laws of physics is that the total momentum of a system is conserved whenever the net external force on the system is zero. Describe all the external forces acting on this system and explain what the net external force is. Is it zero or approximately zero? The external forces acting on this system is kinetic frictional force of the surface with the moving carts. The magnitude of this force is not zero, but approximately zero.

    (Eq.1)