Skill 2.C Comparing Changes

There are several excellent ways to address this question. 

For a math approach: 

1. Identify that the velocity is what is changing, and we want to know how this relates to the acceleration and displacement. 

2. models that relate the quantities are the definition of acceleration (a=Δv/t) and we can choose between the definition of average velocity (average velocity = Δx/t). 

3. discuss functional dependence: The definition of acceleration shows a direct relationship between acceleration and the change in velocity per unit of time and since trial 1 experiences a small change in velocity (around 0.5 m/s) in the 2.5 seconds graphed as compared to trial 2 (changed by around 3 m/s) that means trial 2 has a greater acceleration.
   The definition of average velocity shows a direct relationship between displacement per unit of time and average velocity. The average velocity for trial 1 is greater than trial 2 (since the velocity of each point of trial 1 is greater than every single velocity for trial 2), so that means that the displacement of trial 1 in 2.5 seconds is greater than the displacement of trial 2 in the same time interval.  

For a graphical approach we can find the acceleration as the slope of the best fit line and the displacement as the area under the best fit lines and come to the same conclusions. 

To follow our tips for a change scenario we

1. Clearly identify what quantities are being changed and what quantity is being asked about.
   We are increasing the mass at the bottom of the ramp and being asked about how that affects the distance traveled in the projectile section. 

2. Identify a model or derive an expression that links the change being made and the quantity being asked about. Most importantly, verify that any other factors in this relationship are the same.
   We can break this into 3 connected Physics events as shown in the lettered positions at right. The slide from A to B is not altered by the change in mass. The collision between B and C is governed by conservation of linear momentum (total initial momentum = final total momentum) and the final velocity is affected by the mass change. The projectile section will have the same flight time, but the distance traveled, while not obviously dependent on the mass change, is related because it depends on the velocity after the collision (at point C). We can combine these two by solving the collision equation shown in the diagram for v_c = 2m v_b / (2m + mass of the lower block) and substituting that into the Δx equation for CD to find the distance traveled Δx = 2m v_b / (2m + mass of the lower block) * t 

3. Explain the functional dependence (if one part of the equation goes up, the other will . . .) of the change and its effect.
   In the equation we derived, Δx = 2m v_b / (2m + mass of the lower block) * t  the 2m, v_b, and t are the same between trial 1 and 2, so the only change will be an increase in the mass of the lower block, which is dividing the right side of the equation and will result in a smaller calculated value for Δx. Therefore, increasing the mass of the lower block will decrease the distance to the landing point. 

The velocity graph generated is very useful. Following the steps above we use strategy 1 from above to break the motion into chunks. In this case based on a change in slope (at points B and C) or direction of travel (at point D).

Now we can evaluate the acceleration for each time interval by finding the slope between the points, being sure to properly find the slope, for example for segment 2 you would use (position at c - position at b)/(time c - time b). You can then use these accelerations in the kinematics equations, or you can find displacements using the area under each section of the graph. Be sure if you use areas that you recognize the negative areas (interval 4) and take into account that segment 2 cannot be described as a triangle, but as either a trapezoid or a triangle + rectangle. 

You could compare the average velocity for each segment (2>3>1>4)*, the magnitude of acceleration (4=3>1>2)*, magnitude of displacement during the interval (4<1<2=3)*.
*The answers in parentheses are based on my estimates with my graph but may be different from yours if you use the interactive to generate your own graphs.

For this scenario combining the 2 techniques of first chunking this into 3 sections: 1) traveling down the ramp, 2) the frictionless horizontal section, and 3) the horizontal section with friction. Then organizing the information about starting and stopping with a velocity of zero on the diagram. 

With the goal of the question to compare which distance was greater, you can use a kinematics and forces approach. Since you already the starting and ending velocity is zero, that means that both segment 1 and 3 have to experience the same magnitude of change in speed. You can take a shortcut, knowing that v^2 = v_0^2 + 2a(Δx), that whichever one has the greater acceleration will take less distance. The accelerations are determined by a=ΣF/m and the masses are the same, so we can simply compare the net force on the ramp with the net force on the ground. On the ramp our free-body diagram will show an unbalanced mgsinθ = mgsin60 = 0.866mg and the only horizontal force is friction on the ground, equal to μFn = μmg = 0.5mg in this situation. Since 0.5mg<0.866mg, we can conclude that the block will slide a longer distance in segment 3 than in segment 2. 

You could make a similar argument using energy and work done.