The Cart and the Brick Activity
Begin Activity
View a Sample Analysis
In this activity, you will analyze a collision in order to determine the total momentum of a system before and after the collision. Your goal is to gather evidence that supports the law of conservation of momentum. The collision involves the collision between a stationary brick that is dropped upon a moving cart. The brick lands upon the cart and travels at the same speed as the cart after the collision. A ticker tape is attached to the cart and pulled along with the cart through a ticker tape timer. Ticks (dots) are placed upon the tape every 1/60-th of a second, leaving a trace of the cart's position over the course of time. Analysis of the ticker tape allows one to determine the velocity of the cart before and after the collision.
Sample Analysis
Given:The ticks (dots) are left by the cart every 1/60-th of a second. The spacing between every 6th dot (for example, from the 1st dot to the 7th dot) is equal to the distance traveled in a time of 0.10 seconds. This information can be used to determine the before- and after-collision velocities.
slow
Pre-Collision Speed:
3.0
Mass of Brick (kg):
= 1.0 cm
The goal is to determine the total system momentum before and after the collision. The masses are known. So the first step is to determine the velocities before and after the collision.
Continue
Mass of Cart (kg):
SCALE:
4.0
To determine the before-collision velocity, one must use the fact that the ticks on the ticker tape were left every 1/60-th of a second. For this reason, the spacing before every sixth tick is equivalent to the distance traveled in 6/60-ths second or 0.10 seconds. The work is shown below the graphic.
The distance circled on the left represents the distance traveled in 0.30 seconds - a time equivalent to 18 spaces. Using the scaled centimeter, this distance is equivalent to approximately 10.7 cm. So the before-collision velocity is 10.7 cm/0.30 s = 35.6666… cm/s = 35.7 cm/s. The before-collision momentum of the cart is (4.0 kg)•(35.6666… cm/s) = 142.6666 kg•cm/s = 143 kg•cm/s.
The distance circled on the right represents the distance traveled in 0.30 seconds - a time equivalent to 18 spaces. Using the scaled centimeter, this distance is equivalent to approximately 6.0 cm. So the after-collision velocity of the cart and the brick is 6.0 cm/0.30 s = 20.0 cm/s. The after-collision momentum of the cart and brick is (7.0 kg)•(20.0 cm/s) = 140 kg•cm/s.
Exit Sample Analysis
So for this sample collision, it has been shown that the 4.0-kg cart has a before-collision momentum of 143 kg•cm/s. Since the cart is the only object moving before the collision, the total system momentum is 143 kg•cm/s.  After the collision, it has been shown that the combination of the 4.0-kg cart with the 3.0-kg brick on top of it has a momentum of 140 kg•cm/s. This is the total system momentum after the collision. One would expect there to be small amounts of error inherent in the process of measuring the velocity of the moving cart. The distances traveled during the 0.30 seconds were best estimates and thus subject to small amounts of error. Within a reasonable margin of error - two significant digits - one can confidently say that the total system momentum is the same before the collision as it is after the collision. For this reason, one can conclude that momentum is conserved in this collision.
medium
fast
slow
Select Experimental Conditions
You have a choice of trials that you can run. The trials differ from one another in terms of the mass of the cart, the mass of the brick, and the pre-collision speed of the cart. Click on one of the set of conditions below to run the trial. Click on the More Conditions button for more possibilities.
More Conditions
2.0-kgcart
3.0-kgcart
1.5-kg brick
4.0-kgcart
2.5-kg brick
Previous Conditions
You have a choice of trials that you can run. The trials differ from one another in terms of the mass of the cart, the mass of the brick, and the pre-collision speed of the cart. Click on one of the set of conditions below to run the trial. Click on the Previous Conditions button for other possibilities.
Trial A1 Conditions
Analyze the ticker tape diagram (dot diagram) left by the cart to determine the pre-collision and post-collision velocities. Combine this velocity information with the mass values to determine the total momentum of the system before and after the collision. Is momentum conserved by the system?
2.0
Given:The ticks (dots) are left by the cart every 1/60-th of a second. So every 10 dots is equal to a time of 0.10 seconds.The large divisions on the distance scale below the ticker tape are 1.0-cm apart.
1.5
Repeat Animation
Change Conditions
Analyze Data
Given:The ticks (dots) are left by the cart every 1/60-th of a second. So the spacing between every 6th dot (for example, from the 1st dot to the 7th dot) is equal to the distance traveled in a time of 0.10 seconds.
Analyze the ticker tape diagram (dot diagram) left by the cart to determine the pre-collision and post-collision velocities. Combine this velocity information with the mass values to determine the total momentum of the system before and after the collision. Can you gather such data and present it as evidence to support the principle that momentum is conserved by the system?
medium
Trial A2 Conditions
Analyze the ticker tape diagram (dot diagram) left by the cart to determine the pre-collision and post-collision velocities. Combine this velocity information with the mass values to determine the total momentum of the system before and after the collision. Can you gather such data and present it as evidence to support the principle that momentum is conserved by the system?
fast
Trial A3 Conditions
Trial B1 Conditions
Trial B2 Conditions
Trial B3 Conditions
Fast
Trial C2 Conditions
Trial C1 Conditions
Trial C2 Conditions
Trial C3 Conditions
2.5
Trial D1 Conditions
Trial D2 Conditions
Trial D3 Conditions
Trial E1 Conditions
Trial E2 Conditions
Trial E3 Conditions
Trial F1 Conditions
Trial F2 Conditions
Trial F3 Conditions
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