Materials & Prep
Prepare one low-friction cart, a second identical cart or added masses totaling about 1 kg, a spring scale, and a clear floor or tabletop path. Mark a starting line. Set up a board space for the relationships a = Fnet/m, force held constant, and mass held constant. Confirm the cart demonstration follows local classroom safety procedures, including keeping the path clear and using masses safely.
Opening prediction
Display two identical carts. Leave one empty and place 1 kg of added mass on the other. Ask: “If I pull each cart with the same net force, which cart will have the greater acceleration, or will they accelerate equally?” Students silently choose A, B, or C and write one reason. Require every student to commit before discussion.
Have students recall from memory what acceleration means and what Fnet = ma tells us about acceleration. Do not correct the prediction yet. Turn and talk: “My prediction is because .” Listen for the common claim that the heavier cart has more force acting on it simply because it has more mass.
Demo and direct instruction
ShowPull the empty cart and the loaded cart along the same path, briefly matching the spring scale reading at about 6 N for each. Keep the pull direction and approximate force the same. Students observe which cart changes speed more quickly. State that this demonstration is evidence about the pattern, not a precise acceleration measurement.
ConnectThe observation is that the lighter cart speeds up more quickly under the same pull. The inference is that, for the same net force, less mass produces greater acceleration. Mass does not create a larger net force by itself.
ExplainRearrange Fnet = ma to get a = Fnet/m. Acceleration increases when net force increases while mass stays constant. Acceleration decreases when mass increases while net force stays constant. If the net force is zero, acceleration is zero regardless of mass.
Worked example
A 3 kg object has a net force of 12 N.
First case: a = Fnet/m = 12 N/3 kg = 4 m/s².
Now double the mass to 6 kg while keeping the net force at 12 N.
Second case: a = 12 N/6 kg = 2 m/s².
The acceleration is half as large, not unchanged. The mistake to watch for is plugging in the new mass but assuming that a heavier object must receive a greater force. The force was held at 12 N, so the denominator increased while the numerator stayed the same.
ShowKeep the 3 kg mass and double the net force from 12 N to 24 N. The first acceleration is 12/3 = 4 m/s², and the new acceleration is 24/3 = 8 m/s². Doubling the net force doubles the acceleration when mass stays constant.
Guided practice and check
Display this table for students to complete
| Case | Net force | Mass | Acceleration |
|---|---|---|---|
| A | 10 N | 2 kg | |
| B | 10 N | 4 kg | |
| C | 20 N | 2 kg | |
| D | 0 N | 4 kg |
Have students calculate each acceleration using a = Fnet/m, then write one comparison: “When the force stays at 10 N and the mass doubles from 2 kg to 4 kg, the acceleration changes from to .”
Check for understandingAsk students to hold up one finger for “doubles,” two for “is cut in half,” and three for “stays the same.” “A 5 kg object has a net force of 15 N. If its mass becomes 10 kg and the force remains 15 N, what happens to its acceleration?” Expected response: two fingers. Follow by asking students to calculate both values: 15/5 = 3 m/s² and 15/10 = 1.5 m/s². If students choose “stays the same,” return to the denominator in a = Fnet/m and have them compare the two calculations.
ScaffoldGive students this three-column reasoning frame: “What stays constant? ; What changes? ; Original and new calculations: ; Therefore, acceleration .” Students may use arrows beside the equation: same F, mass up, acceleration down.
Independent prediction and calculation
Have students solve each problem and explain the change in words, not only with a number.
- A 4 kg cart experiences a net force of 20 N. Its mass increases to 8 kg while the net force stays 20 N. Find both accelerations and predict the change.
- A 3 kg object experiences a net force of 9 N. The net force changes to 18 N while the mass stays 3 kg. Find both accelerations and predict the change.
- A 6 kg object experiences zero net force. Its mass changes to 12 kg. Find the acceleration before and after the mass change. Explain why this case does not show the usual “more mass means less acceleration” pattern.
ExpectedProblem 1 gives 20/4 = 5 m/s² and 20/8 = 2.5 m/s², so acceleration is halved. Problem 2 gives 9/3 = 3 m/s² and 18/3 = 6 m/s², so acceleration doubles. Problem 3 gives 0/6 = 0 m/s² and 0/12 = 0 m/s², so acceleration stays zero because the net force is zero.
CirculateRequire students to circle the quantity held constant before calculating. Correct the predictable error of saying “the heavier object has more acceleration” by asking, “What happened to the numerator, the net force?”
ExtensionAsk ready students to create a second situation in which the mass doubles but the acceleration stays the same. They must state what must happen to the net force and verify it with numbers. A valid example is changing from 4 kg and 12 N, which gives 3 m/s², to 8 kg and 24 N, which also gives 3 m/s².
Closing check
Exit ticket“An object has a mass of 2 kg and a net force of 16 N. The mass is doubled to 4 kg, and the net force stays 16 N. Predict the new acceleration before calculating. Then calculate the original and new accelerations and write one sentence explaining the result.”
Expected responseThe acceleration will be cut in half. Original: 16/2 = 8 m/s². New: 16/4 = 4 m/s². Because the net force stayed constant while the mass doubled, the acceleration was halved. Collect the response to determine whether students can both predict and support the prediction with calculation.