1.Using your measurements from Table 1, calculate the total kinetic energy of the two carts before collision for each trial (this is the initial kinetic energy, K). Show your calculations below and record your results in Table 2.
- Using your measurements from Table 1, calculate the total kinetic energy of the two carts after collision for each trial (this is the final kinetic energy, Kʹ). Show your calculations below and record your results in Table 2.
- Calculate the % change in kinetic energy during the collision for each trial. Show your calculations below and record your results in Table 2.
% change = final − initia/ initial × 100
4.You will probably find that the total momentum changed slightly during each trial. Explain what might have caused the total momentum of your system to change during the collisions.
5.You should have found that the collision reduced the total kinetic energy of your system in each trial. Where did the kinetic energy lost during the collision go? In other words, what forms of energy might the lost kinetic energy have been converted to?
The Correct Answer and Explanation is:
1–3. Kinetic Energy Calculations and % Change
Mass of cart 1 (m₁): 294.1 g = 0.2941 kg
Mass of cart 2 (m₂): 294.9 g = 0.2949 kg
Kinetic Energy Formula:
KE=12mv2KE = \frac{1}{2}mv^2
Trial 1
Initial KE: KEinitial=12m1v12+12m2v22=12(0.2941)(0.307)2+12(0.2949)(0)2=0.0138 JKE_{initial} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2}(0.2941)(0.307)^2 + \frac{1}{2}(0.2949)(0)^2 = 0.0138\, \text{J}
Final KE: KEfinal=12m1v1′2+12m2v2′2=12(0.2941)(0.153)2+12(0.2949)(0.153)2=0.0069 JKE_{final} = \frac{1}{2} m_1 v_1’^2 + \frac{1}{2} m_2 v_2’^2 = \frac{1}{2}(0.2941)(0.153)^2 + \frac{1}{2}(0.2949)(0.153)^2 = 0.0069\, \text{J}
% Change: 0.0069−0.01380.0138×100=−50.0%\frac{0.0069 – 0.0138}{0.0138} \times 100 = -50.0\%
Trial 2
Initial KE: KEinitial=12(0.2941)(0.249)2=0.0091 JKE_{initial} = \frac{1}{2}(0.2941)(0.249)^2 = 0.0091\, \text{J}
Final KE: KEfinal=12(0.2941)(0.087)2+12(0.2949)(0.087)2=0.0022 JKE_{final} = \frac{1}{2}(0.2941)(0.087)^2 + \frac{1}{2}(0.2949)(0.087)^2 = 0.0022\, \text{J}
% Change: 0.0022−0.00910.0091×100=−75.8%\frac{0.0022 – 0.0091}{0.0091} \times 100 = -75.8\%
Trial 3
Initial KE: KEinitial=12(0.2941)(0.287)2=0.0121 JKE_{initial} = \frac{1}{2}(0.2941)(0.287)^2 = 0.0121\, \text{J}
Final KE: KEfinal=12(0.2941)(0.076)2+12(0.2949)(0.076)2=0.0017 JKE_{final} = \frac{1}{2}(0.2941)(0.076)^2 + \frac{1}{2}(0.2949)(0.076)^2 = 0.0017\, \text{J}
% Change: 0.0017−0.01210.0121×100=−85.9%\frac{0.0017 – 0.0121}{0.0121} \times 100 = -85.9\%
Table 2: Kinetic Energy and % Change
| Trial # | Initial KE (J) | Final KE (J) | % Change |
|---|---|---|---|
| 1 | 0.0138 | 0.0069 | -50.0% |
| 2 | 0.0091 | 0.0022 | -75.8% |
| 3 | 0.0121 | 0.0017 | -85.9% |
4. Why Momentum Might Not Be Conserved
Ideally, momentum is conserved in inelastic collisions. However, in a real-world setting, slight changes in total momentum can occur due to external forces. For this lab, friction between the carts and the track, as well as air resistance, could slightly affect motion. Additionally, the motion sensors and human interaction (e.g., catching carts) may introduce unintentional forces that alter the system’s momentum. Misalignment in the track or non-uniform magnetic sticking could also contribute.
5. Where the Lost Kinetic Energy Went
In inelastic collisions, some of the initial kinetic energy is transformed into other forms. The loss in kinetic energy during these trials likely converted into:
- Thermal energy: Due to friction and internal deformation during collision.
- Sound energy: Audible impact noise from the carts colliding.
- Internal energy: Slight warping or vibration in the carts and magnets.
These transformations are consistent with the physical principle of energy conservation: although mechanical (kinetic) energy is not conserved in inelastic collisions, total energy (including other forms) remains conserved.
