2-D Collision Lab Report

SPH4UE – 2-D Collision Lab Report Student Number (n = 5)​ Small Ball Mass: 0.55 kg​ Large Ball Mass: 1.45 kg​ Framerate: 15.0 Hz​ Time Between Frames: 1/15 s = 0.0667 s​ Scale: 1.00 m = 10.0 cm → 1.0 cm = 0.10 m Notation: ●​ si = small ball, initial (before the collision)​ ●​ sf = small ball, final (after the collision)​ ●​ li = large ball, initial (before the collision)​ ●​ lf = large ball, final (after the collision) Question 1: Initial and Final Velocity Vectors: Step-by-step: 1.​ Measure displacement between adjacent positions in the strobe photo.​ 2.​ Convert displacement to meters using the scale (1 cm = 0.10 m).​ 3.​ Use the formula:​ v = Δd / Δt​ 4.​ Determine the angle from the diagram for vector direction. Measured and calculated values: ●​ Small ball (initial):​ Displacement = 1.10 cm → 0.11 m​ v = 0.11 / 0.0667 = 1.65 m/s​ Direction: [E9°S]​ ●​ Small ball (final):​ Displacement = 0.08 m​ v = 0.08 / 0.0667 = 1.20 m/s​ Direction: [E28°N]​ ●​ Large ball (initial):​ Displacement = 0.03 m​ v = 0.03 / 0.0667 = 0.45 m/s​ Direction: [E17°N]​ ●​ Large ball (final):​ Displacement = 0.04 m​ v = 0.04 / 0.0667 = 0.60 m/s​ Direction: [E4°S]​ Final velocity vectors (2 sig figs): ●​ v_si = 1.7 m/s [E9°S]​ ●​ v_sf = 1.2 m/s [E28°N]​ ●​ v_li = 0.45 m/s [E17°N]​ ●​ v_lf = 0.60 m/s [E4°S] Question 2: Momentum Vectors and Kinetic Energies Before the Collision Small ball (before):​ Mass = 0.55 kg​ Speed = 1.7 m/s​ p_si = 0.55 × 1.7 = 0.935 kg·m/s​ KE_si = 0.5 × 0.55 × (1.7)^2 = 0.79475 ≈ 0.79 J Large ball (before):​ Mass = 1.45 kg​ Speed = 0.45 m/s​ p_li = 1.45 × 0.45 = 0.6525 kg·m/s​ KE_li = 0.5 × 1.45 × (0.45)^2 = 0.1468 ≈ 0.15 J Question 3: Total Initial Momentum and Kinetic Energy Resolve components: ●​ Small ball (E9°S):​ p_si_x = 0.935 × cos(9°) = 0.923​ p_si_y = -0.935 × sin(9°) = -0.146​ ●​ Large ball (E17°N):​ p_li_x = 0.6525 × cos(17°) = 0.624​ p_li_y = 0.6525 × sin(17°) = 0.191​ Total initial momentum components:​ p_total_ix = 0.923 + 0.624 = 1.547 kg·m/s​ p_total_iy = -0.146 + 0.191 = 0.045 kg·m/s Total initial KE:​ KE_total_i = 0.79 + 0.15 = 0.94 J Question 4: Momentum Vectors and Kinetic Energies After the Collision Small ball (after):​ Mass = 0.55 kg​ Speed = 1.2 m/s​ p_sf = 0.55 × 1.2 = 0.66 kg·m/s​ KE_sf = 0.5 × 0.55 × (1.2)^2 = 0.396 J ●​ p_sf_x = 0.66 × cos(28°) = 0.582​ ●​ p_sf_y = 0.66 × sin(28°) = 0.310​ Large ball (after):​ Mass = 1.45 kg​ Speed = 0.60 m/s​ p_lf = 1.45 × 0.60 = 0.87 kg·m/s​ KE_lf = 0.5 × 1.45 × (0.60)^2 = 0.261 J ●​ p_lf_x = 0.87 × cos(4°) = 0.869​ ●​ p_lf_y = -0.87 × sin(4°) = -0.061 Question 5: Total Final Momentum and Kinetic Energy Total final momentum components:​ p_total_fx = 0.582 + 0.869 = 1.451 kg·m/s​ p_total_fy = 0.310 - 0.061 = 0.249 kg·m/s Total final KE:​ KE_total_f = 0.396 + 0.261 = 0.657 J Question 6: Is Momentum Conserved? Component Initial (kg·m/s) Final (kg·m/s) Difference x 1.547 1.451 0.096 y 0.045 0.249 0.204 Momentum is nearly conserved in the x-direction. There is a larger deviation in the y-direction. Possible sources of error: ●​ Friction or air resistance​ ●​ Calibration inaccuracies in the photo​ ●​ Measuring error in distance or angles​ ●​ Slight deformation of the balls during collision Question 7: Is Kinetic Energy Conserved? Initial KE = 0.94 J​ Final KE = 0.657 J Ratio conserved = 0.657 / 0.94 = 0.699​ Approximately 70% of kinetic energy is conserved. Conclusion:​ The collision is inelastic. Energy is not fully conserved. Question 8: Velocity When Both Objects Are in Contact Use total momentum and total mass: p_total = sqrt(1.547^2 + 0.045^2) = 1.548 kg·m/s​ Total mass = 0.55 + 1.45 = 2.00 kg v_contact = p_total / mass = 1.548 / 2.00 = 0.774 m/s KE_contact = 0.5 × 2.00 × (0.774)^2 = 0.599 ≈ 0.60 J Question 9: Where Is the Remainder of the Energy? Initial KE = 0.94 J​ KE at contact = 0.60 J​ Energy lost or stored:​ 0.94 - 0.60 = 0.34 J This energy is stored or dissipated as: ●​ Thermal energy (E_th)​ ●​ Sound energy​ ●​ Elastic potential energy (E_s) due to deformation​ Notation:​ E_th + E_s = 0.34 J