Lab 6 | Physics homework help

Question 2: 1 point

Introduction to necessary physical concept

Kinetic Energy: Energy of a body associated with its motion:

K = 1

2 mϑ Potential Energy: Energy of a body associated with its position: U = mgh Total mechanical energy: E= K + U Conservation of total mechanical energy: If non-conservative forces do no work, total mechanical energy of system is conserved: Ef = Ei → Uf + Kf = Ui + Ki (f − final , i−innitial)

Linear momentum: ⃗p = mϑ⃗ Conservation of linear momentum:

The total momentum of an isolated system (⟨ F⃗net ext ⟩ = 0 ) remains constant. For a 2-body system, this implies: p⃗f tot = p⃗i tot → m 1 ϑ⃗ 1 f + m2 ϑ⃗2 f = m1 ϑ⃗ 1i + m2 ϑ⃗2 i Totally elastic collision If both the total momentum and total kinetic energy of the two-ball system are conserved before and after collision, i.e., P⃗i tot (before collision)= P⃗f tot (after collision) and Ki tot (before collision) = Kf tot (after collision) Totally inelastic collision If the total momentum but not the total kinetic energy of the two-ball system is conserved before and after collision, i.e., P⃗i tot (before collision)= P⃗f tot (after collision) but Ki tot (before collision) ≠ Kf tot (after collision) A B Figure 1 d T mg h mg Photogate #1photogate #2 2 1

1 A

T guide bar Figure 2 Collision apparatus to 850 interface to 850 interface B Illustrations for Experimental Set-up Release ball #1 from A Collision occurs at B d → diameter of each ball m → mass of each ball h → height of drop Explanation of Dimensions in Table 1 d → diameter of 1 ball d1 → dimension of (1 ball + 1 velcro pad) = d + thickness of 1 velcro pad d2 → dimension of (2 balls + 2 velcro pads) = 2 d1 Explanation of Time Measurements in Table 2 ⟨t 1(elastic)⟩ → Time needed for ball #1 to pass through photogate #1 before elastic collision ⟨t 2(elastic)⟩ → Time needed for ball #2 to pass through photogate #2 after elastic collision ⟨t 1(inelastic )⟩ → Time needed for (ball #1 + 1 velcro pad) to pass through photogate #1 before inelastic collision ⟨t 2(inelastic)⟩ → Time needed for (ball #1 + ball #2 + 2 velcro pads) to pass through photogate #2 after inelastic collision Computation of velocities in table 3

Key Idea:

velocity = dimensionof object passing through photosensor time takenby object while passing through photosensor

Velocityϑ 1 → velocity of 1st ball before elastic collision :

ϑ 1 = d ⟨t 1(elastic)⟩

Velocityϑ ' 2 → velocity of ball #2 after elastic collision:

ϑ ' 2 = d ⟨t 2(elastic)⟩ Computation of velocities in table 3 (contd.)

Velocityϑ 1(i n)→ velocity of ball #1 before inelastic collision:

ϑ 1(i n) = d 1 ⟨t 1(inelastic )⟩

Velocityϑ→ velocity of 2-ball system after inelastic collision:

ϑ = d 2 ⟨t 2(inelastic)⟩ Tables 4 through 8 Table 4: Verify total energy conservation for ball #1 between point A and point B just before elastic collision. Tables 5 & 7 : Verify if the total momentum of 2-ball system is conserved. Tables 6 & 8 : Verify if the total kinetic energy of 2 -ball system is conserved. To verify conservation, we compute % difference between the total magnitude of each quantity before and after collision.

The % difference between quantitiesa andb is given by:

% difference = |a − b| 1 2 (a + b) ×100% End of Theory Provided data for Exp 6 and instructions for data analysis and lab report