Experimental report department of general physics experiment 1 measurement of basic length

111Equation Chapter 1 Section 1 HANOI UNIVERSITY OR SCIENCE AND TECHNOLOGY SCHOOL OF ENGINEERING PHYSICS    EXPERIMENTAL REPORT Department of General Physics Instructor Assoc Prof Ngô Đức Quân Stud[.]

111Equation Chapter Section HANOI UNIVERSITY OR SCIENCE AND TECHNOLOGY SCHOOL OF ENGINEERING PHYSICS -    - EXPERIMENTAL REPORT Department of General Physics Instructor: Assoc.Prof Ngô Đức Quân Student: Full name: Lê Việt Hoàng ID Number: 20185258 Class: 723667 Ha Noi, 2022 MỤC LỤC Experiment 1…………………………………………………………………3 Experiment 3…………………………………………………………………7 Experiment 4………………………………………………………………15 Experiment 5………………………………………………………………32 Experiment 6………………………………………………………………37 Experiment 1: MEASUREMENT OF BASIC LENGTH Full name: Lê Việt Hoàng Student ID: 20185258 Class number: 723667 Group: 03 Verification of instructor I Experimental results Hollow cylinder Trials Steel ball Trials D (mm) 34.78 34.80 34.82 34.84 34.80 d (mm) 28.54 28.50 28.48 28.52 28.46 h (mm) 9.10 9.12 9.18 9.06 9.14 Diameter ( D b = n × 0.5 + m × 0.01 ) n (mm) 23 23 23 23 23 D b (mm) m (mm) 49.1 49.7 49.4 49.3 48.9 11.99 12.00 11.99 11.98 11.98 II Data treatment Metal hollow cylinder (mass: m=26.70 g = 26.70 × 10−3 kg) D D = ∑ i ≈ 34.80 (mm) i=1 5 d d = ∑ i ≈ 28.50 (mm) i=1 s.d △D = S.D = = √5 √∑ i =1 0.01(mm) s.d △d = S.D = = √5 √∑ i =1 ( Di−D)2 ≈ √ 5∗√5 (d i−d)2 ≈ 0.01 √ 5∗√5 (mm) s.d △h = S.D = = √5 h h = ∑ i ≈ 9.12 (mm) i=1 √∑ i =1 (hi −h)2 ≈ 0.01 (mm) √5∗√ Volume of hollow cylinder: π 2 3.14 V = ( D −d ) h = ( 34.80 2−28.502 ) ×9.12 = 2.855 × 103 (mm3 ¿=¿2.85 × 10−6 (m3 ¿ 4 △V √ √ 2 2 △ ( D 2−d ) △π △h △π V ( ) +( ) +( ) =V ( ) +( π h π D 2−d √ = 2.85 × 103× ( 0.01 ) +(2 3.14 √( )( √( = ) ( ) 2 △D △d △h + × 2 ) +( ) D d h D −d ) 0.01 0.01 0.01 + × ) +( ) 34.80 28.50 9.12 34.802−28.502 ≈ 10.5 (mm3 ¿= 0.01 × 10−6 (m ¿ Therefore: V = V ± △ V = (2.85 ± 0.01)× 10−6 (m3) Density of hollow cylinder: ρ= m 26.70× 10−3 3 = −6 = 9.37 × 10 (kg/m ¿ V 2.85× 10 △ ρ = ρ √ ¿ ¿ = 9.37 × 103× √ ¿ ¿ = 0.03 × 103 (kg/m3 ¿ Therefore: Steel ball ρ = ρ ± △ ρ = (9.37 ± 0.03)× 103 (kg/m3 ) D D b = ∑ bi ≈ 11.99 (mm) i=1 s.d △ D b = S.D = = √5 √∑ i =1 (Dbi −Db ) ≈ 0.003 (mm) √ 5∗√ 1 V b = π × Db3= × 3.14 × 11.993=902 ( mm3 )=0.902 ×10−6 (m 3) 6 √ √ △ V b=V b ( △ π ) +¿ ¿=902 × ( 0.01 ) + ¿ ¿≈2.951(mm3 ¿ ≈ 0.003 × 10−6 (m3 ) π 3.14 Therefore: V b = V b ± △ V b = (0.902 ± 0.003)× 10−6 ( m) Experimental Report MOMENTUM AND KINETIC IN ELASTIC AND INELASTIC COLLISIONS Name: Lê Việt Hoàng Student ID: 20185258 Class: 723667 Verification of the instructors Group: 03 II/Experimental results 1) Elastic collision * m1 = 400.3 g = 0.4 kg , Trials 10 m2 = 799.4g = 0.8 kg t1 0,160 0,156 0,157 0,160 0,159 0,158 0,157 0,156 0,155 0,155 t1’ 0,673 0,664 0,664 0,666 0,666 0,668 0,665 0,664 0,670 0,671 t2’ 0,269 0,262 0,262 0,265 0,263 0,264 0,265 0,262 0,263 0,263 2) Inelastic collision * m1 = 798.6 g = 0.8 kg, Trials 10 m2 = 387.1 g = 0.4 kg t1 0,239 0,249 0,243 0,240 0,238 0,232 0,232 0,236 0,230 0,236 t1'= t2' 0,588 0,589 0,590 0,586 0,588 0,580 0,591 0,585 0,581 0,581 III/Data processing 1) Elastic collision a.Velocity *V1: S = 85 mm = 0.085 m - The uncertainty of t: ∑ t =¿ 0.157 ( s ) ¿ 10 i=1 i |∑ (t 1−t ) 10 ⩟ t 1= 10 t 1= | 10 =0.002(s ) - The uncertainty of v: v1 = √ ∆ v 1=v ( ∆t ) =0.007 (m/s) t v ( )=v 1± Δv => ( ) s m m =0.541( ) t s s = 0.541± 0.007(m/s) ' ¿ V 1: - The uncertainty of t: t '1= 10 ∑ t =¿ 0.677 ( s ) ¿ 10 i=1 i ⩟ t 1' = √ 10 ∑ (t '1 i−t1 )2 i=1 10∗10 =0.001(s ) -The uncertainty of v'1 = ' ∆ v =v => ' √ ' ( ∆t1 t ' ) =0.001( v '( )=v' 1± Δv ' ( ) s m m =0.126( ) ' s t1 s m ) s ± =0.126 0.001 (m/s) ¿ V '2 : The uncertainty of t: t '2= 10 ∑ t =¿ 0.264 ( s ) ¿ 10 i=1 i ⩟ t 2' = The uncertainty of √ => ∑ (t '2 i−t 2)2 i=1 10∗10 =0.006(s) ( ) v '2= s m m =0.322( ) ' s t2 s ' ' √ ∆ v =v ( v ' ( )=v' ±Δv '2 10 ' ∆ t2 t ' 2 ) =0.007( m ) s ± = 0.322 0.007 (m/s) b.Momentum The total momentum before collision are: P1 = m a ⃗ v a + mb ⃗ vb1 P1=m v 1=0.4 × 0.541=0.216 (kg ∆ P1=P √( )( ) √( ∆ m1 ∆ v + =(0.216) m1 v1 )( ) m ) s ( ) 0.001 0.007 kgm + =0.003 0.4 0.541 s => P1=( 0.216 ± 0.003 ) ( kgm ) s The total momentum after collision are: P2 = m a ⃗ v a + mb ⃗ vb2 ' ' P2=m2 v 2−m1 v 1=0.8 × 0.322−0.4 × 0.126=0.207( ∆ P2=P × √( )( )( )( ) ' 2 ' ∆ m1 ∆ v1 ∆ m2 ∆ v2 + ' + + ' =0.207 × m1 m2 v1 v2 ¿> P2 =(0.207 ± 0.005)( √( kgm ) s )( )( 2 )( ) 2 0.001 0.001 0.001 0.007 kg + + + =0.005( 0.8 0.126 0.4 0.322 kgm ) s -There is a slightly change of the total momentum before and after collisions due to some external forces or the collisions is not completely elastic * The percent change in momentum C %=¿ P2 −P 1∨ ¿ ¿ ¿ 0.207−0.216∨ ¿ =4 % ¿ P1 = 0.216 c.Kinetic energy The total Kinetic energy before collision are 1 K 1= m1 ( v1 )2= × 400× 10−3 ×(0.541)2=0.059 J 2 ¿> ∆ K 1=K => √( )( ) ∆ m1 ∆ v1 + =( 0.059 ) m1 v1 √( ) ( ) 0.001 0.007 +4 =0.002( J ) 0,4 0.541 K 1=0.059± 0.002(J ) * The total Kinetic energy after collision are '2 '2 K 2= m v + m v =0.043 ( J ) 2 ¿> ∆ K 2=K √( ' ∆ t2 ' t2 ) +¿ ¿ => K 2=( 0.043± 0.001 ) (J ) before and after -There is also a change of kinetic energy collision but the kinetic energy after collision loses way more than momentum The percent change in kinetic energy C %=¿ K 2−K 1∨ ¿ =¿ 0.043−0.059∨ ¿ =27.1% ¿ ¿ K1 0.059 2) Inelastic collision a.Velocity -V1: The uncertainty of t: t 1= 10 ∑ t =0.238( s)⩟ t 1= 10 i=1 i The uncertainty of v: √∑ 10 i=1 ¿¿¿¿ s v1 = (m/s)=0.357 m/ s t1 ∆ v 1=v ' √ ( ∆t t ❑ ) =0.006( m ) s => v ( )=v 1± Δv =0.357± 0.006(m/s) V 2' : The uncertainty of t : t 2= 10 ∑ t =0.586( s) Δt 2= 10 i=1 i √∑ 10 i=1 ¿¿¿¿ The uncertainty of v 2= ( ) s m =0.145 m/ s ❑ t2 s 10 Experimental Report PENDULUM OSCILLATION WITH PC INTERFACE Name: Lê Việt Hoàng Student ID: 20185258 Class: 723667 Group: 03 1)Pendulum with vertical oscillation plan Trial L1= 0.400 (m) L2= 0.500 (m) L3= 0.600 (m) T1 (s) 1.295 1.302 1.292 1.300 1.303 T2 (s) 1.478 1.476 1.467 1.481 1.478 T3 (s) 1.632 1.637 1.638 1.638 1.627 T 2=1.476 s T 3=1.634 s T 1=1.298 s Δ T 1= √ ∑ (T ¿ ¿ i1−T 1)2 ⅈ=1 √ ∑ (T ¿ ¿ i2−T ) ≈ 0.004(s)¿ ∆ T 2= 5 ⅈ=1 √ (T ¿ ¿ i3−T ) ∑ ˙ Δ T 3=)¿ ≈ 0.005(s l =1 ≈ 0.004 (s)¿ * Determination of the gravitational acceleration as a function of the pendulum length: Pendulum with vertical oscillation plane: T =2 π √ ( ) l (s) → g=l π ( m ∕ s2 ) T g a.L1=0.400 (m) g1=L1 ( ) ( ) 2π ×3.141 =0.400 =9.37 ( m ∕ s ) T1 1.298 17 Δ g 1=g × √( √( ) ∆T 0.004 2 =9.37 =0.05 (m ∕ s ) T1 1.298 ) Then, g1=9.37 ±0.05(m ∕ s ) Hence g1=9.37 ±0.05(m ∕ s 2) b.L2=0.5 g2=L2 ( ) 00 (m) ( ) 2π 2 ×3.141 2 =0.500 =9.06 ( m ∕ s ) T2 1.476 Δ g 2=g2 × √( ) √( ∆ T2 0.005 =9.06 =0.06( m ∕ s 2) T2 1.476 ) Then, g2=9.06 ±0.06 (m ∕ s 2) Hence g2=9.06 ±0.06 (m ∕ s 2) c.L3=0.600 (m) g3=L3 ( ) ( ) 2π 2 ×3.141 2 =0.600 =8.87 ( m ∕ s ) T3 1.634 Δ g 3=g3 × √( ) √( ∆ T3 0.004 2 =8.87 =0.04(m ∕ s ) T3 1.634 ) Then, g3=8.87 ±0.04 (m ∕ s2 ) Hence g3=8.87 ±0.04 (m ∕ s ) 18 2)Pendulum with inclined oscillation plan: Tri al θ1=0° θ2=10° θ3 =20° θ 4=30 ° θ5 =60° T1 (s) 0.752 0.760 0.755 0.751 0.758 T 1=0.755 ( s ) T2 (s) 0.788 0.783 0.785 0.781 0.780 T 2=0.783 ( s ) T3 (s) 0.802 0.804 0.807 0.803 0.804 T 3=0.804 ( s ) T4 (s) 0.854 0.850 0.850 0.852 0.853 T 4=0.852 ( s ) T5 (s) 0.915 0.917 0.913 0.915 0.914 T 5=0.915 ( s ) Δ T 1= √ √ ∑ (T ¿ ¿ i1−T 1) ⅈ=1 Δ =0.003 T 2= ⅈ=1(s) ¿ √ ∑ (T ¿ ¿ i2−T ) ∑ (T ¿ ¿ i3−T ) Δ =0.003(s T 3= ⅈ=1 )¿ 5 √ ⅈ=1 Δ T =0.002(s)¿ 4= *Determination of the gravitational acceleration as a function of the inclination of the pendulum force: Pendulum with inclined oscillation plane:T =2 π √ √ ∑ (T ¿ ¿ i 4−T ) ∑ (T ¿ ¿i 5−T )2 ⅈ=1 Δ 5==0.002(s) ¿ ( ) l l 2π ( (s)→ g= m∕s ) g cos θ cos θ T l=0.140 ( m ) a.θ1=0o g1= ( ) ( ) l 2π 0.140 ×3.141 = =9.69(m ∕ s ) o cos θ T 0.755 cos O Δ g 1=g × √( ) √( ∆ T1 0.003 2 −2 =9.69 =0.077(m ∕ s ) T1 0.755 ) Then, g1=9.69± 0.077( m ∕ s 2) Hence g1=9.69± 0.077( m ∕ s ) b θ2=10o 19 = g2= ( ) ( ) l 2π 0.140 ×3.141 = =9.15(m ∕ s ) o cos θ T 0.783 cos 10 Δ g 2=g2 × √( ) √( ∆ T2 0.003 −2 =9.15 =0.07( m ∕ s 2) T2 0.783 ) Then, g2=9.15± 0.07 (m ∕ s 2) Hence g2=9.15± 0.07 (m ∕ s ) c θ3 =20 o ( ) ( ) l π 0.140 2× 3.141 g3= = =9.10(m ∕ s 2) o cos θ T 0.804 cos 20 Δ g 3=g3 × √( −2 ) √( ∆T3 0.002 =9.10 =0.04(m ∕ s 2) T3 0.804 ) Then, g3=9.10 ±0.04 (m ∕ s2 ) Hence g3=9.10 ±0.04 (m ∕ s2 ) d θ 4=3 o ( ) ( ) l π 0.140 2× 3.141 2 g4 = = =8.79(m ∕ s ) o cos θ4 T 0.852 cos 30 Δ g 4=g × √( −2 √( ) ∆T4 0.002 2 =8.79 =0.04 (m ∕ s ) T4 0.852 ) Then, g4 =8.79 ± 0.04(m ∕ s ) 20 ... Trials T (s) 1. 135 1. 136 1. 130 1. 134 1. 127 13 Solid sphere T (s) 2 .13 3 2 .13 7 2 .13 6 2 .13 5 2 .13 1 Trials II Calculating The rod 1. 1 The value of vibration period Ti T 1= ∑ = 2.6 31 (s) i =1 s.d △T =... Trial L1= 0.400 (m) L2= 0.500 (m) L3= 0.600 (m) T1 (s) 1. 295 1. 302 1. 292 1. 300 1. 303 T2 (s) 1. 478 1. 476 1. 467 1. 4 81 1.478 T3 (s) 1. 632 1. 637 1. 638 1. 638 1. 627 T 2 =1. 476 s T 3 =1. 634 s T 1= 1.298... t v ( )=v 1? ? Δv => ( ) s m m =0.5 41( ) t s s = 0.5 41? ? 0.007(m/s) '' ¿ V 1: - The uncertainty of t: t ''1= 10 ∑ t =¿ 0.677 ( s ) ¿ 10 i =1 i ⩟ t 1'' = √ 10 ∑ (t ''1 i−t1 )2 i =1 10? ?10 =0.0 01( s ) -The