פתרון בגרות מכניקה — שאלון 36361

x_0,1 = 0 m ; x_0,2 = 30 m ; 0 ≤ t ≤ 15 s
v₁ (0 ≤ t ≤ 6 s) = 25 m/s ; v₁ (15 s) = 0
v₂ (0) = 0 ; v₂ (15 s) = 33 m/s

0.00 s

0.00 s

25.00 m/s

0.00 m/s

0.0 m

30.0 m

0.0 m

0.00.00 = — m/s 0.00.00 = — m/s

a = ΔvΔt

a₂ = 33 − 015 − 0 = 3315 = 2.2 m/s²

v₂ (15) = 0 + 2.2 · 15 = 33 m/s

a = v₂ − v₁t₂ − t₁

a₁ = 0 − 2515 − 6 = −259 = −2.78 m/s²

v₁ (15) = 25 − 259 · (15 − 6) = 25 − 25 = 0

x (t) = x₀ + A

A₁ = 25 · 6 = 150 m A₂ = 12 · 9 · 25 = 112.5 m x₁ (15) = 0 + 150 + 112.5 = 262.5 m

A = 12 · 15 · 33 = 247.5 m x₂ (15) = 30 + 247.5 = 277.5 m

Δx = 277.5 − 262.5 = 15 m

x₂ = 30 + 12 · 2.2 · 15² = 30 + 1.1 · 225 = 277.5 m 25 + 02 = 12.5 m/s ; Δx = 12.5 · 9 = 112.5 m x₁ (15) = 150 + 112.5 = 262.5 m

x₂ − x₁ = 30 − 0 = 30 m

x₁ (6) = 25 · 6 = 150 m x₂ (6) = 30 + 1.1 · 6² = 30 + 39.6 = 69.6 m x₁ − x₂ = 150 − 69.6 = 80.4 m

x₂ − x₁ = 277.5 − 262.5 = 15 m

25t = 30 + 1.1t² ; 1.1t² − 25t + 30 = 0 t = 25 − √4932.2 = 1.27 s ; x = 25 · 1.271 = 31.8 m 112u² − 531u − 3618 = 0 ; u = 8.53 s ; t = 6 + 8.53 = 14.53 s x = 30 + 1.1 · 14.529² = 262.2 m

v = ΔxΔt

v₂ = 247.515 = 16.5 m/s v₁ = 262.515 = 17.5 m/s

v₂ = 0 + 332 = 16.5 m/s

v₂ (6) = 2.2 · 6 = 13.2 m/s

v₂ (t) = 2.2t v₁ (t) = 25 − 259 · (t − 6)

2.2t = 25 − 259 · (t − 6) 19.8t = 225 − 25 (t − 6) = 375 − 25t 44.8t = 375 ; t = 37544.8 = 8.37 s v = 2.2 · 8.3705 = 18.4 m/s

v = 25 − 259 · (8.3705 − 6) = 25 − 6.585 = 18.415 m/s

A · 0 – 2 s · 50 kg | B · 2 – 4 s · 52.5 kg | C · 4 – 10 s · 50 kg
D · 10 – 12 s · 47.5 kg | E · 12 – 14 s · 50 kg
g = 10 m/s² ; m = 50 kg ; m·g = 500 N

מקטע A · מנוחה

0.00 s

50 kg

0.5 m/s²

תרחיש הבגרות · מסה 50 kg · תאוצה 0.5 m/s² · עלייה

0.00

A

50.0

500

0

0.0

N = m·(g + a) ΣF = N − m·g a =N − m·gm a_B =525 − 50050= 0.5 m/s² a_D =475 − 50050= −0.5 m/s²

N − m·g = 0 ; N = m·g m = 50 kg ; m·g = 50 · 10 = 500 N

ΣF = m·a ; N − m·g = m·a a =N − m·gm

N = 50 · 10 = 500 N a =500 − 50050= 0 m/s²

N_B = 52.5 · 10 = 525 N ; ΣF_B = 525 − 500 = +25 N a_B =525 − 50050=2550= 0.5 m/s²

N_D = 47.5 · 10 = 475 N ; ΣF_D = 475 − 500 = −25 N a_D =475 − 50050=−2550= −0.5 m/s²

Nm·g= 1 +ag a_B = 10 · (1.05 − 1) = 0.5 m/s² ; a_D = 10 · (0.95 − 1) = −0.5 m/s²

v = v₀ + a·t = 0 + 0.5 · 2 = +1.0 m/s

Δv_B = 0.5 · 2 = +1.0 m/s ; Δv_D = −0.5 · 2 = −1.0 m/s Δv_B + Δv_D = +1.0 − 1.0 = 0

0 ≤ t ≤ 2 : v = 0 ; 2 ≤ t ≤ 4 : v = 0.5 · (t − 2) 4 ≤ t ≤ 10 : v = 1.0 m/s ; 10 ≤ t ≤ 12 : v = 1.0 − 0.5 · (t − 10) 12 ≤ t ≤ 14 : v = 0

B:12· 2 · 1.0 = 1.0 m ; C: 1.0 · 6 = 6.0 m ; D:12· 2 · 1.0 = 1.0 m 1.0 + 6.0 + 1.0 = 8.0 m

x_B =12· 0.5 · 2² = 1.0 m x_C = 1.0 · 6 = 6.0 m x_D = 1.0 · 2 −12· 0.5 · 2² = 2.0 − 1.0 = 1.0 m

n         | 1    | 2    | 3    | 4    | 5
T (s)    | 0.370 | 0.275 | 0.210 | 0.175 | 0.160
ω (1/s) | ?     | ?     | ?     | ?     | ?
ω² (1/s²) | ?     | ?     | ?     | ?     | ?
m_A = 0.02 kg ; M = 0.1 kg ; g = 10 m/s²

3

0.163 m

0.020 kg

0.10 kg

3.823 °

0.1626 m

3.000 N

2.9933 N

30.336 1/s

920.2 1/s²

0.2071 s

4.9337 m/s

306.75 1/s²

0

0.000 s

T = n·M·g R = L·cos θ F_c = n·M·g·cos θ ω² = M·gm_A·L · n

cos θ = RL

F_t − n·M·g = 0 ; F_t = n·M·g

F_c = F_t·cos θ = n·M·g·cos θ

m_A·ω²·R = m_A·ω²·L·cos θ m_A·ω²·L·cos θ = n·M·g·cos θ

ω² = M·gm_A·L · n 4π²T² = M·gm_A·L · n

ω = 2πT ; ω² = 4π²T² ω = 6.28320.370 = 16.98 1/s ω² = 16.98² = 288 1/s²

n = 1 ; T = 0.370 s ; ω = 16.98 ; ω² = 288 n = 2 ; T = 0.275 s ; ω = 22.85 ; ω² = 522 n = 3 ; T = 0.210 s ; ω = 29.92 ; ω² = 895 n = 4 ; T = 0.175 s ; ω = 35.90 ; ω² = 1289 n = 5 ; T = 0.160 s ; ω = 39.27 ; ω² = 1542

ω² = 39.4780.160² = 39.4780.0256 = 1542 1/s²

k = M·gm_A·L ; L = M·gm_A·k

k = 307.0 1/s² L = 0.1 · 100.02 · 307.0 = 16.14 = 0.163 m

k = 1542 − 2885 − 1 = 12544 = 313.4 1/s² L = 10.02 · 313.4 = 0.160 m

m = 2 kg ; h = 1.25 m ; BD = 4 m ; g = 10 m/s²
F(0) = 0 ; F(4 m) = 8 N ; F(x) = 2·x [N]
x = 0 (B) ; x = 4 m (D) ; v₀ = 6 m/s

6.0 m/s

−3.4000 m

0.0000 m

6.0000 m/s

0.0000 N

0.0000 m/s²

36.0000 J

0.0000 J

36.0000 J

0.0000 J

0.0000 s

v_min = √2·g·h F(x) = 2·x W_F = − 12 ·d·F(d) = −d² a(x) = F(x)m = x

12·m·v_min² + 0 = 0 + m·g·h v_min = √2·g·h = √2 · 10 · 1.25 = √25 = 5 m/s

W = −m·g·h = −2 · 10 · 1.25 = −25 J ΔE_k = 0 − 12·2·v_min² = −v_min² = −25 ; v_min = 5 m/s

E_k(A) = 12·2·6² = 36 J E_p(B) = m·g·h = 2 · 10 · 1.25 = 25 J E_k(B) = 36 − 25 = 11 J v_B = √2 · 112 = √11 ≈ 3.32 m/s

W_F = E_k(C) − E_k(B) = 0 − 11 = −11 J

E_m(A) = 36 + 0 = 36 J E_m(C) = 0 + 25 = 25 J W_F = 25 − 36 = −11 J

k = 8 − 04 − 0 = 2 N/m ; F(x) = 2·x

|W_F| = 12·d·F(d) = 12·d·2·d = d² −d² = −11 ; d = √11 ≈ 3.32 m

W_F = −∫ 2·x dx = −x² ; −d² = −11 F_avg = 0 + 2·d2 = d ; |W| = d·d = d² = 11

a(x) = F(x)m = 2·x2 = x a_C = √11 ≈ 3.32 m/s² ; a_B = 0

W_F = +d² = +11 J ; v_B = √11 ≈ 3.32 m/s

v(x) = √11 − x² ; v(√11) = 0 ; v(0) = √11

m₁ = 60 gr = 0.06 kg · m₂ = 30 gr = 0.03 kg · v₁ = 0.3 m/s · v₂ = 0
m₁·v₁ = m₁·u₁ + m₂·u₂ 12m₁·v₁² = 12m₁·u₁² + 12m₂·u₂²

הדיסק הפוגע ממשיך קדימה בכיוון החיובי

u₂ גדולה מן המהירות v₁ — הדיסק הנייח קל מן הפוגע

הפעל את המתג של תרשים II ולחץ על עקומה A או על עקומה C כדי לראות מדוע היא נפסלת

60 g

30 g

0.30 m/s

×1.0

0.100 m/s

0.400 m/s

0.0180 kg·m/s

2.700 mJ

0.300 mJ

2.400 mJ

11.1 %

88.9 %

0.0120 N·s

1.200 N

התנע נשמר והאנרגיה הקינטית נשמרת — ההתנגשות אלסטית

u₁ = (m₁ − m₂)·v₁m₁ + m₂ u₂ = 2·m₁·v₁m₁ + m₂

m₁ = 60 gr = 0.06 kg ; m₂ = 30 gr = 0.03 kg ; v₁ = 0.3 m/s ; v₂ = 0

m₁·v₁ = m₁·u₁ + m₂·u₂ 0.06 · 0.3 = 0.06·u₁ + 0.03·u₂ = 0.018 12m₁·v₁² = 12m₁·u₁² + 12m₂·u₂² 0.0027 = 0.03·u₁² + 0.015·u₂²

m₁·(v₁ − u₁) = m₂·u₂ m₁·(v₁ − u₁)·(v₁ + u₁) = m₂·u₂² v₁ + u₁ = u₂ ⇒ u₂ − u₁ = v₁ m₁·(2v₁ − u₂) = m₂·u₂ ⇒ 2·m₁·v₁ = (m₁ + m₂)·u₂ u₂ = 2·m₁·v₁m₁ + m₂ ; u₁ = u₂ − v₁ = (m₁ − m₂)·v₁m₁ + m₂

u₁ = (0.06 − 0.03) · 0.30.06 + 0.03 = 0.0090.09 = 0.1 m/s

u₂ = 2 · 0.06 · 0.30.06 + 0.03 = 0.0360.09 = 0.4 m/s

0.06 · 0.3 = 0.018 ; 0.06 · 0.1 + 0.03 · 0.4 = 0.006 + 0.012 = 0.018 kg·m/s 0.0027 J ; 0.0003 + 0.0024 = 0.0027 J u₂ − u₁ = 0.4 − 0.1 = 0.3 = v₁

m₁·v₁ = m₁·u₁ + m₂·u₂ ⇒ m₁·(v₁ − u₁) = m₂·u₂ m₁·v₁² = m₁·u₁² + m₂·u₂² ⇒ m₁·(v₁ − u₁)·(v₁ + u₁) = m₂·u₂²

v₁ + u₁ = u₂ ⇒ u₁ = u₂ − v₁ m₁·(v₁ − u₂ + v₁) = m₂·u₂ ⇒ 2·m₁·v₁ = (m₁ + m₂)·u₂ u₂ = 2·m₁·v₁m₁ + m₂ u₁ = (m₁ − m₂)·v₁m₁ + m₂

u₂ = v₁ · 2m₁m₁ + m₂ u₂ < v₁ ⟺ 2m₁m₁ + m₂ < 1

2m₁ < m₁ + m₂ ⟺ m₁ < m₂

v₁ − u₂ = v₁ · m₁ + m₂ − 2m₁m₁ + m₂ = v₁ · m₂ − m₁m₁ + m₂

|F₁₂(t)| = |F₂₁(t)| ∀t

a₁ = Fm₁ ; a₂ = Fm₂

Δp₁ = 0.06 · (0.1 − 0.3) = −0.012 kg·m/s Δp₂ = 0.03 · 0.4 = +0.012 kg·m/s

G = 6.67·10⁻¹¹ N·m²/kg² · M_E = 5.98·10²⁴ kg · R_E = 6.37·10⁶ m
G·M_E = 3.98866·10¹⁴ m³/s² r³ = G·M_E·T²4π²

גיאוסטציונרי — הלוויין נעול מעל הנקודה A

בסקאלה לוגריתמית כפולה מתקבל קו ישר ששיפועו שני שלישים — זהו החוק השלישי של קפלר

r³ = G·M_E·T²4π² a = G·M_Er² v = 2π·rT

24.0 h

14.0 h

×3000

86400 s

42250.5 km

35880.5 km

0.2234 m/s²

3072.5 m/s

7.2722 ·10⁻⁵ rad/s

463.2 m/s

7.2722 ·10⁻⁵ rad/s

6.633

50400 s

29496.9 km

23126.9 km

3677.3 m/s

0.4584 m/s²

1.0103 ·10¹³ m³/s²

9.830 m/s²

T = 2πω T = 24 h = 24 · 3600 = 86400 s

G·M_E·mr² = m·v²r

v = 2π·rT ; G·M_Er² = 4π²·rT² r³ = G·M_E·T²4π²

G·M_E = 6.67·10⁻¹¹ · 5.98·10²⁴ = 3.9887·10¹⁴ m³/s² T² = (86400)² = 7.46496·10⁹ s² ; 4π² = 39.478 r³ = 3.9887·10¹⁴ · 7.46496·10⁹39.478 = 7.5421·10²² m³ r = 4.225·10⁷ m

h = r − R_E = 4.2250·10⁷ − 0.637·10⁷ = 3.588·10⁷ m

T = 2π·√r³G·M_E = 2π · 13751 = 86400 s

m·a = G·M_E·mr² ; a = G·M_Er²

r² = (4.225·10⁷)² = 1.7851·10¹⁵ m² a = 3.9887·10¹⁴1.7851·10¹⁵ = 0.223 m/s²

ω = 2π86400 = 7.2722·10⁻⁵ rad/s ; a = ω²·r = 0.2234 m/s² v = 2π·4.225·10⁷86400 = 3073 m/s ; a = v²r = 0.2234 m/s²

g = 3.9887·10¹⁴(6.37·10⁶)² = 9.83 m/s² ; 9.830.2234 = 44.0 = r²R_E²

T₂ = 14 h = 14 · 3600 = 50400 s

r₁³T₁² = r₂³T₂² ; r₂ = r₁ · T₂T₁^2/3 5040086400 = 1424 = 0.58333 ; 0.58333^2/3 = 0.69814 r₂ = 4.225·10⁷ · 0.69814 = 2.9497·10⁷ m

h₂ = 2.9497·10⁷ − 0.637·10⁷ = 2.313·10⁷ m

r₂³ = 3.9887·10¹⁴ · 2.54016·10⁹39.478 = 2.5664·10²² m³ ; r₂ = 2.9497·10⁷ m v₂ = 3677 m/s ; v₁ = 3073 m/s

v = ω·r ; v_A = 7.2722·10⁻⁵ · 6.37·10⁶ = 463 m/s v = 7.2722·10⁻⁵ · 4.225·10⁷ = 3073 m/s ; 3073463 = 6.63 = rR_E

ω_A = ω = 2π86400 = 7.27·10⁻⁵ rad/s

פתרון בגרות · מכניקה