Tiêu đề 4. P1C4 (Newtonian Mechanics) With Solve.pdf ✅

wbDUwbqvb ejwe` ̈v  Varsity Practice Sheet....................................................................................................................... 1 weMZ mv‡j DU-G Avmv cÖkœvejx 1. m GKwU 2m f‡ii `ywU AvqZvKvi ev· GKwU Nl©Ynxb Abyf‚wgK c„‡ô GKwU `wo Øviv mshy3| F gvÎvi GKwU m¤§yL ej Øviv fvix ev·wU‡K Wvbw`‡K Uvbv n‡”Q| d‡j, nvjKv ev·wU `wo Øviv Uvb Abyfe K‡i| `wowU‡Z Uvb KZ? [DU 23-24] F F 6 F 3 F 2 DËi: F 3 e ̈vL ̈v: anet = F 3m  T = manet = mF 3m = F 3 T F m 2m 2. 0.50 kg f‡ii GKwU KYv X Aÿ eivei x(t) = – 13.00 + 2.00t + 4.0022t2 – 3.005t3 mgxKiY Abyhvqx Pj‡Q, †hLv‡b x Gi GKK wgUvi Ges t Gi GKK †m‡KÛ| t = 2.0 s G KYvi Ici jwä ej KZ? [DU 22-23] – 28 N – 14 N 8 N 36 N DËi: – 14 N e ̈vL ̈v: †eM, v = dx dt = d dt (– 13.00 + 2.00t + 4.0022t2 – 3.005t3 ) = 2 + 8t – 9t2 Avevi, Z¡iY, a = dv dt = d dt (2 + 8t – 9t2 ) = 8 – 18t GLb, F = ma = 0.50  (8 – 18t)  F(t = 2s) = 0.50  (8 – 18  2) = – 14 N 3. w ̄’i Ae ̄’vq _vKv GKwU e ̄‘ we‡ùvwiZ n‡q m1 I m2 f‡ii `yBwU e ̄‘‡Z cwiYZ n‡q h_vμ‡g v1 I v2 †e‡M wecixZ w`‡K Pjgvb| v1 v2 Gi AbycvZ KZ? [DU 19-20, 17-18] m1 m2 – m1 m2 m2 m1 m2 m1 DËi: m2 m1 e ̈vL ̈v: m1u1 + m2u2 = m1v1 + m2v2  m1  0 + m2  0 = m1v1 + m2  (–v2)  0 = m1v1 – m2v2  m1v1 = m2v2  v1 v2 = m2 m1 4. 10 kg f‡ii GKwU e ̄‘i Dci 2F gv‡bi ej cÖ‡qvM Kivi d‡j e ̄‘wUi Z¡iY nq 60 ms–2 | M f‡ii GKwU e ̄‘i Dci 5F gv‡bi ej cÖ‡qvM Kivi d‡j hw` e ̄‘wUi Z¡iY 50 ms–2 nq, Z‡e fi M KZ? [DU 18-19] 3.3 kg 4.8 kg 21 kg 30 kg DËi: 30 kg e ̈vL ̈v: cÖ_‡g †ÿ‡Î, F1 = m1  a1  2F = 10  60  F = 300 N Avevi, wØZxq †ÿ‡Î, F2 = m2  a2  5F = M  50  5  300 = M  50  M = 30 kg 5. 5.0 N Gi GKwU Avbyf~wgK ej GKwU 0.50 kg f‡ii AvqZvKvi e ̄‘‡K GKwU Djø¤^ †`Iqv‡j av°v w`‡”Q| e ̄‘wU Avw`‡Z w ̄’i wQj| hw` w ̄’wZ I MZxq Nl©Y ̧Yv1⁄4 h_vμ‡g s = 0.6 Ges k = 0.8 nq, Z‡e ms–2 GK‡K e ̄‘wUi Z¡iY KZ? [DU 18-19] 1.8 2.0 6.0 8.0 DËi: 1.8 e ̈vL ̈v: Fk = kR = 0.8  5 = 4 N R = 5 N F = 5 N W Nl©‡Yi d‡j g›`b, a = 4 0.5 = 8 ms–2  MwZkxj Ae ̄’vq Z¡iY, a = g – a = 9.8 – 8 = 1.8 ms–2 6. †KŠwYK fi‡e‡Mi GKK †KvbwU? [DU 18-19] kgm2 s –1 kgms–2 kgms–1 km2 s –2 DËi: kgm2 s –1 e ̈vL ̈v: †KŠwYK fi‡eM = ˆiwLK fi‡eM  e ̈vmva© = mvr  GKK  kgm2 s –1