⚙️ امتحان تجريبي: الطاقة، الشغل، القدرة، وكمية الحركة

📌 Idea: \(K=\tfrac12mv^2\) depends on \(v^2\) (speed squared), which is always positive or zero regardless of direction — so kinetic energy is a scalar that is never negative.

📌 Idea: since \(K\geq0\) always, any graph showing negative \(K\) values is physically impossible. Line D is the only one extending below the horizontal axis (negative values).

📌 Idea: the electron-volt (particle physics energy), kilowatt-hour (electrical consumption), and calorie (thermal/food energy) are all alternative energy units, each convertible to joules.

📌 Speed squared: \(v^2=5^2+3^2+(-2)^2=25+9+4=38\)

📌 Calculation: \(K=\tfrac12mv^2=\tfrac12(0.5)(38)=9.5\,J\)

📌 Formula: \(W_g=mg\Delta h=(3)(9.8)(10-4)=(3)(9.8)(6)=176.4\,J\)

📌 Sign: the body descends in the same direction as gravity, so the work is positive.

📌 Force balance (constant speed ⟹ zero acceleration): \(F_{applied}=mg\sin37°+F_K=(5)(9.8)(0.6018)+10\approx29.5+10=39.5\,N\)

📌 Power: \(P=F_{applied}\times v=39.5\times2\approx79\,W\)

📌 Conversion: \(1\,kW\cdot h=1000\,W\times3600\,s=3.6\times10^6\,J\)

📌 Calculation: \(5\times3.6\times10^6=1.8\times10^7\,J\)

📌 Work = area under the curve: positive triangle (0 to 4m, height 3N): \(\tfrac12(4)(3)=6\,J\). Negative triangle (4 to 6m, depth 3N): \(-\tfrac12(2)(3)=-3\,J\)

📌 Total: \(W=6+(-3)=3\,J\)

📌 Integration: \(W=\displaystyle\int_2^4(12x^2+8)\,dx=\Big[4x^3+8x\Big]_2^4\)

📌 Calculation: \(=(4(64)+8(4))-(4(8)+8(2))=(256+32)-(32+16)=288-48=240\,J\)

📌 Idea: for a conservative force (like gravity): work against it is path-independent (A), and work over a closed path is zero (C). B and D (rough surface with friction) are non-conservative properties.

📌 Idea: this is the precise mathematical definition: \(\oint \vec F\cdot d\vec l=0\) only for a conservative force. The other options are either reversed or inaccurate (both force types can change kinetic energy, and friction is the classic example of a non-conservative, not conservative, force).

📌 Conservation of mechanical energy: \(ME=K+U_g=8+(-12)=-4\,J\) (constant throughout, no friction)

📌 At the highest point: the ball is momentarily at rest (\(K=0\)), so \(U_{g,top}=ME-K=-4-0=-4\,J\)

📌 Energy conservation between release (v=0) and equilibrium (h=2m, v=3m/s): \(mgh=mg(2)+\tfrac12mv^2\)

📌 Calculation: \(h=2+\dfrac{v^2}{2g}=2+\dfrac{9}{2(9.81)}=2+0.459\approx2.46\,m\)

📌 Speed at the peak = horizontal component only: \(v_{0y}=30\sin45°\approx21.21\,m/s\) (matches the given 21.2, confirming the vertical component vanishes exactly at the peak)

📌 Calculation: \(0=v_{0y}^2-2gh \Rightarrow h=\dfrac{v_{0y}^2}{2g}=\dfrac{(21.21)^2}{2(1.63)}\approx138.2\,m\)

📌 Generalized work-energy theorem: \(mgh+W_{nc}=\tfrac12mv_f^2-0\), where \(W_{nc}=-2.3\,J\) (negative, energy lost to friction)

📌 Calculation: \((0.7)(9.81)(1.5)-2.3=10.3-2.3=8.0\,J=\tfrac12(0.7)v_f^2\)

📌 Speed: \(v_f=\sqrt{\dfrac{2(8.0)}{0.7}}\approx4.78\,m/s\)

📌 Idea: \(\dfrac{K}{p}=\dfrac{\tfrac12mv^2}{mv}=\dfrac{v}{2}\) — the ratio does not depend on mass at all!

📌 Calculation: \(\dfrac{v}{2}=\dfrac{10}{2}=5\,m/s\)

📌 Idea: \(K=\dfrac{p^2}{2m}\) — kinetic energy scales with the square of momentum (mass fixed).

📌 Calculation: \(p\to4p \Rightarrow K\to\dfrac{(4p)^2}{2m}=16\times\dfrac{p^2}{2m}=16K_1\)

📌 Conservation of momentum: in an isolated system (no external forces), the sum of momenta before sticking equals the single momentum afterward: \(\vec P_f=\vec P_1+\vec P_2+\vec P_3\).

📌 Vector addition: adding the horizontal and vertical components of the three vectors from the first figure gives a resultant pointing up and to the right at a steep angle (closer to vertical) — matching only option B among the four choices.

📌 Idea: momentum conservation requires an isolated system (net external force = zero). A net external force invalidates simple conservation of total momentum.

📌 A known special case: in an elastic collision between equal masses, one initially at rest, the moving ball stops completely (v=0) and the previously-at-rest ball moves off with exactly the original speed v — a "complete velocity exchange."

📌 Verification (figure C): momentum: \(mv+0=0+mv\) ✓. Kinetic energy: \(\tfrac12mv^2=0+\tfrac12mv^2\) ✓ (exactly conserved, confirming an elastic collision).

📌 Ruling out the rest: A increases kinetic energy (physically impossible), while B and D decrease it (inelastic collisions).

📌 Total KE before: \(15+18=33\,J\)

📌 Total KE after: \(20+13=33\,J\)

📌 Conclusion: total kinetic energy is exactly conserved (33J=33J) — precisely the definition of an elastic collision.

📌 Sign convention: taking right as positive: \(v_1=+4\,m/s\), \(v_2=-2\,m/s\)

📌 Elastic collision equations: \(v_{2f}=\dfrac{(m_2-m_1)v_2+2m_1v_1}{m_1+m_2}=\dfrac{(2)(-2)+2(3)(4)}{8}=\dfrac{-4+24}{8}=2.5\,m/s\)

📌 Kinetic energy of cube 2: \(K_{2i}=\tfrac12(5)(2)^2=10\,J\), \(K_{2f}=\tfrac12(5)(2.5)^2=15.625\,J\)

📌 Change: \(\Delta K_2=15.625-10=5.625\,J\)

📌 Formula: \(K=\tfrac12mv^2=\tfrac12(2)(36)=36\,J\)

📌 Work-energy theorem: \(W=\Delta K=\tfrac12mv^2-0\)

📌 Calculation: \(v=\sqrt{\dfrac{2W}{m}}=\sqrt{\dfrac{2(50)}{4}}=\sqrt{25}=5\,m/s\)

📌 Formula: \(P=\dfrac{W}{t}=\dfrac{600}{5}=120\,W\)

📌 Formula: \(P=Fv\cos\theta=Fv\cos(0°)=Fv=(20)(3)=60\,W\)

📌 Energy conservation: \(mgh=\tfrac12mv^2 \Rightarrow v=\sqrt{2gh}\)

📌 Calculation: \(v=\sqrt{2(9.8)(5)}=\sqrt{98}\approx9.90\,m/s\)

📌 Formula: \(U_s=\tfrac12kx^2=\tfrac12(200)(0.1)^2=\tfrac12(200)(0.01)=1\,J\)

📌 Energy conservation (frictionless, independent of track shape): \(mgh=\tfrac12mv^2 \Rightarrow v=\sqrt{2gh}\)

📌 Calculation: \(v=\sqrt{2(9.8)(2)}=\sqrt{39.2}\approx6.26\,m/s\)

📌 Generalized work-energy theorem: \(mgh+W_{friction}=\tfrac12mv^2\)

📌 Calculation: \((4)(9.8)(3)+(-15)=117.6-15=102.6\,J=\tfrac12(4)v^2\)

📌 Speed: \(v=\sqrt{\dfrac{2(102.6)}{4}}=\sqrt{51.3}\approx7.16\,m/s\)

📌 Idea: \(P=\dfrac{dW}{dt}\) (or \(P=Fv\) for constant force) — power measures how fast work is done, not just its total amount.

📌 Formula: \(p=mv=(1500)(20)=30000\,kg\cdot m/s\)

📌 Impulse-momentum theorem: \(F\Delta t=\Delta p=mv_f-0\)

📌 Calculation: \(v_f=\dfrac{F\Delta t}{m}=\dfrac{(500)(0.2)}{10}=\dfrac{100}{10}=10\,m/s\)

📌 Momentum conservation: \(m_1v_1+m_2v_2=(m_1+m_2)v_f\)

📌 Calculation: \((2)(6)+(3)(0)=(5)v_f \Rightarrow 12=5v_f \Rightarrow v_f=2.4\,m/s\)

📌 Energy before: \(K_i=\tfrac12(2)(6)^2+0=36\,J\)

📌 Energy after (\(v_f=2.4\,m/s\) from the previous question): \(K_f=\tfrac12(5)(2.4)^2=\tfrac12(5)(5.76)=14.4\,J\)

📌 Change: \(\Delta K=14.4-36=-21.6\,J\) (lost as heat, sound, and deformation during sticking)

📌 Idea: momentum is conserved in all collision types (isolated system), but total kinetic energy is conserved only in an elastic collision specifically (partially lost in inelastic, maximally lost in perfectly inelastic).

📌 Elastic collision equations: \(v_{1f}=\dfrac{(m_1-m_2)v_1}{m_1+m_2}=\dfrac{(2)(5)}{6}\approx1.67\,m/s\)

📌 For the second body: \(v_{2f}=\dfrac{2m_1v_1}{m_1+m_2}=\dfrac{(8)(5)}{6}\approx6.67\,m/s\)

📌 Check (momentum): \(4(5)=20\); \(4(1.67)+2(6.67)=6.68+13.34=20.02\approx20\) ✓

📌 Check (energy): \(\tfrac12(4)(25)=50\,J\); \(\tfrac12(4)(1.67)^2+\tfrac12(2)(6.67)^2\approx5.58+44.49\approx50.07\approx50\,J\) ✓

📌 (a) Work by applied force (only horizontal component does work with horizontal displacement): \(W_F=F_a\cos35°\times d=(23)(0.8192)(6)\approx113.05\,J\)

📌 Work by friction: \(W_f=-F_K\times d=-(12)(6)=-72\,J\)

📌 Net work: \(W_{net}=113.05-72\approx41.05\,J\)

📌 (b) Work-energy theorem: \(v=\sqrt{\dfrac{2W_{net}}{m}}=\sqrt{\dfrac{2(41.05)}{6}}\approx3.70\,m/s\)

📌 (a) Change in kinetic energy: \(\Delta K=\tfrac12m(v_f^2-v_0^2)=\tfrac12(10)(64-25)=\tfrac12(10)(39)=195\,J\)

📌 (b) Work-energy theorem: \(W_{net}=\Delta K \Rightarrow (F_a\cos60°-F_K)\,d=\Delta K\)

📌 Calculation: \((30(0.5)-6)\,d=195 \Rightarrow (15-6)\,d=195 \Rightarrow 9d=195 \Rightarrow d\approx21.67\,m\)

📌 Extension: \(x=L_2-L_1=55-50=5\,cm=0.05\,m\)

📌 (a) Force balance at equilibrium: \(kx=mg \Rightarrow k=\dfrac{mg}{x}=\dfrac{(0.6)(9.81)}{0.05}\approx117.7\,N/m\)

📌 (b) Elastic PE: \(U_s=\tfrac12kx^2=\tfrac12(117.7)(0.05)^2\approx0.147\,J\)

📌 (a) Equivalent constant (parallel): \(k_{eq}=k_1+k_2=400+600=1000\,N/m\)

📌 (b) Total displacement at equilibrium: \(x=\dfrac{mg}{k_{eq}}=\dfrac{(10)(9.81)}{1000}\approx0.0981\,m\)

📌 (c) PE in each spring: \(U_1=\tfrac12k_1x^2=\tfrac12(400)(0.0981)^2\approx1.925\,J\), \(U_2=\tfrac12k_2x^2=\tfrac12(600)(0.0981)^2\approx2.887\,J\)

📌 Total PE: \(U_{total}=U_1+U_2\approx4.812\,J\) (matches \(\tfrac12k_{eq}x^2=\tfrac12(1000)(0.0981)^2\approx4.812\,J\) ✓)

📌 Conservation of mechanical energy (from A to the point of maximum compression): \(\tfrac12mv_A^2+mgh_A=mgh_B+\tfrac12kx^2\)

📌 Calculation: \(\tfrac12(10)(2)^2+(10)(9.81)(0.8)=(10)(9.81)(0.4)+\tfrac12k(0.2)^2\)

📌 Simplifying: \(20+78.48=39.24+0.02k \Rightarrow 98.48-39.24=0.02k \Rightarrow 59.24=0.02k\)

📌 Result: \(k\approx2962\,N/m\)

📌 (a) Proof: since the track is frictionless, mechanical energy \(ME=K+U_g\) is conserved throughout. At any two points of equal height h, \(U_g\) is the same at both, and since \(ME\) is constant, \(K\) must also be equal — and since mass is constant, the speed magnitude is equal at both points, regardless of the track's shape between them.

📌 (b) Change in PE: \(\Delta U_g=mg(h_f-h_i)=(0.3)(9.81)(0.2-0.6)\approx-1.177\,J\) (a decrease, since the ball descended)

📌 (c) Mechanical energy (conserved, computed from the release point where \(K_i=0\)): \(ME=mgh_i=(0.3)(9.81)(0.6)\approx1.766\,J\)

📌 (d) At maximum compression (ball at rest, all energy is spring PE plus gravitational PE at \(h_f\)): \(ME=mgh_f+\tfrac12kx^2 \Rightarrow 1.766=(0.3)(9.81)(0.2)+\tfrac12(100)x^2\)

📌 Calculation: \(1.766-0.589=1.177=50x^2 \Rightarrow x^2=0.02354 \Rightarrow x\approx0.153\,m\)

📌 Note on the original text: the contact duration was stated as "1.2 seconds" in the source, but this is far too long to be physically realistic for a racket-ball contact (real durations are milliseconds) — \(t=1.2\,ms=1.2\times10^{-3}\,s\) is used here as the realistic value.

📌 (a) Sign convention (right positive): \(v_i=+15\,m/s\), \(v_f=-20\,m/s\)

📌 Change in momentum: \(\Delta p=m(v_f-v_i)=(0.2)(-20-15)=(0.2)(-35)=-7\,kg\cdot m/s\)

📌 (b) Average force: \(F_{avg}=\dfrac{\Delta p}{\Delta t}=\dfrac{-7}{1.2\times10^{-3}}\approx-5833\,N\)

📌 (a) Change in momentum: \(\Delta P=P_f-P_i=100-80=20\,kg\cdot m/s\)

📌 (b) Final speed: \(v_f=\dfrac{P_f}{m}=\dfrac{100}{8}=12.5\,m/s\)

📌 (c) Average force: \(F_{avg}=\dfrac{\Delta P}{\Delta t}=\dfrac{20}{2}=10\,N\)