1. Problem: A particle is projected vertically upward from point O with speed $u$ m/s. The greatest height reached is 62.5 m. 1.a Find $u$. Step 1: Use the formula for maximum height under gravity: $$h = \frac{u^2}{2g}$$ where $g = 9.8$ m/s². Step 2: Substitute $h = 62.5$ m: $$62.5 = \frac{u^2}{2 \times 9.8}$$ Step 3: Solve for $u^2$: $$u^2 = 62.5 \times 2 \times 9.8 = 1225$$ Step 4: Find $u$: $$u = \sqrt{1225} = 35 \text{ m/s}$$ Answer: $u = 35$ m/s. 1.b Find total time for which particle is 50 m or more above O. Step 1: Use height formula at time $t$: $$s = ut - \frac{1}{2}gt^2$$ Step 2: Set $s = 50$ m: $$50 = 35t - 4.9t^2$$ Step 3: Rearrange: $$4.9t^2 - 35t + 50 = 0$$ Step 4: Solve quadratic: $$t = \frac{35 \pm \sqrt{35^2 - 4 \times 4.9 \times 50}}{2 \times 4.9} = \frac{35 \pm \sqrt{1225 - 980}}{9.8} = \frac{35 \pm \sqrt{245}}{9.8}$$ Step 5: Calculate roots: $$t_1 = \frac{35 - 15.65}{9.8} = 2.0 \text{ s}, \quad t_2 = \frac{35 + 15.65}{9.8} = 5.2 \text{ s}$$ Step 6: Time above 50 m is $t_2 - t_1 = 5.2 - 2.0 = 3.2$ s. Answer: $3.2$ s. 2. Problem: Find coefficient of friction $mu$ for a 10 kg mass with given forces. Step 1: Weight $W = mg = 10 \times 9.8 = 98$ N. Step 2: Given normal reaction $R = 118$ N. Step 3: Friction force $F$ acts horizontally; friction force $F = \mu R$. Step 4: The force $20\sqrt{2}$ N acts at 45° to horizontal, so horizontal component: $$F_H = 20\sqrt{2} \times \cos 45^\circ = 20\sqrt{2} \times \frac{\sqrt{2}}{2} = 20 \text{ N}$$ Step 5: Since acceleration is along the plane, friction force $F = 20$ N. Step 6: Calculate coefficient of friction: $$\mu = \frac{F}{R} = \frac{20}{118} \approx 0.17$$ Note: The problem states $mu=0.13$; possibly friction force $F$ is less or other forces considered. 3. Problem: Stone projected vertically upwards from O, point A is 3.6 m above O, speed at A is 11.2 m/s upwards. 3.a Find maximum height above O. Step 1: Use energy or kinematic equation: $$v^2 = u^2 - 2gs$$ where $v=0$ at max height, $u=11.2$ m/s at A, $s$ is height from A to max height. Step 2: Solve for $s$: $$0 = 11.2^2 - 2 \times 9.8 \times s \Rightarrow s = \frac{11.2^2}{2 \times 9.8} = \frac{125.44}{19.6} = 6.4 \text{ m}$$ Step 3: Total max height above O: $$3.6 + 6.4 = 10 \text{ m}$$ Answer: 10 m. 3.b Find total time from projection to return to O. Step 1: Find initial speed $u$ from O to A using: $$v^2 = u^2 - 2gs$$ $$11.2^2 = u^2 - 2 \times 9.8 \times 3.6$$ Step 2: Calculate: $$125.44 = u^2 - 70.56 \Rightarrow u^2 = 196 \Rightarrow u = 14 \text{ m/s}$$ Step 3: Total time to max height: $$t_{up} = \frac{u}{g} = \frac{14}{9.8} = 1.43 \text{ s}$$ Step 4: Total time to return: $$t = 2 \times t_{up} = 2.86 \text{ s}$$ Answer: 2.86 s. 3.c Sketch velocity-time graph: velocity decreases linearly from $14$ m/s to $-14$ m/s over $2.86$ s. 4. Problem: Lorry moves with constant acceleration, passes A at speed $u$ m/s, 10 s later passes B at 34 m/s, distance AB = 240 m. 4.a Find $u$. Step 1: Use $v = u + at$: $$34 = u + 10a$$ Step 2: Use $s = ut + \frac{1}{2}at^2$: $$240 = 10u + 50a$$ Step 3: From first equation: $$a = \frac{34 - u}{10}$$ Step 4: Substitute into second: $$240 = 10u + 50 \times \frac{34 - u}{10} = 10u + 5(34 - u) = 10u + 170 - 5u = 5u + 170$$ Step 5: Solve for $u$: $$5u = 70 \Rightarrow u = 14 \text{ m/s}$$ Answer: $u = 14$ m/s. 4.b Time to midpoint of AB (120 m from A). Step 1: Use $s = ut + \frac{1}{2}at^2$ with $s=120$ m. Step 2: Substitute $a = \frac{34 - 14}{10} = 2$ m/s²: $$120 = 14t + \frac{1}{2} \times 2 \times t^2 = 14t + t^2$$ Step 3: Rearrange: $$t^2 + 14t - 120 = 0$$ Step 4: Solve quadratic: $$t = \frac{-14 \pm \sqrt{14^2 + 4 \times 120}}{2} = \frac{-14 \pm \sqrt{196 + 480}}{2} = \frac{-14 \pm \sqrt{676}}{2}$$ Step 5: Calculate roots: $$t = \frac{-14 + 26}{2} = 6 \text{ s}$$ Answer: 6 s. 5. Problem: Pebble projected upward at 21 m/s from 32 m above ground. 5.a Find speed when it strikes ground. Step 1: Use energy or kinematic equation: $$v^2 = u^2 + 2gs$$ where $s=32$ m downward. Step 2: Calculate: $$v^2 = 21^2 + 2 \times 9.8 \times 32 = 441 + 627.2 = 1068.2$$ Step 3: Find $v$: $$v = \sqrt{1068.2} \approx 32.7 \text{ m/s}$$ Answer: 33 m/s (approx). 5.b Time pebble is more than 40 m above ground. Step 1: Height above ground is $s = 40$ m. Step 2: Height from projection point is $h = 40 - 32 = 8$ m. Step 3: Use $s = ut - \frac{1}{2}gt^2$: $$8 = 21t - 4.9t^2$$ Step 4: Rearrange: $$4.9t^2 - 21t + 8 = 0$$ Step 5: Solve quadratic: $$t = \frac{21 \pm \sqrt{441 - 156.8}}{9.8} = \frac{21 \pm \sqrt{284.2}}{9.8}$$ Step 6: Calculate roots: $$t_1 = \frac{21 - 16.86}{9.8} = 0.44 \text{ s}, \quad t_2 = \frac{21 + 16.86}{9.8} = 3.87 \text{ s}$$ Step 7: Time above 40 m is $3.87 - 0.44 = 3.43$ s. Answer: 3.4 s. 5.c Sketch velocity-time graph: velocity decreases linearly from 21 m/s to negative value at impact. 6. Problem: Forces $(2ai + 2bj)$ N, $(-5bi + 3aj)$ N, and $(-11i - 7j)$ N act on object in equilibrium. Step 1: Sum forces in $i$ and $j$ directions must be zero: $$2a - 5b - 11 = 0$$ $$2b + 3a - 7 = 0$$ Step 2: Solve system: From first: $$2a - 5b = 11$$ From second: $$3a + 2b = 7$$ Step 3: Multiply second by 5 and first by 2: $$10a - 25b = 55$$ $$15a + 10b = 35$$ Step 4: Multiply first by 2 and second by 5: $$4a - 10b = 22$$ $$15a + 10b = 35$$ Step 5: Add: $$19a = 57 \Rightarrow a = 3$$ Step 6: Substitute $a=3$ into second original: $$3(3) + 2b = 7 \Rightarrow 9 + 2b = 7 \Rightarrow 2b = -2 \Rightarrow b = -1$$ Answer: $a=3$, $b=-1$. 7. Problem: Forces $F_1 = (-3i + 7j)$ N, $F_2 = (i - j)$ N, $F_3 = (pi + qj)$ N on particle in equilibrium. 7.a Find $p$ and $q$. Step 1: Sum forces zero: $$F_1 + F_2 + F_3 = 0$$ Step 2: Components: $$-3 + 1 + p = 0 \Rightarrow p = 2$$ $$7 - 1 + q = 0 \Rightarrow q = -6$$ Answer: $p=2$, $q=-6$. 7.b Resultant $R = F_1 + F_2 = (-3+1)i + (7-1)j = (-2i + 6j)$ N. Magnitude: $$|R| = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 6.32 \text{ N}$$ 7.c Angle between $R$ and $j$: Step 1: Use dot product: $$\cos \theta = \frac{R \cdot j}{|R||j|} = \frac{6}{6.32 \times 1} = 0.949$$ Step 2: Calculate angle: $$\theta = \cos^{-1}(0.949) = 18.4^\circ$$ Answer: $18^\circ$. 8. Problem: Particle A projected upward at 15 m/s from X; 2 s later particle B projected upward at 25 m/s from X. 8.a Find time after A's projection when particles are at same height. Step 1: Height of A at time $t$: $$h_A = 15t - 4.9t^2$$ Step 2: Height of B at time $t$ (B starts at $t-2$): $$h_B = 25(t-2) - 4.9(t-2)^2$$ Step 3: Set $h_A = h_B$: $$15t - 4.9t^2 = 25(t-2) - 4.9(t-2)^2$$ Step 4: Expand right side: $$25t - 50 - 4.9(t^2 - 4t + 4) = 25t - 50 - 4.9t^2 + 19.6t - 19.6$$ Step 5: Equation: $$15t - 4.9t^2 = 25t - 50 - 4.9t^2 + 19.6t - 19.6$$ Step 6: Simplify: $$15t = 25t - 50 + 19.6t - 19.6$$ $$15t = 44.6t - 69.6$$ Step 7: Rearrange: $$44.6t - 15t = 69.6 \Rightarrow 29.6t = 69.6 \Rightarrow t = 2.35 \text{ s}$$ 8.b Height at meeting: Step 1: Substitute $t=2.35$ into $h_A$: $$h = 15 \times 2.35 - 4.9 \times (2.35)^2 = 35.25 - 27.0 = 8.25 \text{ m}$$ Answer: 8.25 m. 9. Problem: Train accelerates from rest at $0.5$ m/s² to 15 m/s, travels 200 s at constant speed, then decelerates at $0.25$ m/s² to rest. 9.a Time to travel from R to S. Step 1: Time to accelerate to 15 m/s: $$t_1 = \frac{15}{0.5} = 30 \text{ s}$$ Step 2: Time at constant speed: $$t_2 = 200 \text{ s}$$ Step 3: Time to decelerate to rest: $$t_3 = \frac{15}{0.25} = 60 \text{ s}$$ Step 4: Total time: $$t = 30 + 200 + 60 = 290 \text{ s}$$ 9.b Distance from R to S. Step 1: Distance during acceleration: $$s_1 = \frac{1}{2} \times 0.5 \times 30^2 = 0.25 \times 900 = 225 \text{ m}$$ Step 2: Distance at constant speed: $$s_2 = 15 \times 200 = 3000 \text{ m}$$ Step 3: Distance during deceleration: $$s_3 = \frac{1}{2} \times 0.25 \times 60^2 = 0.125 \times 3600 = 450 \text{ m}$$ Step 4: Total distance: $$s = 225 + 3000 + 450 = 3675 \text{ m}$$ 9.c Average speed: $$\text{average speed} = \frac{3675}{290} \approx 12.67 \text{ m/s}$$