1. Problem 9: A body of mass 10 kg moves with constant velocity 5 m/s on a horizontal surface with coefficient of sliding friction 0.2. (i) Find the horizontal force to maintain constant motion. (ii) Find the retardation if the applied force is removed. (iii) Find the time taken to come to rest. 2. Problem 10: A block slides down an inclined plane at 30° with coefficient of kinetic friction 0.2. Find acceleration. 3. Problem 11: An iron block of mass 10 kg rests on a wooden plane inclined at 30°. Least force parallel to plane to slide up is 100 N. Find coefficient of sliding friction. 4. Problem 12: A ball of mass 0.5 kg hits a vertical wall at 12 m/s and bounces back at 8 m/s. Collision lasts 0.1 s. Find average force on ball. --- ### Step-by-step solutions: **Problem 9:** 1. Given mass $m=10$ kg, velocity $v=5$ m/s, coefficient of friction $\mu=0.2$, gravity $g=9.8$ m/s$^2$. 2. Friction force $F_f = \mu mg = 0.2 \times 10 \times 9.8 = 19.6$ N. 3. (i) To maintain constant velocity, applied force $F = F_f = 19.6$ N. 4. (ii) If force removed, only friction acts causing retardation $a$: $$a = \frac{F_f}{m} = \frac{19.6}{10} = 1.96 \approx 2 \text{ m/s}^2$$ 5. (iii) Time to rest from $v=5$ m/s with retardation $a=2$ m/s$^2$: $$t = \frac{v}{a} = \frac{5}{2} = 2.5 \text{ s}$$ --- **Problem 10:** 1. Mass $m$ cancels out, incline angle $\theta=30^\circ$, $\mu=0.2$, $g=9.8$ m/s$^2$. 2. Acceleration down incline: $$a = g(\sin\theta - \mu \cos\theta)$$ 3. Calculate: $$\sin 30^\circ = 0.5, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$ $$a = 9.8 (0.5 - 0.2 \times 0.866) = 9.8 (0.5 - 0.1732) = 9.8 \times 0.3268 = 3.2 \text{ m/s}^2$$ (rounded to 3.3 m/s$^2$) --- **Problem 11:** 1. Mass $m=10$ kg, incline $\theta=30^\circ$, least force $F=100$ N to slide up. 2. Forces along incline: Applied force $F$ balances component of weight down incline plus friction: $$F = mg \sin\theta + \mu mg \cos\theta$$ 3. Rearranged to find $\mu$: $$\mu = \frac{F - mg \sin\theta}{mg \cos\theta}$$ 4. Calculate: $$mg = 10 \times 9.8 = 98$$ $$mg \sin 30^\circ = 98 \times 0.5 = 49$$ $$mg \cos 30^\circ = 98 \times 0.866 = 84.87$$ $$\mu = \frac{100 - 49}{84.87} = \frac{51}{84.87} = 0.6$$ --- **Problem 12:** 1. Mass $m=0.5$ kg, initial velocity $u=12$ m/s, final velocity $v=-8$ m/s (opposite direction), collision time $\Delta t=0.1$ s. 2. Change in momentum: $$\Delta p = m(v - u) = 0.5 (-8 - 12) = 0.5 \times (-20) = -10 \text{ kg m/s}$$ 3. Average force: $$F = \frac{\Delta p}{\Delta t} = \frac{-10}{0.1} = -100 \text{ N}$$ Magnitude is 100 N. --- **Final answers:** (i) 19.6 N (ii) 2 m/s$^2$ (iii) 2.5 s Problem 10 acceleration: 3.3 m/s$^2$ Problem 11 coefficient of friction: 0.6 Problem 12 average force: 100 N