Grade 11 : Physics - Laws of Motion (Set 1)

Explanation:Newton's First Law states that a body continues in its state of rest or uniform motion unless acted upon by a net external force. This property of an object to resist changes in its state of motion is defined as inertia.

Explanation:Using Newton's Second Law: $$F = ma \implies a = \frac{F}{m} = \frac{25\text{ N}}{5\text{ kg}} = 5\text{ m/s}^2$$

Explanation:Linear momentum $p$ is given by the product of mass and velocity: $$p = mv = 4\text{ kg} \times 8\text{ m/s} = 32\text{ kg}\cdot\text{m/s}$$

Explanation:Impulse $J$ delivered by a constant force is defined as the product of force and time interval: $$J = F \Delta t = 15\text{ N} \times 4\text{ s} = 60\text{ N}\cdot\text{s}$$

Explanation:By the law of conservation of linear momentum: $$m_b v_b + M_g v_r = 0 \implies v_r = -\frac{m_b v_b}{M_g} = -\frac{0.02 \times 300}{4} = -1.5\text{ m/s}$$ Speed is the magnitude, 1.5 m/s.

Explanation:When accelerating upward, the normal reaction (apparent weight) is: $$N = m(g + a) = 60\text{ kg} \times (10 + 2)\text{ m/s}^2 = 60 \times 12 = 720\text{ N}$$

Explanation:When accelerating downward, the normal reaction force is: $$N = m(g - a) = 50\text{ kg} \times (10 - 3)\text{ m/s}^2 = 50 \times 7 = 350\text{ N}$$

Explanation:During free fall, acceleration $a = g$. The normal force is: $$N = m(g - a) = m(g - g) = 0\text{ N}$$ This state is referred to as weightlessness.

Explanation:The maximum static friction force is: $$f_{s,\max} = \mu_s N = \mu_s mg = 0.4 \times 10\text{ kg} \times 10\text{ m/s}^2 = 40\text{ N}$$

Explanation:Kinetic friction force $f_k$ is calculated as: $$f_k = \mu_k N = \mu_k mg = 0.25 \times 6\text{ kg} \times 10\text{ m/s}^2 = 15\text{ N}$$

Explanation:The angle of repose $\theta$ is related to the coefficient of static friction by: $$\mu_s = \tan \theta = \tan 30^\circ = \frac{1}{\sqrt{3}}$$

Explanation:To keep the block in equilibrium along the incline, the holding force $F$ must balance the component of gravity down the incline: $$F = mg \sin 30^\circ = 8 \times 10 \times 0.5 = 40\text{ N}$$

Explanation:Considering both blocks as a single system of total mass $M = m_1 + m_2$: $$a = \frac{F}{m_1 + m_2} = \frac{20}{3 + 2} = \frac{20}{5} = 4\text{ m/s}^2$$

Explanation:The tension $T$ in the string accelerates only the trailing mass $m_1 = 3kg$: $$T = m_1 a = 3\text{ kg} \times 4\text{ m/s}^2 = 12\text{ N}$$

Explanation:The formula for acceleration in an Atwood machine is: $$a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g = \left(\frac{5 - 3}{3 + 5}\right) \times 10 = \frac{2}{8} \times 10 = 2.5\text{ m/s}^2$$

Explanation:Tension $T$ in an Atwood machine is given by: $$T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 3 \times 5 \times 10}{3 + 5} = \frac{300}{8} = 37.5\text{ N}$$

Explanation:Centripetal force formula: $$F_c = \frac{m v^2}{r} = \frac{1000 \times 10^2}{50} = \frac{100000}{50} = 2000\text{ N}$$

Explanation:Maximum safe speed on a flat level curve is given by: $$v_{\max} = \sqrt{\mu_s r g} = \sqrt{0.4 \times 25 \times 10} = \sqrt{100} = 10\text{ m/s}$$

Explanation:For a banked road, balancing horizontal component of normal force ($N \sin \theta = m v^2 / r$) and vertical component ($N \cos \theta = mg$) yields: $$\tan \theta = \frac{v^2}{rg}$$

Explanation:Action-reaction pairs must act on two different bodies and be of identical physical origin. Earth pulling book downward (gravitational) and book pulling Earth upward (gravitational) constitute a true action-reaction pair.