Grade 11 : Physics - Laws of Motion (Set 4)

Explanation: System acceleration $a = \frac{m_2 g}{m_1 + m_2} = \frac{6 \times 10}{4 + 6} = 6m/s^2$. Tension on horizontal block $T = m_1 a = 4 \times 6 = 24\text{ N}$.

Explanation: Downslope gravity force $W_{||} = mg \sin 37^\circ = 5 \times 10 \times 0.6 = 30\text{ N}$. Maximum static friction $f_{s,\max} = \mu_s mg \cos 37^\circ = 0.5 \times 50 \times 0.8 = 20\text{ N}$. Since $W_{||} > f_{s,\max}$, the block slips. Kinetic friction acts: $f_k = \mu_k mg \cos 37^\circ = 0.4 \times 50 \times 0.8 = 16\text{ N}$.

Explanation: For downward acceleration $a$, normal reaction $N = m(g - a) = 70 \times (9.8 - 2) = 70 \times 7.8 = 546\text{ N}$.

Explanation: Common acceleration $a = \frac{F}{m_A + m_B} = \frac{25}{3 + 2} = 5m/s^2$. Contact force on $B$ is $N = m_B \cdot a = 2 \times 5 = 10\text{ N}$.

Explanation: Impulse $J = \int_0^3 (6t - 2) dt = [3t^2 - 2t]_0^3 = 27 - 6 = 21\text{ N}\cdot\text{s}$. By impulse-momentum theorem, $m v_f = 21 \Rightarrow 2 v_f = 21 \Rightarrow v_f = 10.5m/s$.

Explanation: Vertical equilibrium requires $T \cos\theta = mg \Rightarrow T(0.8) = 0.5 \times 10 = 5 \Rightarrow T = \frac{5}{0.8} = 6.25\text{ N}$.

Explanation: Tension in string $T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 3 \times 5 \times 10}{8} = 37.5\text{ N}$. Force on support clamp $F = 2T = 2 \times 37.5 = 75.0\text{ N}$.

Explanation: Both statements are true and represent core physical concepts: rolling friction involves negligible sliding area, and friction inherently opposes relative surface motion.

Explanation: Thrust force $F_{\text{thrust}} = u \left(\frac{dm}{dt}\right) = 500 \times 10 = 5000\text{ N}$.

Explanation: Optimum speed $v_0 = \sqrt{R g \tan\theta} = \sqrt{25 \times 10 \times 0.4} = \sqrt{100} = 10m/s$.

Explanation: System acceleration $a = \frac{F}{m_1 + m_2 + m_3} = \frac{12}{1 + 2 + 3} = 2m/s^2$. Tension $T_1 = m_1 a = 1 \times 2 = 2\text{ N}$.

Explanation: By conservation of momentum: $m_g v_g + m_p v_p = 0 \Rightarrow 4 v_g + (0.02 \times 200) = 0 \Rightarrow 4 v_g = -4 \Rightarrow |v_g| = 1.0m/s$.

Explanation: Vertical equilibrium requires friction $f_s = mg = 20\text{ N}$. Maximum static friction is $\mu_s F \ge 20 \Rightarrow 0.4 F \ge 20 \Rightarrow F \ge 50\text{ N}$.

Explanation: From $F = m \cdot a$, for Trial 4: $X = 5kg \times 8m/s^2 = 40\text{ N}$.

Explanation: In the accelerating frame, pseudo-force ma acts horizontally. $\tan\theta = \frac{ma}{mg} = \frac{a}{g} = \frac{3}{10} = 0.3 \Rightarrow \theta = \tan^{-1}(0.3)$.

Explanation: Maximum friction force on $A$ is $f_{\max} = \mu_s m_A g = 0.3 \times 2 \times 10 = 6\text{ N}$. Maximum acceleration without slipping $a_{\max} = \frac{f_{\max}}{m_A} = 3m/s^2$. Maximum force on combined mass $F = (m_A + m_B) a_{\max} = 6 \times 3 = 18\text{ N}$.

Explanation: Maximum speed on a flat curve $v_{\max} = \sqrt{\mu_s g R} = \sqrt{0.5 \times 10 \times 20} = \sqrt{100} = 10m/s$.

Explanation: Impulse $\Delta p = \int_0^2 6t^2 dt = [2t^3]_0^2 = 16\text{ N}\cdot\text{s}$. Velocity $v = \frac{\Delta p}{m} = \frac{16}{2} = 8m/s$.

Explanation: Apparent weight $N = m(g + a) \Rightarrow 60 \times 10 = 50(10 + a) \Rightarrow 600 = 500 + 50a \Rightarrow a = 2m/s^2$.

Explanation: Conservation of momentum gives $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \Rightarrow \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$. Magnitude $p_3 = \sqrt{(2 \times 3)^2 + (2 \times 4)^2} = \sqrt{36 + 64} = 10kg\cdotm/s$. Speed $v_3 = \frac{10}{2} = 5m/s$.