Grade 11 : Physics - Laws of Motion (Set 3)

Explanation: The total mass of the system is $M = 2 + 3 + 5 = 10kg$. The common acceleration is $a = \frac{F}{M} = \frac{50}{10} = 5m/s^2$. The force pushing block $m_3$ is the contact force from $m_2$, given by $N_{23} = m_3 \times a = 5 \times 5 = 25\text{ N}$.

Explanation: Normal force $N = mg \cos 37^\circ = 5 \times 10 \times 0.8 = 40\text{ N}$. Maximum static friction is $f_s = \mu_s N = 0.5 \times 40 = 20\text{ N}$. The component of gravity down the incline is $mg \sin 37^\circ = 5 \times 10 \times 0.6 = 30\text{ N}$. To move up the plane, applied force $F = mg \sin 37^\circ + f_s = 30 + 20 = 50\text{ N}$.

Explanation: Maximum friction force on block A provides its acceleration: $f_{max} = \mu m_A g = 0.4 \times 2 \times 10 = 8\text{ N}$. Maximum acceleration without slipping is $a_{max} = \frac{f_{max}}{m_A} = \mu g = 4m/s^2$. Total system maximum force is $F_{max} = (m_A + m_B) a_{max} = (2 + 3) \times 4 = 20\text{ N}$.

Explanation: Change in momentum $\Delta p = m(v - u) = 0.2 \times (-15 - 25) = 0.2 \times (-40) = -8\text{ N s}$. Impulse magnitude is 8 N s. Average force $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{8}{0.01} = 800\text{ N}$.

Explanation: Force equals rate of change of momentum: $F = n \cdot m \cdot v$, where $n = 20\text{ bullets/s}$, $m = 0.03kg$, $v = 400m/s$. $F = 20 \times 0.03 \times 400 = 240\text{ N}$.

Explanation: Applying Newton's second law upward: $T - mg = ma \implies T = m(g + a) = 1000 \times (10 + 2) = 12000\text{ N}$.

Explanation: Vertical equilibrium requires $T \cos \theta = mg$. Given $\theta = 60^\circ$ and $\cos 60^\circ = 0.5$, $T \times 0.5 = 0.5 \times 10 = 5\text{ N} \implies T = 10\text{ N}$.

Explanation: Centripetal force is supplied by static friction: $\frac{mv^2}{R} \le \mu_s mg \implies v_{max} = \sqrt{\mu_s g R} = \sqrt{0.5 \times 10 \times 20} = \sqrt{100} = 10m/s$.

Explanation: In the accelerated frame of the wedge, pseudo force ma acts horizontally. Component up the incline $ma \cos \theta$ balances gravity component $mg \sin \theta$. Thus $ma \cos 45^\circ = mg \sin 45^\circ \implies a = g = 10m/s^2$.

Explanation: Acceleration of the rope is $a = \frac{F}{M} = \frac{18}{3} = 6m/s^2$. The segment of rope behind the point $x=2m$ has length $L - x = 4m$ and mass $m' = \frac{4}{6} \times 3 = 2kg$. Tension pulls this mass: $T = m' a = 2 \times 6 = 12\text{ N}$.

Explanation: Thrust force in a variable mass system is given by $F_{thrust} = v_{rel} \cdot \frac{dm}{dt} = 800 \times 50 = 40,000\text{ N} = 40\text{ kN}$.

Explanation: Impulse equals area under $F-t$ graph: $J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 30 = 60\text{ N s}$. From impulse-momentum theorem: $m v - 0 = J \implies v = \frac{60}{3} = 20m/s$.

Explanation: Tension while accelerating upward is $T = m(g + a)$. Setting $T \le 500\text{ N}$, $40(10 + a) = 500 \implies 400 + 40a = 500 \implies 40a = 100 \implies a = 2.5m/s^2$.

Explanation: Acceleration of Atwood machine is $a = \frac{m_1 - m_2}{m_1 + m_2} g = \frac{3 - 2}{3 + 2} \times 10 = \frac{1}{5} \times 10 = 2m/s^2$.

Explanation: For friction-free banked turn, $v = \sqrt{g R \tan \theta} = \sqrt{10 \times 40 \times \tan 45^\circ} = \sqrt{400 \times 1} = 20m/s$.

Explanation: At constant velocity, net force along incline is zero: $mg \sin 30^\circ = f_k = \mu_k mg \cos 30^\circ \implies \mu_k = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$.

Explanation: Deceleration $a = \frac{F}{m} = \frac{15}{5} = 3m/s^2$. Time to stop $t = \frac{u}{a} = \frac{12}{3} = 4\text{ s}$.

Explanation: For static equilibrium, $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \implies \vec{F}_3 = -(\vec{F}_1 + \vec{F}_2) = -((6-2)\hat{i} + (8-5)\hat{j}) = -(4\hat{i} + 3\hat{j})\text{ N}$. Magnitude $|\vec{F}_3| = \sqrt{4^2 + 3^2} = 5\text{ N}$.

Explanation: Apparent weight during downward acceleration is $W_{app} = m(g - a) = 6 \times (10 - 3) = 6 \times 7 = 42\text{ N}$.

Explanation: Normal reaction $N = 200\text{ N}$. Weight pulling down $W = mg = 10 \times 10 = 100\text{ N}$. For equilibrium, static friction $f_s = 100\text{ N}$. Since $f_s \le \mu_s N \implies 100 \le \mu_s (200) \implies \mu_s \ge 0.50$.