Grade 11 : Physics - Laws of Motion (Set 2)

Explanation:By Newton's First Law, the lower part of the passenger's body moves forward with the bus while the upper body tends to remain at rest due to inertia of rest, causing the passenger to fall backward.

Explanation:Using Newton's Second Law: $$F = ma \implies a = \frac{F}{m} = \frac{15\text{ N}}{3\text{ kg}} = 5\text{ m/s}^2.$$

Explanation:Linear momentum $p$ is defined as the product of mass and velocity: $$p = m \cdot v = 4\text{ kg} \times 8\text{ m/s} = 32\text{ kg}\cdot\text{m/s}.$$

Explanation:By conservation of linear momentum: $$M v_r + m v_b = 0 \implies v_r = -\frac{m v_b}{M} = -\frac{0.02 \times 400}{4} = -2\text{ m/s}.$$ The negative sign indicates velocity opposite to the bullet's motion.

Explanation:Impulse is given by $I = F \cdot \Delta t$: $$I = 20\text{ N} \times 0.5\text{ s} = 10\text{ N}\cdot\text{s}.$$

Explanation:For an elevator accelerating upward, normal reaction $N = m(g + a)$: $$N = 50\text{ kg} \times (10 + 2)\text{ m/s}^2 = 50 \times 12 = 600\text{ N}.$$

Explanation:The maximum static friction is $f_{s,\text{max}} = \mu_s N = \mu_s m g$: $$f_{s,\text{max}} = 0.4 \times 10 \times 10 = 40\text{ N}.$$

Explanation:Action and reaction forces are equal in magnitude and opposite in direction, but they always act on two different interacting objects simultaneously.

Explanation:The angle of friction $\theta$ is related to the static friction coefficient by $\tan\theta = \mu_s$. Here $\tan\theta = 1 \implies \theta = 45^\circ$.

Explanation:The component of weight down the incline is $F = m g \sin\theta$: $$F = 6 \times 10 \times \sin(30^\circ) = 60 \times 0.5 = 30\text{ N}.$$

Explanation:Acceleration of connected masses over a pulley is given by: $$a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right)g = \left(\frac{5 - 3}{5 + 3}\right) \times 10 = \frac{2}{8} \times 10 = 2.5\text{ m/s}^2.$$

Explanation:System acceleration $a = \frac{F}{m_A + m_B} = \frac{18}{4+2} = 3m/s^2$. The contact force on $B$ produces its acceleration: $$F_{AB} = m_B \cdot a = 2 \times 3 = 6\text{ N}.$$

Explanation:The formula for optimum banking angle is $\tan\theta = \frac{v^2}{rg}$: $$\tan\theta = \frac{20^2}{40 \times 10} = \frac{400}{400} = 1 \implies \theta = 45^\circ.$$

Explanation:Centripetal force is calculated as $F_c = \frac{mv^2}{r}$: $$F_c = \frac{2 \times 4^2}{0.5} = \frac{2 \times 16}{0.5} = 64\text{ N}.$$

Explanation:Kinetic frictional force is $f_k = \mu_k N = \mu_k mg$: $$f_k = 0.2 \times 5 \times 10 = 10\text{ N}.$$

Explanation:A pseudo force vector is given by $\vec{F}_{pseudo} = -m\vec{a}_0$, acting opposite to the direction of acceleration of the non-inertial frame.

Explanation:For upward acceleration, tension $T - mg = ma \implies T = m(g + a)$: $$T = 4 \times (10 + 3) = 4 \times 13 = 52\text{ N}.$$

Explanation:Taking initial direction as positive, $u = +10m/s$ and $v = -10m/s$: $$\Delta p = m(v - u) = 0.1 \times (-10 - 10) = -2\text{ kg}\cdot\text{m/s}.$$ Magnitude of change is $2kg\cdotm/s$.

Explanation:Rocket propulsion relies on the continuous expulsion of hot gases downward, generating an upward momentum change of equal magnitude by conservation of linear momentum.

Explanation:The resultant of two perpendicular forces is $R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ N}$. For equilibrium, the third force must equal this resultant in magnitude (10 N) and act in the opposite direction.