Laws of Motion

Concept: According to Newton’s First Law, if a body is at rest or moving with constant velocity (zero acceleration), the net force acting on it is zero ($F_{net} = ma = 0$).

(a) Rain drop falling with constant speed:
Since $v = \text{constant}$, acceleration $a = 0$.
Net Force = 0.

(b) Cork of mass 10 g floating on water:
The cork is stationary (floating). The weight is balanced by the buoyant force.
Net Force = 0.

(c) Kite skillfully held stationary:
Stationary implies $v = 0$ and $a = 0$.
Net Force = 0.

(d) Car moving with constant velocity of 30 km/h:
Constant velocity implies $a = 0$.
Net Force = 0.

(e) High-speed electron in empty space:
Far from material objects (no gravity) and free of electric/magnetic fields means no external forces act on it.
Net Force = 0.

Given: $m = 0.05 \text{ kg}$, $g = 10 \text{ m/s}^2$.
In all cases below, neglecting air resistance, the only force acting on the pebble is Gravity.

(a) During upward motion:
Force $F = mg = 0.05 \times 10 = \mathbf{0.5 \text{ N vertically downwards}}$.

(b) During downward motion:
Force $F = mg = \mathbf{0.5 \text{ N vertically downwards}}$.

(c) At highest point:
Velocity is zero momentarily, but acceleration is still $g$.
Force $F = mg = \mathbf{0.5 \text{ N vertically downwards}}$.

If thrown at 45°:
The gravitational force depends only on mass, not the direction of motion. The answer remains 0.5 N vertically downwards.

Mass $m = 0.1 \text{ kg}$.

(a) Dropped from stationary train:
Once dropped, only gravity acts.
$F = mg = 0.1 \times 10 = \mathbf{1.0 \text{ N downwards}}$.

(b) Dropped from constant velocity train:
Horizontal velocity is constant (inertia), but no horizontal force acts. Only gravity acts.
$F = \mathbf{1.0 \text{ N downwards}}$.

(c) Dropped from accelerating train ($a=1$):
Once it leaves the hand, the stone is no longer pushed by the train. The only force is gravity.
$F = \mathbf{1.0 \text{ N downwards}}$.

(d) Lying on floor of accelerating train ($a=1 \text{ m/s}^2$):
The stone is at rest relative to the train, meaning it accelerates with the train.
$F = ma = 0.1 \times 1 = \mathbf{0.1 \text{ N horizontal}}$ (in direction of train).

Options: (i) $T$, (ii) $T – \frac{mv^2}{l}$, (iii) $T + \frac{mv^2}{l}$, (iv) $0$.

Correct Alternative: (i) $T$

Given: $m = 20 \text{ kg}$, $u = 15 \text{ m/s}$, $F = -50 \text{ N}$ (retarding), $v = 0$.

$a = \frac{F}{m} = \frac{-50}{20} = -2.5 \text{ m/s}^2$.

Using $v = u + at$:
$0 = 15 – 2.5t \implies 2.5t = 15$
$t = \mathbf{6 \text{ s}}$.

Given: $m = 3.0 \text{ kg}$, $u = 2.0 \text{ m/s}$, $v = 3.5 \text{ m/s}$, $t = 25 \text{ s}$.

Acceleration $a = \frac{v – u}{t} = \frac{3.5 – 2.0}{25} = \frac{1.5}{25} = 0.06 \text{ m/s}^2$.

Force $F = ma = 3.0 \times 0.06 = \mathbf{0.18 \text{ N}}$.

Direction: Since speed increases, force acts in the direction of motion.

$m = 5 \text{ kg}$. Forces $F_1 = 8 \text{ N}$, $F_2 = 6 \text{ N}$ (perpendicular).

Resultant Force $F = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$.

Acceleration $a = \frac{F}{m} = \frac{10}{5} = \mathbf{2 \text{ m/s}^2}$.

Direction: $\tan \beta = \frac{6}{8} = 0.75 \Rightarrow \beta = \tan^{-1}(0.75) \approx 37^\circ$ with the 8 N force.

$u = 36 \text{ km/h} = 10 \text{ m/s}$. $v = 0$. $t = 4.0 \text{ s}$.

Acceleration $a = \frac{v-u}{t} = \frac{0-10}{4} = -2.5 \text{ m/s}^2$.

Total Mass $M = \text{Vehicle} + \text{Driver} = 400 + 65 = 465 \text{ kg}$.

Retarding Force $F = M|a| = 465 \times 2.5 = \mathbf{1162.5 \text{ N}}$.

Given: Mass $m = 20,000 \text{ kg}$. Acc $a = 5.0 \text{ m/s}^2$ (upwards). $g=10 \text{ m/s}^2$.

$T = m(g + a) = 20000(10 + 5)$
$T = 20000 \times 15 = \mathbf{3.0 \times 10^5 \text{ N}}$.

$m=0.4 \text{ kg}$. $u=10 \text{ m/s}$ (North). Force $F=-8.0 \text{ N}$ (South). Applied for $0 \le t \le 30 \text{ s}$.

Acceleration $a = F/m = -8/0.4 = -20 \text{ m/s}^2$.

(i) At t = -5 s: Before force starts.
Uniform motion. $x = ut = 10(-5) = \mathbf{-50 \text{ m}}$.

(ii) At t = 25 s: During acceleration.
$x = ut + \frac{1}{2}at^2 = 10(25) + \frac{1}{2}(-20)(25)^2$
$x = 250 – 10(625) = 250 – 6250 = \mathbf{-6000 \text{ m}}$.

(iii) At t = 100 s: Force acts till $t=30 \text{ s}$, then uniform motion.
State at $t=30$:
$v_{30} = u + at = 10 – 20(30) = -590 \text{ m/s}$.
$x_{30} = 10(30) – 10(30)^2 = 300 – 9000 = -8700 \text{ m}$.
Motion from $t=30$ to $100$ (70s):
$x_{100} = x_{30} + v_{30}(70) = -8700 + (-590 \times 70)$
$x_{100} = -8700 – 41300 = \mathbf{-50000 \text{ m}} = \mathbf{-50 \text{ km}}$.

Truck Motion: $u=0, a=2 \text{ m/s}^2$.
At $t=10 \text{ s}$, horizontal velocity $v_x = u+at = 0+2(10) = 20 \text{ m/s}$.

Stone at t=11s (1 sec after drop):
Horizontal velocity $v_x = 20 \text{ m/s}$ (constant).
Vertical velocity $v_y = gt’ = 10(1) = 10 \text{ m/s}$.

(a) Net Velocity: $v = \sqrt{20^2 + 10^2} = \sqrt{500} \approx \mathbf{22.4 \text{ m/s}}$.

(b) Acceleration: Once dropped, only gravity acts.
$a = \mathbf{10 \text{ m/s}^2 \text{ vertically downwards}}$.

(a) At extreme position:
Velocity of bob is zero momentarily. If cut, it falls purely under gravity.
Trajectory: Vertically downwards.

(b) At mean position:
Velocity is maximum ($1 \text{ m/s}$) and horizontal. If cut, it acts like a horizontal projectile.
Trajectory: Parabolic path.

Mass $m = 70 \text{ kg}$. Reading $R$ is the Normal Force.

(a) Moving up with uniform speed ($a=0$):
$R = mg = 70 \times 10 = \mathbf{700 \text{ N}}$.

(b) Moving down with acceleration $5 \text{ m/s}^2$:
Equation: $mg – R = ma \implies R = m(g-a)$.
$R = 70(10-5) = 70(5) = \mathbf{350 \text{ N}}$.

(c) Moving up with acceleration $5 \text{ m/s}^2$:
Equation: $R – mg = ma \implies R = m(g+a)$.
$R = 70(10+5) = 70(15) = \mathbf{1050 \text{ N}}$.

(d) Free fall ($a=g$):
$R = m(g-g) = \mathbf{0 \text{ N}}$ (Weightlessness).

(a) Force:
$t < 0$: $x=0$ (rest) $\rightarrow a=0, F=0$.
$0 < t < 4$: $x$ increases linearly (const velocity) $\rightarrow a=0, F=0$.
$t > 4$: $x$ constant (rest) $\rightarrow a=0, F=0$.

(b) Impulse:
$m=4 \text{ kg}$. Slope $0 < t < 4$ is $v = 3/4 = 0.75 \text{ m/s}$.
At $t=0$: $u=0, v=0.75$. $I = \Delta p = 4(0.75 – 0) = \mathbf{3 \text{ kg m/s}}$.
At $t=4$: $u=0.75, v=0$. $I = \Delta p = 4(0 – 0.75) = \mathbf{-3 \text{ kg m/s}}$.

Masses $m_1 = 10 \text{ kg (A)}, m_2 = 20 \text{ kg (B)}$. Total $M = 30 \text{ kg}$.
Force $F = 600 \text{ N}$. Acceleration $a = 600/30 = 20 \text{ m/s}^2$.

(i) Force applied to A:
Tension pulls mass B alone.
$T = m_2 a = 20 \times 20 = \mathbf{400 \text{ N}}$.

(ii) Force applied to B:
Tension pulls mass A alone.
$T = m_1 a = 10 \times 20 = \mathbf{200 \text{ N}}$.

$m_1 = 8 \text{ kg}, m_2 = 12 \text{ kg}$. $m_2$ moves down.

Acceleration: $a = \frac{12-8}{12+8} \times 10 = \frac{4}{20} \times 10 = \mathbf{2 \text{ m/s}^2}$.

Tension: $T = \frac{2 \times 8 \times 12}{20} \times 10 = \frac{1920}{20} = \mathbf{96 \text{ N}}$.

$m = 0.05 \text{ kg}$. Initial speed $u = 6 \text{ m/s}$. Final speed $v = -6 \text{ m/s}$.

Impulse = Change in Momentum ($\Delta p$).
$\Delta p = m(v – u) = 0.05(-6 – 6) = 0.05(-12) = -0.6 \text{ kg m/s}$.

Magnitude of Impulse = 0.6 Ns.

Shell mass $m = 0.020 \text{ kg}$. Gun mass $M = 100 \text{ kg}$.
Muzzle speed $v = 80 \text{ m/s}$. Recoil speed $V$.

Conservation of Momentum: $P_i = P_f \Rightarrow 0 = mv + MV$.
$V = -\frac{mv}{M} = -\frac{0.020 \times 80}{100} = -\frac{1.6}{100} = \mathbf{-0.016 \text{ m/s}}$.

$m = 0.15 \text{ kg}$. Speed $v = 54 \text{ km/h} = 15 \text{ m/s}$. Angle $\theta = 45^\circ$.

Vector Impulse Formula:
$|\Delta p| = \sqrt{p^2 + p^2 – 2p^2 \cos \theta} = p \sqrt{2(1 – \cos 45^\circ)}$
or using half-angle formula for reflection: $|\Delta p| = 2mv \sin(\theta/2)$.

Calculation:
$|\Delta p| = 2(0.15)(15) \sin(22.5^\circ)$
$|\Delta p| = 4.5 \times 0.3826 \approx \mathbf{1.72 \text{ Ns}}$.

$m = 0.25 \text{ kg}$, $r = 1.5 \text{ m}$, $n = 40 \text{ rpm} = 2/3 \text{ rps}$.
Angular velocity $\omega = 2\pi n = 2\pi(2/3) = \frac{4\pi}{3} \text{ rad/s}$.

Tension:
$T = m\omega^2 r = 0.25 \times (\frac{16\pi^2}{9}) \times 1.5 \approx \mathbf{6.6 \text{ N}}$.

Max Speed ($T_{max} = 200 \text{ N}$):
$T_{max} = \frac{mv_{max}^2}{r} \implies v_{max} = \sqrt{\frac{T_{max} r}{m}}$
$v_{max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{1200} \approx \mathbf{34.6 \text{ m/s}}$.

Correct: (b) The stone flies off tangentially.

Reason: At the instant the string breaks, the centripetal force vanishes. According to Newton’s First Law (Law of Inertia), the stone continues to move in the direction of its instantaneous velocity, which is tangential to the circular path.

(a) Horse in empty space: A horse runs by pushing the ground backwards (Action). The ground pushes the horse forward (Reaction/Friction). In empty space, there is no ground to provide this reaction force.

(b) Passengers thrown forward: Due to Inertia of Motion. When the bus stops, the feet in contact with the floor stop, but the upper body continues to move forward.

(c) Pulling lawn mower is easier:

Pushing: The vertical component of force adds to weight ($mg + F\sin\theta$), increasing normal force and thus friction.
Pulling: The vertical component acts upwards ($mg – F\sin\theta$), reducing normal force and friction.

(d) Cricketer moves hands back:
To increase the time of impact ($\Delta t$). Since Impulse $I = F \times \Delta t$ is fixed for a catch (change in momentum is constant), increasing $\Delta t$ reduces the average Force $F$ exerted on the hands, preventing injury.