Chapter 4: Laws of Motion (Laws of Motion)

In (a)–(d) the body is either at rest or moving with constant velocity, so by the first law its acceleration is zero and the net force is zero (the forces acting balance each other: weight and air drag; weight and buoyancy; weight, wind force and string tension; engine force and friction).

In (e) no force acts at all, since the electron is far from all objects and fields.

Zero in every case.

The only force is gravity: $mg=0.05\times 10=0.5$ N, vertically downwards, in all three cases (even at the top, where velocity is zero but acceleration is still g).

For a throw at 45°, the force is still only gravity, 0.5 N downwards. (At the top the pebble is not at rest then: it keeps its horizontal velocity.)

0.5 N vertically downwards in (a), (b), (c); no change for a 45° throw.

(a), (b), (c): once the stone leaves the train, the only force on it is gravity, $mg=0.1\times 10=1$ N downwards. The train's acceleration in (c) does not matter: force at an instant depends on the situation at that instant, not on its past.

(d) The stone moves with the train, so it has acceleration 1 m s⁻² horizontally. Net force $=ma=0.1\times 1=0.1$ N in the direction of the train's acceleration (provided by friction; weight and normal force cancel).

(a) 1 N down (b) 1 N down (c) 1 N down (d) 0.1 N horizontal, along the train's acceleration

On a smooth horizontal table, weight and normal force cancel. The only horizontal force is the tension, which acts towards the peg and provides the centripetal force: $T=\frac{m{v}^2}{l}$.

(i) T

$a=-\frac{50}{20}=-2.5{\text{m s}}^{-2}$. $v=u+at\Rightarrow 0=15-2.5t$.

6.0 s

$a=\frac{3.5-2.0}{25}=0.06{\text{m s}}^{-2}$; $F=ma=3.0\times 0.06=0.18$ N, in the direction of motion (the speed increases).

0.18 N, along the direction of motion

Resultant $F=\sqrt{8^2+6^2}=10$ N at $\theta ={\tan }^{-1}\frac{6}{8}=37^{\circ }$ to the 8 N force.

$a=\frac{F}{m}=\frac{10}{5}=2{\text{m s}}^{-2}$, along the resultant.

2 m s⁻², at 37° to the 8 N force

$u=36\times \frac{5}{18}=10{\text{m s}}^{-1}$, $a=\frac{0-10}{4.0}=-2.5{\text{m s}}^{-2}$, total mass $=465$ kg.

$F=465\times 2.5=1162.5$ N.

About 1.2 × 10³ N

Thrust F up, weight mg down: $F-mg=ma\Rightarrow F=m(g+a)=20000\times (10+5.0)$.

3.0 × 10⁵ N

Take north as positive. $a=-\frac{8.0}{0.40}=-20{\text{m s}}^{-2}$ for $0\le t\le 30$ s.

t = −5 s: no force before t = 0, uniform motion: $x=10\times (-5)=-50$ m.

t = 25 s: $x=10(25)+\frac{1}{2}(-20)(25{)}^2=250-6250=-6000$ m $=-6$ km.

t = 100 s: up to 30 s: $x_1=10(30)-10(30{)}^2=-8700$ m and $v=10-20(30)=-590{\text{m s}}^{-1}$. After that, uniform motion for 70 s: $x_2=-590\times 70=-41300$ m. Total $x=-50000$ m.

−50 m, −6 km, −50 km

At t = 10 s the truck's speed is $2.0\times 10=20{\text{m s}}^{-1}$; the stone leaves with this horizontal velocity and keeps it (first law). It needs $\sqrt{2\times 6/10}=1.1$ s to fall, so it is still in the air at t = 11 s.

(a) After 1 s: horizontal 20 m s⁻¹, vertical $10\times 1=10{\text{m s}}^{-1}$ downwards. Speed $=\sqrt{20^2+10^2}=\sqrt{500}=22.4{\text{m s}}^{-1}$, at ${\tan }^{-1}\frac{10}{20}=26.6^{\circ }$ below the horizontal.

(b) Once dropped, only gravity acts: 10 m s⁻² vertically downwards.

(a) 22.4 m s⁻¹ at 26.6° below the horizontal (b) 10 m s⁻² downwards

(a) At an extreme position the bob is momentarily at rest. When the string is cut, only gravity acts, so it falls vertically down.

(b) At the mean position it has a horizontal velocity of 1 m s⁻¹. After the cut it moves like a horizontally projected body: a parabola.

(a) Vertical straight line (b) Parabola

The scale reads the normal force N (by the third law, the man presses the scale with N).

(a) Uniform speed, a = 0: $N=mg=700$ N, reading 70 kg.

(b) $mg-N=ma\Rightarrow N=70(10-5)=350$ N, reading 35 kg.

(c) $N-mg=ma\Rightarrow N=70(10+5)=1050$ N, reading 105 kg.

(d) Free fall, a = g: $N=70(10-10)=0$, reading zero.

(a) 70 kg (b) 35 kg (c) 105 kg (d) zero

(a) For t < 0 the particle is at rest; for 0 < t < 4 s it moves with constant velocity $\frac{3}{4}=0.75{\text{m s}}^{-1}$; for t > 4 s it is at rest again. In all three intervals the acceleration, and hence the force, is zero.

(b) Impulse = change of momentum.

At t = 0: $4\times (0.75-0)=3{\text{kg m s}}^{-1}$. At t = 4 s: $4\times (0-0.75)=-3{\text{kg m s}}^{-1}$.

(a) Zero in all three intervals (b) +3 kg m s⁻¹ at t = 0, −3 kg m s⁻¹ at t = 4 s

Common acceleration in both cases: $a=\frac{600}{10+20}=20{\text{m s}}^{-2}$. The string must accelerate the body that is not pulled.

(i) Force on A (10 kg): the string pulls B (20 kg): $T=20\times 20=400$ N.

(ii) Force on B (20 kg): the string pulls A (10 kg): $T=10\times 20=200$ N.

(i) 400 N (ii) 200 N

The 12 kg mass moves down: $12g-T=12a$ and $T-8g=8a$. Adding: $4g=20a\Rightarrow a=\frac{4\times 10}{20}=2{\text{m s}}^{-2}$.

$T=8(g+a)=8\times 12=96$ N.

a = 2 m s⁻², T = 96 N

No external force acts during the break-up, so momentum is conserved. Initial momentum is zero, so $m_1{\vec{v}}_1+m_2{\vec{v}}_2=\vec{0}\Rightarrow {\vec{v}}_2=-\frac{m_1}{m_2}{\vec{v}}_1$.

The minus sign shows that ${\vec{v}}_2$ is opposite to ${\vec{v}}_1$: two momenta can add to zero only if they are equal and opposite.

${\vec{v}}_2=-\frac{m_1}{m_2}{\vec{v}}_1$: opposite directions.

For one ball: $\Delta p=m(v-u)=0.05[(-6)-6]=-0.6{\text{kg m s}}^{-1}$. The other ball gets $+0.6{\text{kg m s}}^{-1}$.

0.6 kg m s⁻¹ each, in opposite directions

Initial momentum is zero: $100V=0.020\times 80\Rightarrow V=0.016{\text{m s}}^{-1}$.

0.016 m s⁻¹ (1.6 cm s⁻¹)

$v=54\times \frac{5}{18}=15{\text{m s}}^{-1}$. As in the NCERT solution, the incoming and outgoing paths of the ball make 45° with each other, so each makes 22.5° with the line bisecting them. The velocity component along this bisector is reversed and the perpendicular component is unchanged.

$|\Delta p|=2mv\cos 22.5^{\circ }=2\times 0.15\times 15\times 0.924=4.2{\text{kg m s}}^{-1}$, directed along the bisector (away from the bat).

4.2 kg m s⁻¹, along the bisector of the initial and final directions

$v=2\pi r\nu =2\pi \times 1.5\times \frac{40}{60}=2\pi =6.28{\text{m s}}^{-1}$.

$T=\frac{m{v}^2}{r}=\frac{0.25\times (6.28{)}^2}{1.5}=6.6$ N.

$T_{\max }=200=\frac{0.25v_{\max }^2}{1.5}\Rightarrow v_{\max }^2=1200\Rightarrow v_{\max }=34.6{\text{m s}}^{-1}$.

T = 6.6 N; v_max ≈ 35 m s⁻¹

When the string breaks, the centripetal force vanishes. By the first law the stone continues with the velocity it had at that instant, which is along the tangent.

(b)

(a) The horse and cart pull each other with equal and opposite forces (third law), which cancel inside the system. On the ground, the horse pushes the ground backwards and the friction of the ground on its hooves pushes it forward. In empty space there is no ground, so no external force acts and the system cannot start moving.

(b) When the bus stops, the lower body (in contact with the seat) stops with it, but the upper body keeps moving forward due to inertia.

(c) The force is applied at an angle to the ground. When pushing, its vertical component presses the mower down, increasing the normal force and hence friction. When pulling, the vertical component lifts it up, reducing the normal force and friction.

(d) Moving the hands back increases the time in which the ball's momentum becomes zero. Since $F=\frac{\Delta p}{\Delta t}$, the force on the hands is smaller and the catch hurts less.

(a) No external force in space (b) inertia of motion (c) pulling reduces the normal force and friction (d) longer stopping time means smaller force