Laws of Motion Class 9 ICSE Important Questions with Answers

Answer: (b) $\text{kg m s}^{-1}$kg m s^-1.

Momentum $p=mv$p=mv, so its unit is $\text{kg}\times\text{m s}^{-1}=\text{kg m s}^{-1}$kg× m s^-1=kg m s^-1 (equivalently $\text{N s}$N s).

Answer: (b) force and inertia.

The first law introduces the idea of inertia (a body resists change in its state) and defines force as the external agent that changes that state.

Answer: (b) $5\text{ m s}^{-2}$5 m s^-2.

$a=\dfrac{F}{m}=\dfrac{10}{2}=5\text{ m s}^{-2}$a=F/m=10/2=5 m s^-2.

Answer: (d) four times.

$F=\dfrac{Gm_1m_2}{r^2}$F=Gm₁m₂/r², so $F\propto\dfrac{1}{r^2}$F∝ 1/r². If $r\to\dfrac{r}{2}$r→ r/2, then $F\to\dfrac{1}{(1/2)^2}F=4F$F→ 1/(1/2)²F=4F.

Answer: (a) The feet move forward with the bus while the upper body stays at rest owing to inertia of rest, so the passenger falls backward; R correctly explains A.

Newton's second law: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force, i.e. $F=ma$F=ma.

One newton is that force which produces an acceleration of $1\text{ m s}^{-2}$1 m s^-2 in a body of mass $1\text{ kg}$1 kg: $1\text{ N}=1\text{ kg}\times1\text{ m s}^{-2}$1 N=1 kg×1 m s^-2.

Change in momentum $=mv-mu=m(v-u)$=mv-mu=m(v-u).

$\Delta p=2\times(8-3)=2\times5=\boxed{10\text{ kg m s}^{-1}}$Δ p=2×(8-3)=2×5=10 kg m s^-1

Static inertia (inertia of rest) is the tendency of a body at rest to stay at rest. Example: when a carpet is beaten with a stick, the carpet moves but the dust stays at rest and falls off.

Dynamic inertia (inertia of motion) is the tendency of a moving body to keep moving with the same velocity. Example: when a moving bus stops suddenly, the passengers lurch forward.

Any three: (i) it keeps the Moon in orbit around the Earth and the planets around the Sun; (ii) it causes tides in the sea (the pull of the Moon and the Sun on the oceans); (iii) it gives every body its weight and makes objects fall to the ground; (iv) it holds the atmosphere around the Earth.

Given $m=500\text{ g}=0.5\text{ kg}$m=500 g=0.5 kg, $u=10\text{ m s}^{-1}$u=10 m s^-1, $v=0$v=0, $t=0.2\text{ s}$t=0.2 s.

(i) $\Delta p=mv-mu=0.5(0-10)=-5\text{ kg m s}^{-1}$Δ p=mv-mu=0.5(0-10)=-5 kg m s^-1, i.e. a change of magnitude $\boxed{5\text{ kg m s}^{-1}}$5 kg m s^-1.

(ii) $F=\dfrac{\Delta p}{t}=\dfrac{-5}{0.2}=-25\text{ N}$F=(Δ p)/t=-5/0.2=-25 N.

The force is $\boxed{25\text{ N}}$25 N opposing the motion.

Given $F=20\text{ N}$F=20 N, $m=4\text{ kg}$m=4 kg, $u=0$u=0, $t=3\text{ s}$t=3 s.

(i) $a=\dfrac{F}{m}=\dfrac{20}{4}=\boxed{5\text{ m s}^{-2}}$a=F/m=20/4=5 m s^-2.

(ii) $v=u+at=0+5\times3=\boxed{15\text{ m s}^{-1}}$v=u+at=0+5×3=15 m s^-1.

(iii) $s=ut+\dfrac12at^2=0+\dfrac12\times5\times3^2=\boxed{22.5\text{ m}}$s=ut+1/2at²=0+1/2×5×3²=22.5 m.

(i) $W=mg=5\times9.8=\boxed{49\text{ N}}$W=mg=5×9.8=49 N. Since $1\text{ kgf}=9.8\text{ N}$1 kgf=9.8 N, $W=\dfrac{49}{9.8}=\boxed{5\text{ kgf}}$W=49/9.8=5 kgf.

(ii)

(i) Any two: they act without any physical contact, across a distance; they can be attractive or repulsive (gravitational force is always attractive); their strength decreases as the distance increases, varying inversely as the square of the distance ($F\propto\dfrac1{r^2}$F∝ 1/r²).

(ii) $F\propto\dfrac1{r^2}$F∝ 1/r², so tripling $r$r makes the force $\dfrac1{3^2}=\dfrac19$1/3²=1/9 of its value: $F=\dfrac{36}{9}=\boxed{4\text{ N}}$F=36/9=4 N.

(i) To every action there is an equal and opposite reaction. Action and reaction act on two different bodies.

(ii) (a) Action: the book pushes down on the table with a force equal to its weight. Reaction: the table pushes up on the book with an equal normal reaction.

(b) Action: the swimmer's hands push the water backward. Reaction: the water pushes the swimmer forward.

(iii) They act on two different bodies, so they never act together on the same body and cannot cancel. Forces can cancel only when they act on the same body.

(iv) Contact forces: friction, tension in a string. Non-contact forces: magnetic force, gravitational force.

Law: Every particle of matter attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

$F=\dfrac{Gm_1m_2}{r^2}$F=Gm₁m₂/r²
where $m_1,m_2$m₁,m₂ are the masses, $r$r is the distance between their centres, and $G$G is the universal gravitational constant.

Calculation: $m_1=60\text{ kg}$m₁=60 kg, $m_2=80\text{ kg}$m₂=80 kg, $r=0.4\text{ m}$r=0.4 m.
$F=\dfrac{6.67\times10^{-11}\times60\times80}{(0.4)^2}=\dfrac{6.67\times10^{-11}\times4800}{0.16}$F=6.67×10^-11×60×80(0.4)²=6.67×10^-11×48000.16
$F=\dfrac{3.2016\times10^{-7}}{0.16}=\boxed{2.0\times10^{-6}\text{ N}}$F=3.2016×10^-70.16=2.0×10^-6 N

This extremely small force shows why gravitation between ordinary objects is not noticed.

$m=150\text{ g}=0.15\text{ kg}$m=150 g=0.15 kg, $u=20\text{ m s}^{-1}$u=20 m s^-1, final velocity $=0$=0.

(i) $p=mu=0.15\times20=\boxed{3\text{ kg m s}^{-1}}$p=mu=0.15×20=3 kg m s^-1.

(ii) $F=\dfrac{\Delta p}{t}=\dfrac{3}{0.5}=\boxed{6\text{ N}}$F=(Δ p)/t=3/0.5=6 N.

(iii) $F=\dfrac{3}{0.1}=\boxed{30\text{ N}}$F=3/0.1=30 N.

(iv) The same change in momentum spread over a longer time requires a smaller force ($F=\Delta p/t$F=Δ p/t). By lowering his hands the player increases the stopping time, reducing the force on his hands and avoiding injury.