Laws of Motion — ICSE Class 9 Physics Important Questions

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=Gm_1m_2/r^2, so $F\propto\dfrac{1}{r^2}$F1/r^2. If $r\to\dfrac{r}{2}$r/2, then $F\to\dfrac{1}{(1/2)^2}F=4F$F1/(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

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.

By conservation of momentum (both at rest initially, total momentum $=0$=0):
$m_bv_b+m_gv_g=0$m_bv_b+m_gv_g=0

$m_b=20\text{ g}=0.02\text{ kg}$m_b=20 g=0.02 kg, $v_b=200\text{ m s}^{-1}$v_b=200 m s^-1, $m_g=5\text{ kg}$m_g=5 kg.

$v_g=-\dfrac{m_bv_b}{m_g}=-\dfrac{0.02\times200}{5}=-\dfrac{4}{5}=\boxed{-0.8\text{ m s}^{-1}}$v_g=-m_bv_b/m_g=-0.02×200/5=-4/5=-0.8 m s^-1

The gun recoils at $0.8\text{ m s}^{-1}$0.8 m s^-1 in the direction opposite to the bullet.

Take the direction of the $3\text{ kg}$3 kg body as positive: $u_1=+4\text{ m s}^{-1}$u_1=+4 m s^-1, $u_2=-6\text{ m s}^{-1}$u_2=-6 m s^-1.

By conservation of momentum, $m_1u_1+m_2u_2=(m_1+m_2)v$m_1u_1+m_2u_2=(m_1+m_2)v.
$3(4)+2(-6)=(3+2)v$3(4)+2(-6)=(3+2)v
$12-12=5v\Rightarrow v=\dfrac{0}{5}=\boxed{0\text{ m s}^{-1}}$12-12=5v v=0/5=0 m s^-1

The combined body is momentarily at rest after the collision.

Newton's third law: To every action there is an equal and opposite reaction; the action and reaction act on two different bodies.

Derivation: Consider two bodies A and B of masses $m_1,m_2$m_1,m_2 moving with velocities $u_1,u_2$u_1,u_2 that collide and then move with $v_1,v_2$v_1,v_2. During the collision A exerts force $F_{AB}$F_AB on B and B exerts $F_{BA}$F_BA on A; by the third law $F_{AB}=-F_{BA}$F_AB=-F_BA for the same contact time $t$t.

Impulse on B $=F_{AB}\,t=m_2v_2-m_2u_2$=F_AB\,t=m_2v_2-m_2u_2.
Impulse on A $=F_{BA}\,t=m_1v_1-m_1u_1$=F_BA\,t=m_1v_1-m_1u_1.

Adding, since $F_{AB}+F_{BA}=0$F_AB+F_BA=0:
$(m_1v_1-m_1u_1)+(m_2v_2-m_2u_2)=0$(m_1v_1-m_1u_1)+(m_2v_2-m_2u_2)=0
$\therefore m_1u_1+m_2u_2=m_1v_1+m_2v_2.$m_1u_1+m_2u_2=m_1v_1+m_2v_2.

So the total momentum before collision equals the total momentum after collision.

Example: The recoil (backward kick) of a gun when a bullet is fired, and the forward motion of a rocket as hot gases are pushed backward.

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=Gm_1m_2/r^2
where $m_1,m_2$m_1,m_2 are the masses, $r$r is the distance between their centres, and $G$G is the universal gravitational constant.

Calculation: $m_1=60\text{ kg}$m_1=60 kg, $m_2=80\text{ kg}$m_2=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.