Laws of Motion MCQ Questions & Answers in Basic Physics | Physics

$$\eqalign{ & {\text{Tension in the wire, }}T = \left( {\frac{{2mM}}{{m + M}}} \right)g \cr & = \frac{{\frac{{{\text{Force}}}}{{{\text{Tension}}}}}}{{{\text{Area}}}} = \frac{{2mM}}{{A\left( {m + M} \right)}}g \cr & = \frac{{2\left( {m \times 2m} \right)g}}{{A\left( {m + 2m} \right)}} = \frac{{4{m^2}}}{{3mA}}g = \frac{{4mg}}{{3A}}\,\left( {M = 2\,m\,{\text{given}}} \right) \cr} $$

$$\eqalign{ & {\text{Let}}\,{m_1} + {m_2} = m = {\text{constant}}\,{\text{and}} \cr & {m_1} = x \Rightarrow {m_2} = m - x \cr & a = \frac{{x - \left( {m - x} \right)}}{m}g \cr & T = \frac{{2x\left( {m - x} \right)}}{m}g \Rightarrow T = \frac{m}{{2g}}\left( {{g^2} - {a^2}} \right) \cr} $$

When force is applied on $${m_1}$$ then $$T = {m_2}a$$   and when force is applied on $${m_2},$$  then $$T = {m_1}a.$$   Thus value of $$T$$ is different for each case. And it depends on whether the force is applied on $${m_1},$$ or $${m_2}.$$

$$\eqalign{ & {\text{Mass}}\left( m \right) = 0.3\,kg \Rightarrow F = m.a = - 15\,x \cr & a = - \frac{{15}}{{0.3}}x = \frac{{ - 150}}{3}x = - 50x \cr & a = - 50 \times 0.2 = 10\,m/{s^2} \cr} $$

Thrust on the satellite is the force with which the satellite moves upwards in space. It is given by
$$F = - u\frac{{dm}}{{dt}}$$
Here, initial velocity
$$u = v,$$   rate of change of mass $$\frac{{dm}}{{dt}} = \alpha v$$
As we know that,
$$F = - v\frac{{dm}}{{dt}} = - v\left( {\alpha v} \right) = - \alpha {v^2}$$
Acceleration $$ = \frac{F}{M} = - \frac{{\alpha {v^2}}}{M}$$

According to Newton's first law of motion, a body continues to be in a state of rest or of uniform motion, unless it is acted upon by an external force to change the state. Hence, Newton’s first law of motion is related to physical independence of force.

The apple will fall slightly away from the hand of his brother in the direction of motion of the train due to inertia of motion. When train is just going to stop, the boy and his brother slows down with train but the apple which is in free fall continue to move with the same speed and therefore, falls slightly away from the hand in the direction of motion of the train.

Due to reaction force by road

Maximum bearable tension in the rope $$T = 25 \times 10 = 250N$$
From the figure, $$T - mg = ma$$
or acceleration $$a = \frac{{T - mg}}{m}$$
Given, mass $$m = 20\,kg,$$
$$g = 10\,m/{s^2},$$
Taking, $$T = 250\,N$$
Hence, $$a = \frac{{250 - 20 \times 10}}{{20}}$$
$$ = \frac{{50}}{{20}} = 2.5\,m/{s^2}$$

Let upthrust of air be $${F_a}$$ then
For downward motion of balloon
$$\eqalign{ & {F_a} = mg - ma \cr & mg - {F_a} = ma \cr} $$
For upward motion $${F_a} - \left( {m - \Delta m} \right)g = \left( {m - \Delta m} \right)a$$
Therefore $$\Delta m = \frac{{2ma}}{{g + a}}$$