Kinematics MCQ Questions & Answers in Basic Physics | Physics

Given, $$v = \beta {x^{ - 2n}}$$
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{{dx}}{{dt}} \cdot \frac{{dv}}{{dx}} \cr & \Rightarrow a = v\frac{{dv}}{{dx}} = \left( {\beta {x^{ - 2n}}} \right)\left( { - 2n\beta {x^{ - 2n - 1}}} \right) \cr & \Rightarrow a = - 2n{\beta ^2}{x^{ - 4n - 1}} \cr} $$

Let the two bodies meet each other at a height $$h$$ after time $$T$$ of the projection of second body. Then before meeting, the first body was in motion for time $$\left( {t + T} \right)$$  whereas the second body was in motion for time $$T.$$
The distance moved by the first body in time $$\left( {t + T} \right)$$
$$ = u\left( {t + T} \right) - \frac{1}{2}g{\left( {t + T} \right)^2}.$$
And the distance moved by the second body in time
$$T = vT - \frac{1}{2}g{T^2} = h\,\,\left( {{\text{supposed above}}} \right).\,......\left( 1 \right)$$
$$\because $$ The two bodies meet each other,
$$\therefore $$ They are equidistant from the point of projection.
$$\eqalign{ & {\text{Hence,}}\,\,\,u\left( {t + T} \right) - \frac{1}{2}g{\left( {t + T} \right)^2} = vT - \frac{1}{2}g{T^2} \cr & {\text{or}}\,\,g{t^2} + 2t\left( {gT - u} \right) + 2\left( {v - u} \right)T = 0\,......\left( 2 \right) \cr} $$
Also from (1) we get,
$$h = vT - \frac{1}{2}g{T^2}$$
$$\therefore \frac{{dh}}{{dT}} = v - gT$$
$$\therefore $$ $$h$$ increases as $$T$$ increases
$$\therefore $$ $$T$$ is minimum when $$h$$ is minimum i.e., when
$$\frac{{dh}}{{dT}} = 0,\,\,{\text{i}}{\text{.e}}{\text{.}}\,{\text{when}}\,\,v - gT = 0\,\,{\text{or}}\,\,T = \frac{v}{g}.$$
Substituting this value of $$T$$ in (2), we get
$$\eqalign{ & g{t^2} + 2t\left( {v - u} \right) + 2\left( {v - u} \right)\left( {\frac{v}{g}} \right) = 0 \cr & {\text{or}}\,\,t = \frac{{2g\left( {u - v} \right) + \sqrt {4{g^2}{{\left( {u - v} \right)}^2} + 8v{g^2}\left( {u - v} \right)} }}{{2{g^2}}} \cr & {\text{or}}\,\,t = \frac{{u - v + \sqrt {{u^2} - {v^2}} }}{g} \cr} $$
neglecting the negative sign which gives negative value of $$t.$$

Given, distance $$x = {\left( {t + 5} \right)^{ - 1}}......\left( {\text{i}} \right)$$
Differentiating Eq. (i) w.r.t. $$t,$$ we get
$$\frac{{dx}}{{dt}} = \left( v \right) = \frac{{ - 1}}{{{{\left( {t + 5} \right)}^2}}}\,......\left( {{\text{ii}}} \right)$$
Again, differentiating Eq. (ii) w.r.t. $$t,$$ we get
$$\frac{{{d^2}x}}{{d{t^2}}} = \left( a \right) = \frac{2}{{{{\left( {t + 5} \right)}^3}}}\,......\left( {{\text{iii}}} \right)$$
Comparing Eqs. (ii) and (iii), we get $$\left( a \right) \propto {\left( v \right)^{\frac{3}{2}}}$$

Speed is a scalar quantity. It gives no idea about the direction of motion of the object. Velocity is a vector quantity, as it has both magnitude and direction. Displacement is a vector as it possesses both magnitude and direction. When an object goes on the path $$ABC$$  (in figure), then the displacement of the object is $$AC.$$  The arrow head at $$C$$ shows that the object is displaced from $$A$$ to $$C.$$

Torque is turning effect of force which is a vector quantity.

$$\eqalign{ & {h_1} = u\sin \theta {t_1} + \frac{1}{2}gt_1^2; \cr & {h_2} = u\sin \theta {t_2} + \frac{1}{2}gt_2^2 \cr & {\text{So,}}\,\,\frac{{{t_1}}}{{{t_2}}} = \frac{{{h_1} + \frac{1}{2}gt_1^2}}{{{h_2} + \frac{1}{2}gt_2^2}} \cr & \Rightarrow {h_1}{t_2} - {h_2}{t_1} = \frac{1}{2}g\left( {{t_1}t_2^2 - t_1^2{t_2}} \right) \cr & {\text{Time of flight}} = \frac{{2u\sin \theta }}{g} = \frac{{{h_1}t_2^2 - {h_2}t_1^2}}{{{h_1}{t_2} - {h_2}{t_1}}}\,\left[ {{\text{Use above eqn to simplify}}} \right] \cr} $$

$$\eqalign{ & {S_n}th = u + \frac{a}{2}\left( {2n - 1} \right) \cr & {\text{so,}}\,\,\frac{{{S_4}}}{{{S_5}}} = \frac{{0 + \frac{g}{2}\left( 7 \right)}}{{0 + \frac{g}{2}\left( 9 \right)}} = \frac{7}{9} \cr} $$

The relation between linear velocity $$v,$$ angular velocity $$\omega $$ and position vector $$r$$ is given by
$$\eqalign{ & v = \omega \times r \cr & = \left( {3\hat i - 4\hat j + \hat k} \right) \times \left( {5\hat i - 6\hat j + 6\hat k} \right) \cr} $$
\[\begin{array}{l} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&{ - 4}&1\\ 5&{ - 6}&6 \end{array}} \right|\\ = \hat i\left| {\begin{array}{*{20}{c}} { - 4}&1\\ { - 6}&6 \end{array}} \right| - \hat j\left| {\begin{array}{*{20}{c}} 3&1\\ 5&6 \end{array}} \right| + \hat k\left| {\begin{array}{*{20}{c}} 3&{ - 4}\\ 5&{ - 6} \end{array}} \right| \end{array}\]
$$\eqalign{ & = \left( { - 24 + 6} \right)\hat i - \left( {18 - 5} \right)\hat j + \left( { - 18 + 20} \right)\hat k \cr & = - 18\hat i - 13\hat j + 2\hat k \cr} $$
Alternative
$$\eqalign{ & v = \omega \times r \cr & = \left( {3\hat i - 4\hat j + \hat k} \right) \times \left( {5\hat i - 6\hat j + 6\hat k} \right) \cr & = \left( {3 \times 5} \right)\left( {\hat i \times \hat i} \right) + \left[ {3 \times \left( { - 6} \right)} \right]\left( {\hat i \times \hat j} \right) + \left( {3 \times 6} \right)\left( {\hat i \times \hat k} \right) + \left( { - 4 \times 5} \right)\left( {\hat j \times \hat i} \right) + \left( { - 4 \times - 6} \right)\left( {\hat j \times \hat j} \right) + \left( { - 4 \times 6} \right)\left( {\hat j \times \hat k} \right) + \left( {1 \times 5} \right)\left( {\hat k \times \hat i} \right) + \left( {1 \times - 6} \right)\left( {\hat k \times \hat j} \right) + \left( {1 \times 6} \right)\left( {\hat k \times \hat k} \right) \cr & {\text{Use}}\,\,\hat i \times \hat j = - \hat j \times \hat i = \hat k \cr & \hat j \times \hat k = - \hat k \times \hat j = \hat i \cr & {\text{and}}\,\,\hat k \times \hat i = - \hat i \times \hat k = \hat j \cr & {\text{Thus,}}\,v = 0 + \left( { - 18} \right)\left( {\hat k} \right) + \left( {18} \right)\left( { - \hat j} \right) + \left( { - 20} \right)\left( { - \hat k} \right) + 0 + \left( { - 24} \right)\left( {\hat i} \right) + \left( 5 \right)\left( {\hat j} \right) + \left( { - 6} \right)\left( { - \hat i} \right) + 0 \cr & = - 18\hat k - 18\hat j + 20\hat k - 24\hat i + 5\hat j + 6\hat i \cr & = - 18\hat i - 13\hat j + 2\hat k \cr} $$

Given : $$u = 0,t = 5\,\sec ,v = 108\,km/hr = 30\,m/s$$
By equation of motion $$v = u + at$$
$$\eqalign{ & {\text{or}}\,\,a = \frac{v}{t} = \frac{{30}}{5} = 6\,m/{s^2}\left[ {\because u = 0} \right] \cr & {S_1} = \frac{1}{2}a{t^2} = \frac{1}{2} \times 6 \times {5^2} = 75\,m \cr} $$
Distance travelled in first $$5\,\sec $$  is $$75\,m.$$
Distance travelled with uniform speed of $$30\,m/s$$  is $${S_2}$$
$$\eqalign{ & 395 = {S_1} + {S_2} + {S_3} \Rightarrow 395 = 75 + {S_2} + 45 \cr & \Rightarrow {S_2} = 275\,m \cr} $$
Time taken to travel $$275\,m = \frac{{275}}{{30}} = 9.2\,\sec $$
For retarding motion, we have
$$\eqalign{ & {0^2} - {32^2} = 2\left( { - a} \right) \times 45,\,{\text{we}}\,{\text{get}}\,a = 10\,m/{s^2} \cr & S = ut + \frac{1}{2}a{t^2} \Rightarrow 45 = 30t + \frac{1}{2}\left( { - 10} \right){t^2} \cr & \Rightarrow 45 = 30t - 5{t^2} \cr} $$
On solving we get, $$t = 3\,\sec $$
Total time taken $$ = 5 + 9.2 + 3 = 17.2\,\sec .$$

$$\eqalign{ & 150 = 100t + \frac{1}{2}10\,{t^2};t = 10\,\sec \cr & x = \frac{{400}}{3}t = \frac{{4000}}{3}m \cr} $$

$$\eqalign{ & {13^2} = {12^2} + {5^2} + 2 \times 12 \times 5\cos \theta \cr & \therefore \theta = {90^ \circ } \cr} $$