Kinematics MCQ Questions & Answers in Basic Physics | Physics

Given, $$\left| {A \times B} \right| = \sqrt 3 \left( {A \cdot B} \right)$$
$$\eqalign{ & \Rightarrow AB\sin \theta = \sqrt 3 AB\cos \theta \cr & \Rightarrow \tan \theta = \sqrt 3 \cr & \therefore \theta = {60^ \circ } \cr} $$

Let $$v$$ be the maximum velocity attained and $$t$$ the total time of journey. $$t'$$ is the duration of acceleration and retardation. Then $$v = 0 + at'.$$
$$\eqalign{ & \therefore L = \frac{1}{2}a{{t'}^2} + v\left( {t - 2t'} \right) + \frac{1}{2}a{{t'}^2} \cr & = a{\left( {\frac{v}{a}} \right)^2} + v\left( {t - \frac{{2v}}{a}} \right) \cr & = \frac{{{v^2}}}{a} + vt - \frac{{2{v^2}}}{a} = vt - \frac{{{v^2}}}{a} \cr & t = \frac{L}{v} + \frac{v}{a} \Rightarrow \frac{{dt}}{{dv}} = - \frac{L}{{{v^2}}} + \frac{1}{a} \cr} $$
When $$t$$ is minimum, $$\frac{{dt}}{{dv}} = 0$$
$$\therefore {v_{\max }} = \sqrt {La} $$

Suppose $$u$$ is the speed of the boat relative to water, then velocity of the flow (w.r.t bank) $${V_0}$$
$${v_x} = \left( {u\cos \theta + {v_0}} \right)$$    and perpendicular to flow will be $${v_y} = u\sin \theta .$$   Time to cross the river, $$t = \frac{b}{{u\sin \theta }}.$$    In the time the distance travelled by the boat in the direction of flow
$$\eqalign{ & a = {v_x}t = \left( {u\cos \theta + {v_0}} \right)\frac{b}{{u\sin \theta }} \cr & {\text{or}}\,au\sin \theta = ub\cos \theta + {v_0}b \cr & \therefore u = \frac{{{v_0}b}}{{\left( {a\sin \theta - b\cos \theta } \right)}}.....\left( {\text{i}} \right)\,u\,{\text{to}}\,{\text{be}}\,{\text{minimum}}\,{\text{duld}} \cr & \theta = 0\,{\text{or}}\,\frac{d}{{d\theta }}\left[ {\frac{{{v_0}b}}{{a\sin \theta - b\cos \theta }}} \right] = 0 \cr & {\text{or}}\,\tan \theta = - \frac{a}{b} \cr & \therefore \cos \theta = \frac{b}{{\sqrt {{a^2} + {b^2}} }} \cr} $$
On substituting these values in equation (i), we get
$${u_{\min }} = \frac{{{V_0}b}}{{\sqrt {{a^2} + {b^2}} }}$$

Given, horizontal range $$R =$$  vertical maximum height $$H$$
Range, $$R = \frac{{{u^2}\left( {2\sin \theta \cos \theta } \right)}}{g}$$
Height, $$H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$$
Hence, $$\frac{{{u^2}\left( {2\sin \theta \cos \theta } \right)}}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$$
$$\eqalign{ & 2\cos \theta = \frac{{\sin \theta }}{2} \cr & \tan \theta = 4 \Rightarrow \theta = {\tan ^{ - 1}}\left( 4 \right) \cr} $$

Initial velocity of car $$\left( u \right) = 0$$
Final velocity of car $$\left( v \right) = 144\,km/hr = 40\,m/s$$
Time taken $$ = 20\,s$$
We know that, $$v = u + at$$
$$\eqalign{ & 40 = a \times 20 \Rightarrow a = 2\,m/{s^2} \cr & {\text{Also,}}\,\,{v^2} - {u^2} = 2as \Rightarrow s = \frac{{{v^2} - {u^2}}}{{2a}} \cr & \Rightarrow s = \frac{{{{\left( {40} \right)}^2} - {{\left( 0 \right)}^2}}}{{2 \times 2}} = \frac{{1600}}{4} = 400\,m. \cr} $$

peed to cover $$1200\,m$$  by scootarist $${v_r} \times 60 = 1200 \Rightarrow {v_r} = 20$$
speed to overtake bus $$v = {v_r} + 10 = 30\,m/s$$

distance travelled $$ = {\text{area}}\,\,\square OABC$$
$$ = \frac{1}{2} \times \left( {OC + AB} \right) \times AD$$

$$\eqalign{ & \vec L = m\left( {\vec r \times \vec v} \right) \cr & \vec L = m\left[ {{v_0}\,\cos \theta \,t\,\hat i + \left( {{v_0}\sin \theta \,t - \frac{1}{2}g{t^2}\,} \right)\hat j} \right] \times \left[ {{v_0}\,\cos \theta \,\hat i + \left( {{v_0}\sin \theta - gt\,} \right)\hat j} \right] \cr & \vec L = m{v_0}\,\cos \theta \,t\left[ { - \frac{1}{2}gt} \right]\hat k \cr & \vec L = - \frac{1}{2}\,mg\,{v_0}\,{t^2}\cos \theta \,\hat k \cr} $$

Let $$\vec c = \hat a + 2\hat b$$   and $$\vec d = 5\hat a - 4\hat b$$
Since $${\vec c}$$ and $${\vec d}$$ are perpendicular to each other
$$\eqalign{ & \therefore \vec c.\vec d = 0 \Rightarrow \left( {\hat a + 2\hat b} \right).\left( {5\hat a - 4\hat b} \right) = 0 \cr & \Rightarrow 5 + 6\hat a.\hat b - 8 = 0\,\,\left( {\because \vec a.\vec a = 1} \right) \cr & \Rightarrow \hat a.\hat b = \frac{1}{2} \Rightarrow \theta = \frac{\pi }{3} \cr} $$

$$\eqalign{ & {\text{For first ball, }}u = 0 \cr & \therefore {s_1} = \frac{1}{2}gt_1^2 = \frac{1}{2} \times g{\left( {18} \right)^2} \cr & {\text{For second ball, initial velocity}} = v \cr & \therefore {s_2} = v{t_2} + \frac{1}{2}g{t^2} \cr & {t_2} = 18 - 6 = 12\;s \cr & \Rightarrow {s_2} = v \times 12 + \frac{1}{2}g{\left( {12} \right)^2} \cr & {\text{Here,}}\,{s_1} = {s_2} \cr & \frac{1}{2}g{\left( {18} \right)^2} = 12v + \frac{1}{2}g{\left( {12} \right)^2} \cr & \Rightarrow v = 74\;m{s^{ - 1}} \cr} $$