Kinematics MCQ Questions & Answers in Basic Physics | Physics

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then $$u\cos {60^ \circ } = v\cos {45^ \circ }.$$
$$\eqalign{ & \Rightarrow 150 \times \frac{1}{2} = v \times \frac{1}{{\sqrt 2 }}\,\,{\text{or}}\,\,\frac{{150}}{{\sqrt 2 }}m/s \cr & {\text{Initially}}:\,{u_y} = u\sin {60^ \circ } = \frac{{150\sqrt 3 }}{{\sqrt 2 }}m/s \cr & {\text{Finally}}:{v_y} = v\sin {45^ \circ } = \frac{{150}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = \frac{{150}}{2}m/s \cr & {\text{But}}\,\,{v_y} = {u_y} + {a_y}t\,\,{\text{or}}\,\,\frac{{150}}{2} = \frac{{150\sqrt 3 }}{2} - 10t \cr & 10t = \frac{{150}}{2}\left( {\sqrt 3 - 1} \right)\,{\text{or}}\,t = 7.5\left( {\sqrt 3 - 1} \right)s \cr} $$

Horizontally after time $$t$$
$$\eqalign{ & v\cos \theta = v\cos \beta \,......\left( {\text{i}} \right)\,\,\left[ {\beta = {\text{angle with horizontal after time}}\,t} \right] \cr & {\text{Vertically}}, \cr & v\sin \theta - gt = v\sin \beta \,.......\left( {{\text{ii}}} \right) \cr} $$
Dividing equation $$\frac{{\left( {{\text{ii}}} \right)}}{{\left( {\text{i}} \right)}}$$ we get
$$\tan \beta = \frac{{v\sin \theta - gt}}{{v\cos \theta }} = \beta = {\tan ^{ - 1}}\left( {\frac{{v\sin \theta - gt}}{{v\cos \theta }}} \right)$$

In scalar triple product of vectors, the positions of dot and cross can be interchanged, i.e.
$$\eqalign{ & A \cdot \left( {B \times A} \right) = \left( {A \times B} \right) \cdot A = \left( {A \times A} \right) \cdot B \cr & {\text{but}}\,A \times A = 0 \cr & \therefore A \cdot \left( {B \times A} \right) = 0 \cr} $$
Alternative
$$\eqalign{ & A \cdot \left( {B \times A} \right) \cr & {\text{Let}}\,\,A \times B = C \cr} $$
The direction of $$C$$ is $$ \bot $$ to $$A$$ and $$B$$ from cross product formula
So, $$A \cdot C = 0$$     (since, $$A$$ and $$C$$ are $$ \bot $$ to each other)

$${v^2} = {u^2} + 2as\,\,\,400 > 2a\left( {100} \right);a < 2$$

In a circular motion, particle repeats its motion after equal intervals of time. So, particle moving on a circular path is periodic but not simple harmonic as it does not execute to and fro motion about a fixed point.

We have $${R^2} = {P^2} + {Q^2} + 2PQ\cos \theta \,......\left( {\text{i}} \right)$$
$${\text{and}}\,\,{S^2} = {P^2} + {Q^2} - 2PQ\cos \theta \,......\left( {{\text{ii}}} \right)$$
Adding equations (i) and (ii), we get
$${R^2} + {S^2} = 2\left( {{P^2} + {Q^2}} \right).$$

Let the ball takes $$T$$ second to reach maximum height $$H.$$
$$v = u - gT$$
put $$v = 0$$  (at height $$H$$ )
$$\eqalign{ & \therefore u = gT \cr & {\text{or}}\,T = \frac{u}{g}\,......\left( {\text{i}} \right) \cr} $$
Velocity attained by the ball in $$\left( {T - t} \right)s$$   is,
$$\eqalign{ & v' = u - g\left( {T - t} \right) = u - gT + gt \cr & = u - g\frac{u}{g} + gt = u - u + gt \cr & v' = gt\,.......\left( {{\text{ii}}} \right) \cr} $$
Hence, distance travelled in last $$t\,\sec$$  of its ascent
$$\eqalign{ & CB = v't - \frac{1}{2}g{t^2} = \left( {gt} \right)t - \frac{1}{2}g{t^2} \cr & = g{t^2} - \frac{1}{2}g{t^2}\,\,\left[ {{\text{From}}\,{\text{Eq}}{\text{.}}\,\left( {{\text{ii}}} \right)} \right] \cr & = \frac{1}{2}g{t^2} \cr} $$

From question,
Horizontal velocity (initial), $${u_x} = \frac{{40}}{2} = 20\,m/s$$
Vertical velocity (initial), $$50 = {u_y}t + \frac{1}{2}g{t^2}$$
$$\eqalign{ & \Rightarrow {u_y} \times 2 + \frac{1}{2}\left( { - 10} \right) \times 4;\,\,{\text{or,}}\,\,{u_y} = \frac{{70}}{2} = 35\,m/s \cr & \therefore \tan \theta = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4} \Rightarrow {\text{Angle}}\,\theta = {\tan ^{ - 1}}\frac{7}{4} \cr} $$

As magnitude of acceleration is same $$\frac{{30 - 15}}{t} = \frac{{20 - {V_Q}}}{t}$$
$$15 = 20 - {V_Q} \Rightarrow {V_Q} = 5\,m/s$$

Suppose the stream velocity is $${v_s} = v,$$  then the velocity of each boat with respect to water is $${v_b} = 1.2\,v.$$   Let each boat travel a distance $$\ell .$$ Then for boat $$A,$$ time of motion
$$\eqalign{ & {\tau _A} = \frac{\ell }{{{v_b} + {v_s}}} + \frac{\ell }{{{v_b} - {v_s}}} \cr & = \left[ {\frac{\ell }{{1.2v + v}} + \frac{\ell }{{1.2v - v}}} \right] = \frac{{60\ell }}{{11v}}\,......\left( {\text{i}} \right) \cr} $$
For the boat $$B,$$ time of motion
$$\eqalign{ & {\tau _B} = \frac{\ell }{{\sqrt {v_b^2 - v_s^2} }} + \frac{\ell }{{\sqrt {v_b^2 - v_s^2} }} = \frac{{2\ell }}{{\sqrt {v_b^2 - v_s^2} }}\,......\left( {{\text{ii}}} \right) \cr & = \frac{{2\ell }}{{\sqrt {{{\left( {1.2\,v} \right)}^2} - {v^2}} }} = \frac{{3.01\ell }}{v} \cr} $$
The ration $$\frac{{{\tau _A}}}{{{\tau _B}}} = \frac{{\left( {\frac{{60\ell }}{{11v}}} \right)}}{{\left( {\frac{{3.01\ell }}{v}} \right)}} \approx 1.8$$