Kinematics MCQ Questions & Answers in Basic Physics | Physics

From equation of motion time taken by ball to reach maximum height $$v = u - gt$$
At maximum height, final speed is zero i.e. $$v = 0$$
$$\eqalign{ & {\text{So,}}\,u = gt \cr & {\text{or}}\,t = \frac{u}{g} \cr & {\text{In}}\,\,2s,\,u = 2 \times 9.8 = 19.6\,m/s \cr} $$
If man throws the ball with velocity of $$19.6\,m/s$$   then after $$2\,s$$ it will reach the maximum height. When he throws 2^nd ball, 1^st is at top. When he throws third ball, 1^st will come to ground and 2^nd will be at the top. Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than $$19.6\,m/s.$$

At the highest point $$v = 0.$$

$$\eqalign{ & {\text{Given,}}\,\left| {A \times B} \right| = \sqrt 3 A \cdot B\,......\left( {\text{i}} \right) \cr & {\text{but}}\,\left| {A \times B} \right| = \left| A \right|\left| B \right|\sin \theta = AB\sin \theta \cr & {\text{and}}\,A \cdot B = \left| A \right|\left| B \right|\cos \theta = AB\cos \theta \cr} $$
Substituting these values in Eq. (i), we get
$$\eqalign{ & AB\sin \theta = \sqrt 3 AB\cos \theta \cr & {\text{or}}\,\tan \theta = \sqrt 3 \cr & \therefore \theta = {60^ \circ } \cr} $$
The addition of vectors $$A$$ and $$B$$ can be given by the law of parallelogram.
$$\eqalign{ & \therefore \left| {A + B} \right| = \sqrt {{A^2} + {B^2} + 2AB\cos {{60}^ \circ }} \cr & = \sqrt {{A^2} + {B^2} + 2AB \times \frac{1}{2}} \cr & = {\left( {{A^2} + {B^2} + AB} \right)^{\frac{1}{2}}} \cr} $$

The situation is depicted in figure.

Here, $${v_1} = 50\,km/h$$   due North
$${v_2} = 50\,km/h$$   due West
From figure it indicates that angle between $${v_1}$$ and $${v_2}$$ is $${90^ \circ }.$$
Now, $$ - {v_1} = 50\,km/h$$    due South.
$$ \Rightarrow $$ Change in velocity $$ = \left| {{v_2} - {v_1}} \right| = \left| {{v_2} + \left( { - {v_1}} \right)} \right|$$
$$\eqalign{ & = \sqrt {v_2^2 + v_1^2} \cr & = \sqrt {{{50}^2} + {{50}^2}} \cr & = 70.7\,km/h \cr} $$
From figure it is clear that change in velocity is in South-West.

Alternative
As two vectors are $$ \bot $$ to each other.
So, resultant is given by $${C^2} = {A^2} + {B^2}$$
i.e., change in velocity $$ = \sqrt {v_1^2 + v_2^2} $$
$$\eqalign{ & = \sqrt {{{50}^2} + {{50}^2}} \cr & = 70.7\,km/h \cr} $$
The direction of resultant will be in South-West direction.

$$\eqalign{ & v = \sqrt {3x + 16} \Rightarrow {v^2} = 3x + 16 \cr & \Rightarrow {v^2} - 16 = 3x \cr} $$
Comparing with $${v^2} - {u^2} = 2aS,$$
we get, $$u$$ = 4 units, $$2a = 3$$  or $$a$$ = 1.5 units

Horizontal distance covered should be same for the time of collision.
$$\eqalign{ & 400\cos \theta = 200 \cr & {\text{or}}\,\,\cos \theta = \frac{1}{2} \cr & {\text{or}}\,\,\theta = {60^ \circ } \cr} $$

$$\eqalign{ & S = AB = \frac{1}{2}gt_1^2 \cr & \Rightarrow 2\,S = AC = \frac{1}{2}g{\left( {{t_1} + {t_2}} \right)^2} \cr & {\text{and}}\,\,3\,S = AD = \frac{1}{2}g{\left( {{t_1} + {t_2} + {t_3}} \right)^2} \cr & {t_1} = \sqrt {\frac{{2\,S}}{{\;g}}} \cr & {t_1} + {t_2} = \sqrt {\frac{{4\,S}}{{\;g}}} ,{t_2} = \sqrt {\frac{{4\;S}}{{\;g}}} - \sqrt {\frac{{2\;S}}{{\;g}}} \cr & {t_1} + {t_2} + {t_3} = \sqrt {\frac{{6\;S}}{{\;g}}} \cr & {t_3} = \sqrt {\frac{{6\;S}}{{\;g}}} - \sqrt {\frac{{4\;S}}{{\;g}}} \cr & {t_1}:{t_2}:{t_3}::1:\left( {\sqrt 2 - \sqrt 1 } \right):\left( {\sqrt 3 - \sqrt 2 } \right) \cr} $$

Acceleration, $$a = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}},$$
Velocity $$v = \frac{{dx}}{{dt}}.$$
The given equation is $$x = a{t^2} - b{t^3}$$
Velocity, $$v = \frac{{dx}}{{dv}} = 2at - 3b{t^2}$$
Acceleration $$a = \frac{{dv}}{{dt}} = 2a - 6bt$$
but $$a = 0$$  (given)
$$\eqalign{ & \therefore 2a - 6bt = 0 \cr & {\text{or}}\,\,6bt = 2a \cr & {\text{or}}\,\,t = \frac{{2a}}{{6b}} = \frac{a}{{3b}} \cr} $$

$$y = b{x^2}.$$   Differentiating w.r.t to $$t$$ an both sides, we get
$$\frac{{dy}}{{dx}} = b2x\frac{{dx}}{{dt}} \Rightarrow {v_y} = 2bx{v_x}$$
Again differentiating w.r.t to $$t$$ on both sides we get
$$\frac{{d{v_y}}}{{dt}} = 2b{v_x}\frac{{dx}}{{dt}} + 2bx\frac{{d{v_x}}}{{dt}} = 2bv_x^2 + 0$$
[$$\frac{{d{v_x}}}{{dt}} = 0,$$   because the particle has constant acceleration along $$y$$-direction]
$$\eqalign{ & {\text{Now,}}\,\,\frac{{d{v_y}}}{{dt}} = a = 2bv_x^2; \cr & v_x^2 = \frac{a}{{2\;b}} \Rightarrow {v_x} = \sqrt {\frac{a}{{2\;b}}} \cr} $$

$$\eqalign{ & \left| {\vec A \times \vec B} \right| = \sqrt 3 .\vec A.\vec B \cr & {\text{or}}\,\,AB\sin \theta = \sqrt 3 AB\cos \theta \cr & \therefore \tan \theta = \sqrt 3 ,\,\,{\text{or}}\,\,\theta = {60^ \circ } \cr & {\text{Thus}}\,\,\left| {\vec A + \vec B} \right| = \sqrt {{A^2} + {B^2} + 2AB\cos {{60}^ \circ }} \cr & = \sqrt {{A^2} + {B^2} + AB} . \cr} $$