Kinematics MCQ Questions & Answers in Basic Physics | Physics

In figure shown, $$A + B = C$$

$$\eqalign{ & {\text{Also,}}\,\left| A \right| = 3,\,\left| B \right| = 4,\,\left| C \right| = 5 \cr & {\text{As,}}\,A + B = C \cr & {\text{and}}\,{\left| C \right|^2} = {\left| A \right|^2} + {\left| B \right|^2} + 2\left| A \right|\left| B \right|\cos \theta \cr & {\text{So,}}\,{5^2} = {3^2} + {4^2} + 2 \cdot 4 \cdot 3\cos \theta \cr & \cos \theta = 0 \Rightarrow \theta = \frac{\pi }{2} \cr} $$
$$ \Rightarrow A$$  is perpendicular to $$B.$$

Let $$u$$  be the speed with which the ball of mass $$m$$  is projected. Then the kinetic energy $$\left( E \right)$$  at the point of projection is-

$$E = \frac{1}{2}m{u^2}\,.....(i)$$
When the ball is at the highest point of its flight, the speed of the ball is $$\frac{u}{{\sqrt 2 }}$$  (Remember that the horizontal component of velocity does not change during a projectile motion).
$$\therefore $$   The kinetic energy at the highest point
$$\eqalign{ & = \frac{1}{2}m{\left( {\frac{u}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}\frac{{m{u^2}}}{2} \cr & = \frac{E}{2}\,\,\left[ {{\text{From (i)}}} \right] \cr} $$

Distance travelled in $${n^{th}}$$ second is given by
$${s_n} = u + \frac{1}{2}a\left( {2n - 1} \right)$$
Here, $$u = 0,\,a = \frac{4}{3}$$
$$\therefore {s_3} = 0 + \frac{1}{2} \times \frac{4}{3} \times \left( {6 - 1} \right) = \frac{{10}}{3}m$$

By the concept of centrifugal force cream is separated from milk. A mass $$m$$ of milk revolving at a distance $$r$$ from the axis of rotation of the centrifuge requires a centripetal force $$mr{\omega ^2},$$  where $$\omega $$ is the angular speed of the centrifuge. If in place of this mass of milk, lighter particles of mass (cream) $$m'\left( {m' < m} \right)$$   are present, then the centripetal force $$\left( {mr{\omega ^2}} \right)$$  on the milk will be greater than the centripetal force $$\left( {m'r{\omega ^2}} \right)$$  on the cream.
As a result, cream move towards the axis of rotation under the effect of the net force $$\left( {m - m'} \right)r{\omega ^2}.$$   When the centrifuge is stopped, the cream is found at the top and milk at the bottom.

Since displacement is zero.

Speed along the shortest path $$ = \frac{1}{{\frac{{15}}{{60}}}} = 4\,km/hr$$

Speed of water $$v = \sqrt {{5^2} - {4^2}} = 3\,km/hr$$

As when they collide
$$\eqalign{ & vt + \frac{1}{2}\left( {\frac{{72{v^2}}}{{25\pi R}}} \right){t^2} - \pi R = vt \cr & \therefore t = \frac{{5\pi R}}{{6v}} \cr} $$
Now, angle covered by $$A = \pi + \frac{{vt}}{R}$$
$${\text{Put}}\,t,\therefore {\text{angle covered by }}A = \frac{{11\pi }}{6}$$

Total time of motion is $$3\,\min \,20\sec = 20\,\sec .$$     As time period of circular motion is $$40\,\sec $$  so in $$20\,\sec $$  athlete will complete 5 revolution i.e., he will be at starting point i.e., displacement = zero.

$$\eqalign{ & a = r\,{w^2} = r\, \times {\left( {\frac{{2\pi }}{T}} \right)^2} \cr & \therefore \frac{{{a_1}}}{{{a_2}}} = \frac{{{r_1}}}{{{r_2}}}\,\,\left[ {\because T{\text{ is same}}} \right] \cr} $$

$${V_h} = V\cos \theta $$
where $$h$$ is the maximum height
$$\eqalign{ & {V_h} \times h = \left( {\upsilon \cos \theta } \right)\left( {\frac{{{\upsilon ^2}{{\sin }^2}\theta }}{{2g}}} \right) \cr & = \frac{{{\upsilon ^3}{{\sin }^2}\theta \cos \theta }}{{2g}} \cr & = \frac{{\sqrt 3 {\upsilon ^3}}}{{16\,g}} \cr} $$