Kinematics MCQ Questions & Answers in Basic Physics | Physics

As the time taken from $$D$$ to $$A = 2\,\sec .$$   and $$D \to A \to B \to C = 10\,\sec \,\left( {{\text{given}}} \right).$$
As ball goes from $$B \to C\left( {u = 0,t = 4\,\sec } \right)$$
$${v_c} = 0 + 4g.$$
As it moves from $$C$$ to $$D,$$ $$s = ut + \frac{1}{2}g{t^2}$$
$$s = 4g \times 2 + \frac{1}{2}g \times 4 = 10\,g.$$

Range of projectile is given by
$$R = \frac{{{u^2}\sin 2\theta }}{g}$$
For range to be maximum, angle $$\theta $$ should be of $${45^ \circ }.$$
$$\eqalign{ & \therefore {R_{\max }} = \frac{{{u^2}\sin 2 \times {{45}^ \circ }}}{g} = \frac{{{u^2}\sin {{90}^ \circ }}}{g} \cr & {\text{or}}\,{R_{\max }} = \frac{{{u^2}}}{g} \cr & {\text{Here,}}\,{R_{\max }} = \frac{{{u^2}}}{g} = 16\,km = 16000\,m \cr & {\text{or}}\,u = \sqrt {16000\,g} = \sqrt {16000 \times 10} = 400\,m{s^{ - 1}} \cr} $$

The distance covered in $$n$$^th second is
$${S_n} = u + \frac{1}{2}\left( {2n - 1} \right)a$$
where $$u$$ is Initial velocity $$\& \,a$$  is acceleration
$$\eqalign{ & {\text{then}}\,26 = u + \frac{{19a}}{2}\,......\left( {\text{i}} \right) \cr & 28 = u + \frac{{21a}}{2}\,......\left( {{\text{ii}}} \right) \cr & 30 = u + \frac{{23a}}{2}\,......\left( {{\text{iii}}} \right) \cr & 32 = u + \frac{{25a}}{2}\,......\left( {{\text{iv}}} \right) \cr} $$
From eqs. (i) and (ii) we get
$$u = 7\,m/\sec ,a = 2\,m/{\sec ^2}$$
$$\therefore $$ The body starts with initial velocity $$u = 7\,m/\sec $$
and moves with uniform acceleration $$a = 2m/{\sec ^2}$$

Given: mass of car $$m = 1000\,kg,u = 30\,m/s;v = 0\,m/s$$
retardation, $$ - a = 5\,m/{s^2}$$
By equation, $${v^2} - {u^2} = 2as$$
$$\eqalign{ & 0 - {\left( {30} \right)^2} = - 2 \times 5 \times d \cr & \therefore d = \frac{{900}}{{10}} = 90\,m \cr} $$
and $$a = \frac{{v - u}}{t}\therefore t = \frac{{v - u}}{a} = \frac{{0 - 30}}{{ - 5}} = 6s$$

$$\eqalign{ & x + {u_2}\cos {\theta _2}t = {u_1}\cos {\theta _1}t \cr & \therefore t = \frac{x}{{{u_1}\cos {\theta _1} - {u_2}\cos {\theta _2}}}\,......\left( {\text{i}} \right) \cr} $$
Also $${u_1}\sin {\theta _1} = {u_2}\sin {\theta _2}\,......\left( {{\text{ii}}} \right)$$
After solving above equations, we get
$$t = \frac{{x\sin {\theta _2}}}{{{u_1}\sin \left( {{\theta _2} - {\theta _1}} \right)}}.$$

$$\eqalign{ & x = 3{t_1} = 5{t_2} \Rightarrow {t_1} = \frac{x}{3}\,\,{\text{and}}\,\,{t_2} = \frac{x}{5} \cr & {\text{Now}}\,\,{h_1} = 4{t_1} - \frac{1}{2}gt_1^2 = \frac{4}{3}x - \frac{{g{x^2}}}{{18}} \cr & {\text{and}}\,\,{h_2} = 4{t_2} - \frac{1}{2}gt_2^2 = \frac{{4x}}{5} - \frac{{g{x^2}}}{{10}} \cr & {\text{Clearly,}}\,\,{h_2} < {h_1}. \cr} $$

$$\eqalign{ & \vec u = 3\hat i + 4\hat j \Rightarrow ux = 3\,\,{\text{and}}\,\,uy = 4 \cr & \vec a = 0.4\hat i + 0.3\hat j \Rightarrow ax = 0.4\,\,{\text{and}}\,\,ay = 0.3 \cr & {\text{So, }}{V_x} = ux + {a_x}t = 3 + 0.4 \times 10 = 3 + 4 = 7 \cr & {V_y} = uy + {a_y}t = 4 + 0.3 \times 10 = 4 + 3 = 7 \cr & \vec V = 7\hat i + 7\hat j \cr & \Rightarrow \left| {\vec V} \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,units \cr} $$

For A:
It goes up with velocity $$u$$ will it reaches its maximum height (i.e. velocity becomes zero) and comes back to $$O$$ and attains velocity $$u.$$
Using $${v^2} = {u^2} + 2as \Rightarrow {v_A} = \sqrt {{u^2} + 2gh} $$

For B : going down with velocity $$u$$
$$ \Rightarrow {v_B} = \sqrt {{u^2} + 2gh} $$
For C : horizontal velocity remains same, i.e. $$u.$$
Vertical velocity $$ = \sqrt {0 + 2gh} = \sqrt {2gh} $$
The resultant $${v_C} = \sqrt {v_x^2 + v_y^2} = \sqrt {{u^2} + 2gh} .$$
Hence $${v_A} = {v_B} = {v_C}$$

For two particles $$A$$ and $$B$$ move with constant velocities $${v_1}$$ and $${v_2}.$$  Such that two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle.
i.e. $$\frac{{{r_1} - {r_2}}}{{\left| {{r_1} - {r_2}} \right|}} \to $$   direction of relative position of 1 w.r.t. 2.
Similarly $$\frac{{{v_1} - {v_2}}}{{\left| {{v_1} - {v_2}} \right|}} \to $$   direction of velocity of 2 w.r.t. 1.
So, for collision of $$A$$ and $$B,$$ we get
$$\frac{{{r_1} - {r_2}}}{{\left| {{r_1} - {r_2}} \right|}} = \frac{{{v_2} - {v_1}}}{{\left| {{v_2} - {v_1}} \right|}}$$
Alternate Method
As resultant displacement of a particle,

$$R = {r_1} - {v_1}t = {r_2} - {v_2}t$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,{r_1} - {r_2} = \left( {{v_2} - {v_1}} \right)t \cr & {\text{So,}}\,\,\frac{{{r_1} - {r_2}}}{{\left| {{r_1} - {r_2}} \right|}} = \frac{{\left( {{v_2} - {v_1}} \right)t}}{{\left| {{v_2} - {v_1}} \right|t}} \cr & \frac{{{r_1} - {r_2}}}{{\left| {{r_1} - {r_2}} \right|}} = \frac{{\left( {{v_2} - {v_1}} \right)}}{{\left| {{v_2} - {v_1}} \right|}} \cr} $$

$$\eqalign{ & s = {t^3} + 5 \Rightarrow {\text{velocity}},\,\,v = \frac{{ds}}{{dt}} = 3{t^2} \cr & {\text{Tangetial acceleration }}{a_t} = \frac{{dv}}{{dt}} = 6t \cr & {\text{Radial acceleration }}{a_c} = \frac{{{v^2}}}{R} = \frac{{9{t^4}}}{R} \cr & {\text{At }}t = 2\,s,\,\,\,\,{a_t} = 6 \times 2 = 12\,m/{s^2} \cr & {a_c} = \frac{{9 \times 16}}{{20}} = 7.2\,\,m/{s^2} \cr & \therefore {\text{Resultant acceleration}} \cr & = \sqrt {a_t^2 + a_c^2} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} \cr & = \sqrt {144 + 51.84} \cr & = \sqrt {195.84} \cr & = 14\,m/{s^2} \cr} $$