Motion In A Straight Line Practice Questions

The juggler throws ${\text{n}}$ balls per second.$\therefore$ Interval between balls $=\;\frac{1}{n}$ seconds At maximum height velocity of first ball $v=0$Juggler throws all balls with same initial velocity $=u$For 1 st ball,$u_1=u, v_1=0, a=-g, S=H_{\max }$Using formula,$v^2=u^{2+}2 a s$ $0=u^{2-}2 g H_{\max }$ $\Rightarrow H_{\max }=\;\frac{u^2}{2 g} . . . . . . \;\text{ (1) }\;$ And using formula,$v=u+a t$ $\Rightarrow 0=u-g t$ $\Rightarrow t=\;\frac{u}{g}$ $\therefore$ Time taken by ball 1 to reach maximum height $(H_{\max }) =\;\frac{u}{g}$According to the question,$t=\;\frac{1}{n}$ $\Rightarrow \;\frac{u}{g}=\;\frac{1}{n}$ $\Rightarrow u=\;\frac{g}{n} . . . . . .(2)$ Putting value of $u$ in equation (1), we get$H_{\max }=\;\frac{ (\frac{g}{n}) ^2}{2 g}$ $=\;\frac{g^2}{2 n^2 g}$ $=\;\frac{g}{2 n^2}$

Based on the situation $h=-u t_1+\;\frac{1}{2} g t_1^2 \to \;\text{ throwing up ...... (i) }\;$ $h=u t_2+\;\frac{1}{2} g t_2^2 \to \;\text{ throwing up ...... (ii) }\;$ $h=\;\frac{1}{2} g t^2 \to \;\text{ dropping }\; . . . . . . . . . \;\text{ (iii) }\;$ $\;\text{ and }\; 0=u(t_1-t_2)-\;\frac{1}{2} g(t_1-t_2)^2 . . . . . . \;\text{ ( iv)}$solving above equations $t=\sqrt{t_1 t_2}$ $\Rightarrow t=\sqrt{6 \times 1.5}=3 {\text{s}}$

For first $\;\frac{h}{2}$ distance,$\;\frac{h}{2}=\;\frac{1}{2} g t_1^2$ $\Rightarrow h=g t_1^2$For total distance h,$h=\;\frac{1}{2} g (t_1+t_2) ^2$ $\Rightarrow g t_1^2=\;\frac{1}{2} g (t_1+t_2) ^2$ $\Rightarrow 2 t_1^2= (t_1+t_2) ^2$ $\Rightarrow {\sqrt{2}} t_1=t_1+t_2$ $\Rightarrow ({\sqrt{2}}-1) t_1=t_2$

Let us assume that they meet at ${\text{t}}= {\text{t}}_0$ $A: 80=\;\frac{1}{2} g t_0^2 . . . . . . \;\text{ (i) }\;$ $B: 80=u (t_0-2) +\;\frac{1}{2} g (t_0-2) ^2$From (i), $t_0=4$ $\Rightarrow 80=2 u+5(2)^2$ $\Rightarrow u=30 {\text{m}} / {\ s}$

${\text{t}}=\sqrt{ {\text{x}}}+4$ $\Rightarrow {\text{x}}=( {\text{t}}-4)^2= {\text{t}}^{2-}8 {\text{t}}+16$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{dt}}}=2 {\text{t}}-8$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{dt}}}{{}/{}|} _{ {\text{t}}=4}=2 \times 4-8=0$

$\;\frac{d v}{d x}=5 {ms}^{-1} / {m}$Acceleration of particle when $v=20 {m} / {s}$ $a=v \;\frac{d v}{d x}=20(5) {m} / {s}^{2}=100 {m} / {s}^{2}$

$|v_{10}|=(100+10 \times 10) {\text{m}} / {\text{s}}$ $v_{10}=-200 {\text{m}} / {\text{s}}$ and $v_0=-100 {\text{m}} / {\text{s}}$from 10 s to 20 s velocity remains zero$\Rightarrow$ from $t=0$ s to $10 {\text{s}}$ velocity increases in magnitude linearly.$\Rightarrow$ graph given in option A fits correctly.

${\text{t}}_{ {\text{a}}}=\;{ {\text{u}}}/{ {\text{g}}}=\;\frac{19.6}{9.8}=2 {\ s}$ ${\text{t}}_{ {\text{d}}}=6-2 {\ s}=\sqrt{\;{2 {\text{h}}_{\max }}/{ {\text{g}}}}$ $\Rightarrow {\text{h}}_{\max }=\;\frac{16 \times 9.8}{2}=\;\frac{392}{5}$

$(\;{ {\text{V}}}/{3})^2= {\text{V}}^{2-}2 {\text{a}}(4) \Rightarrow {\text{a}}=\;{8 {\text{V}}^2}/{9(8)}=\;{ {\text{V}}^2}/{9}$ $0= {\text{V}}^{2-}2 {\text{a}}(4+ {\text{x}})$ $\Rightarrow \;\; {\text{V}}^2=2(\;{ {\text{V}}^2}/{9})(4+ {\text{x}})$ $4.5=4+ {\text{x}}$ ${\text{x}}=0.5$

$X_P=\alpha t+\beta t^2$ $X_Q=f t-t^2$ $\therefore V_P=\alpha +2 \beta t$ $V_Q=f-2 t$ $\because V_P=V_Q$ $\Rightarrow \alpha +2 \beta t=f-2 t$ $\Rightarrow t=\;\frac{f-\alpha }{2(1+\beta )}$

Max. Height $=h=\;\frac{u^2}{2 g}$ $\Rightarrow {\text{u}}=\sqrt{2 {\text{gh}}}$ ${\text{S}}= {\text{ut}}+\;\frac{1}{2} a {\text{at}}^2$ $\;{ {\text{h}}}/{3}={\sqrt{2 g h}} {\text{t}}+\;\frac{1}{2}(- {\text{g}}) {\text{t}}^2$ $\;{ {\text{gt}}^2}/{2}-{\sqrt{2 g h t}}+\;\frac{h}{3}=0 \;\; (.$ Roots are $t_1 \& t_2)$ $\;{ {\text{t}}_2}/{ {\text{t}}_1}=\;{\sqrt{2 {\text{gh}}}+\sqrt{2 {\text{gh}}-4 \times { {\text{g}}}/{2} \times { {\text{h}}}/{3}}}/{\sqrt{2 {\text{gh}}}-\sqrt{2 {\text{gh}}-4 \times \;{ {\text{g}}}/{2} \times \;{ {\text{h}}}/{3}}}=\;{\sqrt{2 {\text{gh}}}+\sqrt{{4 {\text{gh}}}/{3}}}/{\sqrt{2 {\text{gh}}}-\sqrt{\;{4 {\text{gh}}}/{3}}}=\;{{\sqrt{3}}+{\sqrt{2}}}/{{\sqrt{3}}-{\sqrt{2}}}$

$H=\;\frac{1}{2} g t^2$ $19.6=4.9 t^2$ $t=2 {\ sec}$ $D=9 \times \;\frac{5}{18} \times 2=5 {\text{m}}$

Given, height of nozzle from floor, ${H}=9.8 {m}$ Let, $t_{1}=$ time taken by first drop to reach the groundand $t_{2}=$ time taken by 2 nd drop. $t_{2}=\;\frac{t_{1}}{2}$ $g=$ acceleration due to gravity $(9.8 {ms}^{-2})$ $u=$ initial speed $=0$Since, $H=u t+\;\frac{1}{2} g t_{1}^{2}\;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ $\therefore \;\; t_{1}=\sqrt{\;\frac{2 H}{g}}$ $\Rightarrow \;\; t_{1}=\sqrt{\;\frac{2 \times 9.8}{9.8}}=\sqrt{2} {s}$ $\therefore \;\; t_{2}=\;\frac{\sqrt{2}}{2}=\;\frac{1}{\sqrt{2}} {s}$Again from Eq. (i)Distance covered by 2 nd drop, $x=\;\frac{1}{2} g t_{2}^{2}$ $\therefore \;\; x \;\; =\;\frac{1}{2} \times 9.8 \times (\;\frac{1}{\sqrt{2}})^{2}$ $\;\; =\;\frac{1}{2} \times 9.8 \times \;\frac{1}{2}=\;\frac{9.8}{4}=2.45 {m}$and position of second drop from ground,$H_{2}=H_{1}-x=9.8-2.45=7.35 {m}$

From the graph for first line, the slope is negative and intercept is positive. So, equation of line is $v \;\; =-m t+c$ $a_{1} \;\; =\;\frac{d v}{d t}=-m$ Similarly, for second line, the slope is positive and intercept is negative, so equation of line is $v \;\; =m t-c$ $a_{2} \;\; =\;\frac{d v}{d t}=m$ $\therefore$ The corresponding acceleration-time graph as shown below Hence, option (a) is correct.

According to given data; Initial speed of scooter, $u_{1}=0 {ms}^{-1}$ and final speed, $v_{2}=0 {ms}^{-1}$ By using first equation of motion, $\;v_{1} \;\; =u+a_{1} t_{1}$ $\;\Rightarrow \;v_{1} \;\; =a_{1} t_{1}$. . . (i) and for second case, $v_{2} \;\; =v_{1}-a_{2} t_{2}$ $0 \;\; =v_{1}-a_{2} t_{2}$ $v_{1} \;\; =a_{2} t_{2}$ . . . (ii) On dividing Eq. (i) by Eq. (ii), we get $\;v_{1} / v_{1} \;\; =1=\;\frac{a_{1} t_{1}}{a_{2} t_{2}}$ $\Rightarrow \;a_{2} t_{2} \;\; =a_{1} t_{1}$ $\Rightarrow \;t_{1} / t_{2} \;\; =a_{2} / a_{1}$

Object is projected as shown so as per motion under gravity ${S}={ut}+\frac{1}{2} {at}^{2}$ -75=+10 t+\frac{1}{2}(-10) t^{2} \ $\Rightarrow t=5 {sec}$ Object takes ${t}=5 {s}$ to fall on ground Height of balloon from ground ${H}=75+{ut}$ $=75+10 \\times 5=125 {m}$

Given, $m_{1}=m_{2}=m$ $r_{1}=r_{2}=r=5 {\text{cm}}=5 \times 10^{-2} {\text{m}}$ $u_{1}=u_{2}=35 {\text{m}} / {\text{s}}$ $t_{1}=0$ and $t_{2}=3 {\text{s}}$ When both ball will collide, $(\text{S}_{1})_{{\text{vertical} }}=(\text{S}_{2})_{{\text{vertical} }}$ Let after time t both ball will collide. $(\text{S}_{1})_{{\text{vertical} }}=u_{1} t-\frac{1}{2} g t^{2}$ $(\text{S}_{2})_{{\text{vertical} }}=u_{2}(t-3)-\frac{1}{2} g(t-3)^{2}$ From Eq. (i), we have $\Rightarrow u_{1} t-\frac{1}{2} g t^{2}=u_{2}(t-3)-\frac{1}{2} g(t-3)^{2}$ $\Rightarrow 35 t-\frac{1}{2} \times 10 \times t^{2}$ $=35(t-3)-\frac{1}{2} \times 10(t-3)^{2}$ $\Rightarrow 35 t-35(t-3)=5 t^{2}-5(t-3)^{2}$ $\Rightarrow 35(t-t+3)=5[t^{2}-(t-3)^{2}]$ $35 \times 3=5(t+t-3)(t-t+3)$ $\Rightarrow \frac{105}{5}=(2 t-3)(3)$ $\Rightarrow \frac{21}{3}=2 t-3$ $\Rightarrow 7=2 t-3$ $\Rightarrow 2 t=10$ $\Rightarrow t=5 {\text{s}}$ So, the height at which ball will collide is $(\text{S}_{1})_{{\text{vertical} }}=\text{u}_{1} t-\frac{1}{2} g t^{2}$ $=35 \times 5-\frac{1}{2} \times 10 \times (5)^{2}$ $=175-125$ $=50 {\text{m}}$

Given, initial speed of engine $=u$ Speed of engine last compartment $=v$ Let the length of train be $I$ and the speed of mid-point of train be $v^{'}$. Using third equation of motion, $v^{2}=u^{2}+2 a s$ $\Rightarrow \;\; v^{' 2}=u^{2}+2 a l / 2$ $\Rightarrow \;\; v^{' 2}-u^{2}=a l$ . . . (i) $\;\text{ Also, }\; \;\; v^{2}-u^{2}=2 a l \Rightarrow a=\;\frac{v^{2}-u^{2}}{2 l}$ Substituting the value of $a$ in Eq. (i), we get $v^{' 2}-u^{2} \;\; =\;\frac{v^{2}-u^{2}}{2 l} \cdot l=\;\frac{v^{2}-u^{2}}{2}$ $\Rightarrow \;\; v^{'} \;\; =\sqrt{\;\frac{v^{2}-u^{2}}{2}+u^{2}}=\sqrt{\;\frac{v^{2}+u^{2}}{2}}$