Physics 5 Marks Questions

1. When y = 16m:

$\text{t}=\Big(\frac{16}{4}\Big)^{1/2}=2\text{s}$

$\therefore\text{y}=10\times2+(2)^2=24\text{m}$

2. Velocity of the particle is given by:

$\vec{\nu}(\text{t})=8.0\text{t}\hat{\text{i}}+2.0\text{t}\hat{\text{j}}+\vec{u}$

$\text{At t}=2\text{s}$

$\vec{\nu}(2)=8.0\times2\hat{\text{i}}+2.0\times2\hat{\text{j}}+10\hat{\text{j}}$

$=16\hat{\text{i}}=14\hat{\text{j}}$

$\therefore$ Speed of the particle,

$\text{V}=\sqrt{(16)^2+(14)^2}$

$=\sqrt{256+196}=\sqrt{452}$

$=21.26\text{m/s}$

1. $\vec{\nu}(\text{t})=(3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}});\text{a}=-4.0\hat{\text{j}}$

The position of the particle is given by:

$\vec{\text{r}}=3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{J}}+4.0\hat{\text{k}}$

Velocity $\vec{\nu},$ of the particle given as:

$\vec{\nu}=\frac{\text{d}\vec{\text{r}}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{J}}+4.0\hat{\text{k}})$

$\therefore\vec{\nu}=3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}}$

Acceleration, $\vec{\text{a}},$ of the particle is given as:

$\vec{\text{a}}=\frac{\text{d}\vec{\nu}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}})$

$\therefore\vec{\text{a}}=-4.0\hat{\text{J}}$

2. 8.54m/s, 69.45° below the x - axis

We have velocity vector, $\vec{\nu}=3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}}$

$\text{At t}=2.0\text{s}:$

$\vec{\nu}=3.0\hat{\text{i}}-8.0\hat{\text{J}}$

The magnitude of veocity is given by:

$|\vec{\nu}|\sqrt{3^2+(-8)^2}=\sqrt{73}=8.54\text{m/s}$

Direction, $\theta=\tan^{-1}\Big(\frac{\nu_\text{y}}{\nu_\text{x}}\Big)$

$=\tan^{-1}\Big(\frac{-8}{3}\Big)=-\tan^{-1}(2.667)$

$=-69.45^\circ$

The negative sing indicates that the direction of velocity is below the x - axis.

1. Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.
2. Average velocity is given by the relation:

$\text{Average velocity}=\frac{\text{Net Displacement}}{\text{Total time}}$

Since the net displacement of the cyclist is zero, his average velocity will also be zero.

3. Average speed of the cyclist is given by the relation:

$\text{Average speed}=\frac{\text{Total path length}}{\text{Total time}}$

Total path length = OP + PQ + QO

$=1+\frac{1(2\pi\times1)}{4}+1$

$=2+\frac{\pi}{2}$

= 3.570km

Time taken = 10 min $=\frac{10}{60}=\frac{1}{6}\text{h}$

$\therefore$ Average speed = $\frac{3.570}{\frac{1}{6}}=21.42\text{km/h}$

V_PS = 5.00km/ h, to the right

1. Here we must find the magnitude of the vector v_PG, given the magnitude and direction of two other vectors. We find the velocity v_PG by using the relation

v_PG = v_PS + v_SG

Here the vectors are parallel, and so the vector addition is quite simple see figure. We add vectors by adding magnitudes.

v_PG = v_PS + v_SG

= 5.00km/ h + 3.00km/ h

= 8.00km/ h

2. The length of the sidewalk is 135m, and so this is the distance $\Delta\text{x}_\text{G}$ the passenger travels relative to the ground. So, our problem is to find $\Delta\text{t}$ when $\Delta\text{x}_\text{G}=135\text{m}.$ The rate at which this distance along the ground is covered by the passenger is V_PG, where

​​​​​​​$\text{v}_\text{PG}=\frac{\Delta\text{x}_\text{G}}{\Delta\text{t}}$

Therefore,

$\Delta\text{t}=\frac{\Delta\text{x}_\text{G}}{\text{v}_\text{PG}}=\frac{135\text{m}}{8.00\text{km/h}\big(\frac{1.00\text{m/s}}{3.60\text{km/h}}\big)}=60.8\text{s}$

3. The problem here is to determine how much of the sidewalk's surface the passenger moves over. If he was standing still and not walking along the surface, he would cover none of it. Because he is moving relative to the surface at velocity v_PS, he does move some distance $\Delta\text{x}_\text{s}$ relative to the surface. The problem is to find $\Delta\text{x}_\text{s}$ when $\Delta\text{x}=60.8\text{s},$ since we found in part (b) that this is the time interval during which he is on the moving sidewalk. His velocity relative to the sidewalk is $\text{v}_\text{PS}=\frac{\Delta\text{x}_\text{s}}{\Delta\text{t}},$ and so

​​​​​​​$\Delta\text{x}_\text{s}=\text{v}_\text{PS}\Delta\text{t}$​​​​​​​​​​​​​​

$=(5.00\text{km/h})\Big(\frac{1.00\text{m/s}}{3.60\text{km/h}}\Big)(60.8\text{s})$

$=84.4\text{m}$

and $\text{AB}=20\times\sqrt{2}\Big(\frac{\text{u}}{\text{g}}\Big)=20\sqrt{2}\frac{\text{u}}{\text{g}}$ $(\therefore72\text{km/ h}=20\text{ms}^{-1})$

$\therefore\ \frac{\text{u}^2}{\text{g}}=500+20\sqrt{2}\frac{\text{u}}{\text{g}}$

$\text{or }\ \text{u}^2-20\sqrt{2}\text{u}-500\times9.8=0$

$\text{or }\ \text{u}^2-20\sqrt{2}\text{u}-4900=0$

$\text{or }\text{ u}=\frac{20\sqrt{2}\pm\sqrt{400\times4+4\times4900}}{2}\text{ms}^{-1}$

$=(10\sqrt{2}\pm\sqrt{5300})\text{ms}^{-1}$

$=10[\sqrt{2}\pm\sqrt{53}]\text{ms}^{-1}$

$=10[1.414+7.280]\text{ms}^{-1}=86.94\text{ms}^{-1}$

the usual expressions for the range and time of flight of a projectile.

1. t = ? if y = -1.00m; $\text{y}=\frac{-1}{2}\text{gt}^2$ $[\because\ \text{v}_{\text{x}_0}=0]$

$\Rightarrow\ \text{t}=\sqrt{\frac{-2\text{y}}{\text{g}}}=\sqrt{\frac{(-2)(-1.00)}{9.8}}=0.452\text{s}$

2. x = ?, when t = 0.452s

$\therefore\text{x}=\text{v}_{\text{x}_0}\text{t}=1.00\times0.452\text{s}=0.452\text{m}$

3. v = ?, $\theta=?$ at t = 0.452s

The x-component of velocity is constant throughout the motion, $\text{v}_\text{x}=\text{v}_{\text{x}_0}=1.00\text{m/s}$

The y-component of velocity $\text{v}_\text{y}=\text{v}_{\text{y}_0}-\text{gt}$

$=0-9.8\times0452=-4.43\text{m/s}$

$\therefore\ \text{v}=\sqrt{\text{v}^2_\text{x}+\text{v}^2_\text{y}}$

$=\sqrt{(1.00)^2+(-4.43)^2}=4.54\text{m/s}$

$\theta=\tan^{-1}\Big|\frac{\text{v}_\text{y}}{\text{v}_\text{x}}\Big|=\frac{4.43}{1.00}=77.3^\circ$

As the velocity hits the floor, its velocity is 4.54m/ s directed 77.3° below the horizontal.

Magnitude of $\vec{\text{R}}:$ Draw $\text{QN}\perp\text{OP}$ Produced

In right-angled triangle ONQ, we have

$\text{OQ}^2=\text{ON}^2+\text{NQ}^2$

$\Rightarrow\ (\text{OQ})^2=(\text{OP}+\text{PN})^2+(\text{NQ})^2$

$\Rightarrow\ \text{R}^2=(\vec{\text{A}}+\vec{\text{B}}\cos\theta)^2+(\vec{\text{B}}\sin\theta)^2$

$\Rightarrow\ \text{R}^2=\vec{\text{A}^2}+2\overrightarrow{\text{AB}}\cos\theta+\vec{\text{B}^2}(\cos^2\theta+\sin^2\theta)$

$\Rightarrow\ \text{R}=\sqrt{\vec{\text{A}^2}+\vec{\text{B}^2}+2\overrightarrow{\text{AB}}\cos\theta}$

Direction of $\vec{\text{R}}:$ Let resultant vector $\vec{\text{R}}$ makes on angle $\beta$ with the direction of $\vec{\text{A}}.$ Then from right angled $\triangle\text{QNO,}$

$\tan\beta=\frac{\text{QN}}{\text{ON}}=\frac{\text{QN}}{\text{OP}+\text{PN}}$

$=\frac{\vec{\text{B}}\sin\theta}{\vec{\text{A}}+\vec{\text{B}}\cos\theta}$

At the sixth turn motorist is at the starting point A.

Total path length = AB + BC + CD + DE + EF + FA

$=\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\frac{1}{\sqrt{3}}=\tan30^\circ$

$\beta=30^\circ$

$\therefore$ Displacement of the motorist at the end of eighth turn is 866m making an angle 30 with the initial direction of motion.

Total path length = 8 × 500 = 4000m

For upward throw of the ball, we have

u_x, a_x and u_y, a_y

where $\text{u}_\text{x}=\text{u}\cos\theta,\text{ u}_\text{y}=\text{u}\sin\theta,$

$\text{t}=\frac{\text{x (t)}}{\text{u}\cos\theta}\dots(\text{i})$

$\text{u}_\text{y}=\text{u}\sin60^\circ$

$\text{S}=\text{u}_\text{y}\text{t}+\frac{1}{2}\text{gt}^2$

$15=\text{u}\sin60^\circ\text{t}-\frac{1}{2}\times10\text{t}^2$

$15=\frac{\sqrt{3}}{2}\text{ut}-5\text{t}^2$

$15=30\sqrt{3}-5\Big(\frac{60}{\text{u}}\Big)^2$

$\text{u}=\frac{18000}{(30\sqrt{3}-15)}=22.07\text{ m/s}$

$\therefore$ Initial velocity of projectile = 22.07m/ s

From (ii) and (iii), we have

$\text{y}=\text{x}\tan\theta\Big[1-\frac{\text{xg}}{2\text{u}^2\cos^2\theta\tan\theta}\Big]$

Putting, $\text{R}=\frac{2\text{u}^2\sin\theta\cos\theta}{\text{g}}$

we get $\text{y}=\text{x}\tan\theta\Big[1-\frac{\text{xg}}{2\text{u}^2\cos\theta\sin\theta}\Big]$

$\frac{\text{y}}{\text{x}}=\tan\theta\Big(\frac{\text{R}-\text{x}}{\text{R}}\Big)\dots(\text{iv})$

Using trigonometric considerations,

1. The problem here is to find t when x = 32.0m. We can use $(\text{x}=\text{v}_{\text{x}_0}\text{t}),$ if we first find $\text{v}_{\text{x}_0}.$ From figure, we see that $\text{v}_{\text{x}_0}=\text{v}_0\cos\theta_0=(20.0\text{m/s})(\cos30.0^\circ)$

= 17.3m/ s

Using the relation and solve for t.

$\text{x}=\text{v}_{\text{x}_0}\text{t}$

$\text{t}=\frac{\text{x}}{\text{v}_{\text{x}_0}}=\frac{32.0\text{m}}{17.3\text{m/s}}=1.85\text{s}$

2. We want to find y when x = 32.0m, or since we have already found the time in part (a), we can state this, find y when t = 1.85s. Using the relation,

$\text{y}=\text{v}_{\text{y}_0}\text{t}-\frac{1}{2}\text{gt}^2$

where, $\text{v}_{\text{y}_0}=\text{v}_0\sin\theta_0$

$=(20.0\text{m/s})(\sin30.0^\circ)=10.0\text{m/s}$

Thus, $\text{y}=(10.0\text{m/s})(1.85\text{s})-\frac{1}{2}(9.80\text{m/s}^2)(1.85\text{s})^2$

= 1.73m

Since, y = 0 is 2.00m above the ground, this means the ball is 3.73m above the ground as it crosses the goal line too much high to be caught at that point.

$=(\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2)\text{t}$

$\therefore\ \frac{\text{Y}}{\text{X}}=\frac{(\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2)\text{t}}{(\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2)\text{t}}$

$=\frac{\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2}{\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2}=\text{m}$ (constant)

or, $\text{Y}=\text{mX}$

$\therefore\ \frac{\text{mv}^2_2}{\text{r}}\geq\text{mg}$ or $\text{v}_2^2\geq\text{rg}$

from (ii) $\text{v}_2^2=\frac{2\text{mg}(\text{h}-2\text{r})}{\text{m}}=2\text{g}(\text{h}-2\text{r})$

$\therefore\ 2\text{g}(\text{h}-2\text{r})\geq\text{rg}$ or $\text{h}\geq\frac{5\text{r}}{2}$

$\vec{\text{r}}\ \text{ or }\ \vec{\text{r}}_{12}=\vec{\text{r}}_2-\vec{\text{r}}_1=(\text{x}_2-\text{x}_1)\hat{\text{i}}+(\text{y}_2-\text{y}_1)\hat{\text{j}},$

where, $|\vec{\text{r}}|=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$

Accelerataion: If velocity of particle is variable then time rate of change of velocity is the acceleration $\vec{\text{a}}.$ If $\vec{\text{v}}_1$ and $\vec{\text{v}}_2$ be the velocities of the particle at times t₁ and t₂ respectively and acceleration be uniform, then

or, $\theta=60^\circ$

$\therefore$ Angle with the horizontal,