Physics 2 Marks Questions

1. $\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix}\hat{\text{i}} & \hat{\text{j}}&\hat{\text{k}} \\-2 & 3&-4\\3&-4&5 \end{vmatrix}$

$=\hat{\text{i}}\begin{vmatrix}3&-4\\-4&5\end{vmatrix}-\hat{\text{j}}\begin{vmatrix}-2&-4\\3&5\end{vmatrix}+\hat{\text{k}}\begin{vmatrix}-2&3\\3&-4\end{vmatrix}$

$=\hat{\text{i}}(15-16)-\hat{\text{j}}(-10+12)+\hat{\text{k}}(8-9)$

$=-\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$

2. $\vec{\text{A}}.\vec{\text{B}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$

$=-6-12-20=-38$

1. $\vec{\text{A}}.\vec{\text{B}}$ refers to the product of magnitude of $\vec{\text{A}}$ and projection of $\vec{\text{B}}$ on $\vec{\text{A}}.$
2. $\vec{\text{A}}\times\vec{\text{B}}$ refers to the area of the parallelogram formed by the vectors.

1. If angle of projection $\theta=45^\circ+\alpha$ and R = R₁

then, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{i})$

2. If angle of projection $\theta=45^\circ-\alpha$ and R = R₂

$\therefore\ \text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{ii})$

$\tan\theta=4$

$\Rightarrow\ \theta=75^\circ58'$

Because directions are perpendicular to X and Y, it may be interpreted for a circular motion with radius a.

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

$\text{F}=\sqrt{\text{F}^2+\text{F}^2+2\text{F}.\text{F}\cos\theta}$

$\text{F}=\text{F}\sqrt{2(1+\cos\theta)}$

$1=2(1+\cos\theta)$

$\cos\theta=-\frac{1}{2}=\cos120^\circ$

$\theta=120^\circ$

$=\cot60^\circ=\frac{1}{\sqrt{3}}$

As, $\theta=60^\circ,\ \cot60^\circ=\frac{1}{\sqrt{3}}$

$(\text{A}+\text{B}).(\text{A}-\text{B})=0$

$\text{A}.\text{A}-\text{A}.\text{B}+\text{B}.\text{A}-\text{B}.\text{B}=0$

$\text{A}-\text{B}=0\ \ [\because\ \text{A}.\text{B}=\text{B}.\text{A}]$

$\Rightarrow\ \text{A}=\text{B}$

The goal man is 25m away in the direction of the ball, so to catch the ball, he is to cover a distance

= 40.82 - 25 = 15.82m in time 2.886s.

$\therefore$ Velocity of the goal man to catch the ball

$\text{v}=\frac{15.82}{2.886}=5.48\text{m/ s}$

$\therefore$ using $\text{R}^2=\text{P}^2+\text{Q}^2+2\text{PQ}\cos\theta$

we get,

$\text{p}^2=\text{p}^2+\text{p}^2+2\text{p}^2\cos\theta$

$\cos\theta=-\frac{1}{2};\ \theta=120^\circ$

v_R = 5km/ h

v_S = 10km/ h

$\cos\theta=\frac{5}{10}=\frac{1}{2}$

$\theta=60^\circ$

⇒ with the direction of river, the angle would be 120° to reach the opposite point.

The path of the cyclist at R is circular of constant radius 1km with centre O and he is moving with constant speed 10m/ s. So his motion is uniform circular motion at R.

Hence, the R = 1000m, v =10m/ s

$\therefore\ \text{a}_\text{c}=\frac{\text{v}^2}{\text{R}}=\frac{10\times10}{1000}=\frac{1}{10}=0.1\text{m/ s}^2$ along RO.

1. If they act at opposite direction, resultant is zero.
2. For the resultant to be F,

$\text{F}^2=\text{F}^2+\text{F}^2+2\text{F}^2\cos\theta$

$\Rightarrow\ \cos\theta=\frac{-1}{2}\Rightarrow\ \theta=120^\circ$​​​​​​​

1. $\frac{\text{h}_1}{\text{h}_2}=\frac{\frac{\text{v}^2_0\sin^2\alpha}{2\text{g}}}{\frac{\text{v}^2_0\sin^2(90-\alpha)}{2\text{g}}}$

$=\frac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha$

2. Horizontal ranges in both the cases are same, i.e.,

$\text{R}_1=\text{R}_2,\ \frac{\text{R}_1}{\text{R}_2}=1$

1. If they act at opposite direction, resultant is zero.
2. If they act in same direction, R = 2F.
3. For the resultant to be F,

$\text{F}^2=\text{F}^2+\text{F}^2+2\text{F}^2\cos\theta$

$\cos\theta=-\frac{1}{2}$ or $\theta=120^\circ.$

Consider $\vec{\text{OA}}=\vec{\text{a}}$

$\vec{\text{OB}}=\vec{\text{b}}$

$\vec{\text{OC}}=\vec{\text{c}}$

Then $\vec{\text{b}}\times\vec{\text{c}}$ is a vector perpendicular to the plane of $\vec{\text{b}}$ and $\vec{\text{c}}.$

Let $\phi$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}\times\vec{\text{c}}$ and $\hat{\text{n}}$ unit vector along $\vec{\text{b}}\times\vec{\text{c}}.$

$\vec{\text{a}}.(\vec{\text{b}}\times\vec{\text{c}})=$ (area of parallelogram OBDC)

= area of parallelogram OBDC $(\hat{\text{n}}.\vec{\text{a}})$

= area of parallelogram OBDC $|\vec{\text{a}}|\cos\phi$

$(\because\ |\hat{\text{n}}|=1)$

= (area of parallelogram OBDC) (OL)

$[\because\ \text{OA}\cos\phi=\text{OL}]$

= (area of parallelogram OBDC) × (height)

= volume of parallelopiped with edges $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$

$\theta=45^\circ+\alpha.$

Range, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}$

$=\frac{\text{u}^2\cos2\alpha}{\text{g}}$

Case II: When angle of projection.

$\theta=45^\circ-\alpha.$

Range, $\text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}$

$=\frac{\text{u}^2}{\text{g}}\sin(90^\circ-2\alpha)$

$=\frac{\text{u}^2}{\text{g}}\cos^2\alpha=\text{R}_1$

The direction of the vector $(\vec{\text{B}}\times\vec{\text{C}})$ will be perpendicular to the plane containing by vectors $\vec{\text{B}}$ and $\vec{\text{C}}$ by Right-hand thumb or Right-hand grip rule (RHGR).

The direction of the vector $\vec{\text{A}}\times(\vec{\text{B}}\times\vec{\text{C}})$ will be perpendicular to $\vec{\text{A}}$ and in a plane containing $\vec{\text{B}}$ and $\vec{\text{C}}$ by Right-hand grip rule.

or $\text{r}=\Big(\frac{1250}{9}\Big)^2\frac{\sqrt{3}}{1}\times\frac{1}{9.8}$

$\Big[\because5000\text{ km/hr}=\frac{1250}{9}\text{ m/s}\Big]$

or, $\text{r}=3.41\times10^3\text{m}\Big[\because\tan30^\circ=\frac{1}{\sqrt{3}}\Big]$

$\vec{\text{v}_\text{w}}=10\text{ m/s}=\vec{\text{OA}}$

$\vec{\text{v}_\text{r}}=30\text{ m/s}=\vec{\text{OB}}$

$\therefore\ \vec{\text{v}_\text{rw}}=\sqrt{\text{v}_\text{r}^2+\text{v}_\text{w}^2}$

$=\sqrt{(10)^2+(30)^2}=31.6\text{ m/s}$

$\tan\theta=\frac{\text{BD}}{\text{OB}}=\frac{10}{30}=0.33\ (\text{BD}=\text{OA})$

$\theta=18^\circ16'$ with vertical direction.