Learn relative motion examples in two dimensional. Practice problems of kinematics in two dimensional motion and vectors.

- Consider a situation in which a passenger is moving in straight line with constant velocity \(\text v_o\hat {\text i}\) on the platform.
- A train is also moving with constant velocity \(\text v_{\text t} \hat {\text i}\), as shown in figure.
- The velocities shown are in reference frame attached to the ground.
- Now, the reference frame is attached to the passenger. He observes that the velocity of train is not same as before.
- Because, according to passenger, his velocity is zero.
- To make him at rest in frame of reference attached to him, add \(-\vec{\text v}_0\) to all objects.

?\(\vec {\text v}_0=\text v_0\hat {\text i}=\) Velocity of passenger in ground frame

\(\vec {\text v}_{\text t}= {\text v}_{\text t}\hat {\text i}=\) Velocity of train in ground frame

\(\vec {\text v}_{{\text t}/0}=\) Velocity of train observed by passenger

\(\vec {\text v}_{{\text t}/0}=\vec{\text v}_{\text t}-\vec {\text v}_0\)

\(\vec {\text v}_{{\text t}/0}={\text v}_{\text t}\hat {\text i}- {\text v}_0\hat {\text i}\)

- Two men \(\text m_1\) and \(\text m_2\) are moving along the same line, as shown in figure.
- They both have different velocities, \(\vec {\text v}_1=3 \,\text {m/s}\) and \(\vec {\text v}_2=8\,\text{m/s}\) respectively.
- According to man \(\text m_1\), he is at rest and \(\text m_2\) is completely responsible for the change in distance between them, as \(\text m_2\) is moving at \(5\,\hat {\text i}\,\text{m/s}\) .

Mathematically,

\(\vec {\text v}_{2/1}=\vec {\text v}_2-\vec {\text v}_1\)

where,

\(\vec {\text v}_{2/1}\rightarrow\) Velocity of m_{2} as observed by m_{1}

\(\vec {\text v}_{2}\rightarrow\) Velocity of m_{2}

\(\vec {\text v}_{1}\rightarrow\) Velocity of m_{1}

Velocity of object with respect to observer = Velocity of object being observed – Velocity of observer

\(\text v_{2/1}=8-3=5\,\text{m/s}\) in positive x - direction i.e., \(5\,\hat{\text i}\,\text{m/s}\)

Similarly,

- Consider a particle is located at point B and another particle is located at point A, as shown in figure.
- Let \(\vec{r}_{B}\) = Position vector of B with respect to origin or ground

\(\vec{r}_A\) = Position vector of A with respect to origin or ground

\(\vec{r}_{B/A} \) = Position vector of B with respect to observer A

- According to vector addition,

\(\vec{r}_A+\vec{r}_{B/A}=\vec{r}_B\)

\(\Rightarrow\vec{r}_{B/A}=\vec{r}_B-\vec{r}_A\)

After differentiating both sides

\(\Rightarrow \vec{v}_{B/A}=\vec{v}_B-\vec{v}_A\)

A \(tan^{-1}\left(\dfrac{3}{4}\right)\) South of East

B \(tan^{-1}\left(\dfrac{3}{4}\right)\) North of East

C \(tan^{-1}\left(\dfrac{4}{3}\right)\) North of West

D \(tan^{-1}\left(\dfrac{4}{3}\right)\) North of East

- A man is moving with a velocity \(\vec{v}_m\) along x-axis.
- Rain is falling vertically downwards with a velocity \(\vec{v}_r\).

- Due to relative motion, man observes that rain is falling at some angle \(\theta\) with the vertical direction.

\(\vec{v}_{r/m}\) = Velocity of rain with respect to man

\(\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m\)

\(tan\,\theta=\dfrac{|\vec{v}_m|}{|\vec{v}_r|}\)

\(\theta=tan^{-1}\dfrac{|\vec{v}_m|}{|\vec{v}_r|}\)

- Rain is falling at an angle \(\theta=tan^{-1}\dfrac{|\vec{v}_m|}{|\vec{v}_r|}\)with the vertical direction.

A \(\theta=tan^{-1}\left(\dfrac{2}{5}\right)\)

B \(\theta=tan^{-1}\left(\dfrac{3}{2}\right)\)

C \(\theta=tan^{-1}\left(\dfrac{5}{4}\right)\)

D \(\theta=tan^{-1}\left(\dfrac{4}{3}\right)\)

- Consider a man is running along x-axis.
- Rain is falling at an angle \(\phi\) with the vertical direction.

- Due to relative motion, man observes that rain is falling at some different angle \(\theta\) with the vertical direction.

\(\vec{v}_m\) = velocity of man

\(\vec{v}_r\) = velocity of rain

\(\vec{v}_{r/m}\) = velocity of rain with respect to man

- Assuming horizontal direction as x-axis and vertical direction as y-axis.
- Resolve \(\vec{v}_r\) and \(\vec{v}_m\) in component form.
- Calculate \(\vec{v}_{r/m}\), \(\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m\)
- Angle \(tan\,\theta=\dfrac{|(v_{r/m})_x|}{|(v_{r/m})_y|}\)

A \(tan^{-1}(3)\)

B \(tan^{-1}\left(\dfrac{5}{4}\right)\)

C \(tan^{-1}\left(\dfrac{11}{4}\right)\)

D \(tan^{-1}(2)\)

- Consider two particles, projected from point A and point B, as shown in figure.
- These particles will collide, if the relative velocity of one with respect to another is towards itself.
- Taking A as a reference frame.
- Relative position vector of B with respect to A

\(\vec{r}_{B/A}=\vec{r}_B-\vec{r}_A\)

\(\hat{r}_{B/A}=\dfrac{\vec{r}_B-\vec{r}_A}{|\vec r_ B-\vec{r}_A|}\)

- Relative velocity of B with respect to A

\(\vec{v}_{B/A}=\vec{v}_B-\vec{v}_A\)

\(\hat{v}_{B/A}=\dfrac{\vec{v}_B-\vec{v}_A}{|\vec{v}_B-\vec{v}_A|}\)

- If they collide, both unit vectors must be towards each other.

\(\hat{r}_{B/A}= -\hat{v}_{B/A}\)

\(\dfrac{\vec{r}_B-\vec{r}_A}{|\vec{r}_B-\vec{r}_A|} =-\left(\dfrac{\vec{v}_B-\vec{v}_A}{|\vec{v}_B-\vec{v}_A|}\right)\)

A \(\theta=sin^{-1}\left(\dfrac{4}{5}\right)\)

B \(\theta=sin^{-1}\left(\dfrac{3}{5}\right)\)

C \(\theta=sin^{-1}\left(\dfrac{3}{7}\right)\)

D \(\theta=cos^{-1}\left(\dfrac{3}{5}\right)\)

- Consider a man is running on a horizontal road with some velocity.
- Rain is falling with velocity \(v_r\), making an angle \(\theta\) with the vertical direction.
- The rain appears falling vertically to the man.
- Let the velocity of man be \(\vec{v}_m= v_{m} \hat i\) along x-axis.
- Let \(\vec{v}_r= -a \hat i-b\hat j\) where \(\dfrac{b}{a}=tan\theta\)
- Let \(\vec{v}_{r/m}\) = Velocity of rain with respect to man
- For \(\vec{v}_{r/m}\), rain appears vertically to the man, \((v_{r/m})_x=0\)

\(\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m\)

\(=-a \hat i-b\hat j-v_{m} \hat i\)

\(=(-a -v_m)\hat i-b\hat j\)

\(\because(v_{r/m})_x=0\)

\(\Rightarrow -a -v_m = 0\)

\(\Rightarrow v_m=-a\)

\(v_m=-a\hat i\)

A \(-5\;m/s\) toward left

B \(-4\;m/s\) toward left

C \(3\;m/s\) toward left

D \(-2\;m/s\) toward left

- Consider a particle is projected from point A, and other particle is projected from point B, as shown in figure.

- The relative acceleration of particle A with respect to particle B is zero.
- Let \(a_{A/B}\) = Relative acceleration of A with respect to B

\(a_{A/B}=a_A-a_B\)

\(a_{A/B}=g-g\)

\(a_{A/B}=0\)

- Motion of particle under zero acceleration is a straight line.

A Parabola

B Straight Line

C Circular Path

D Triangular Path

- Consider two bodies A and B are d distance apart from each other.

Let

\(\vec{v}_A\) = Velocity of A

\(\vec{v}_B\) = Velocity of B

\(\vec{v}_{A/B}\) = Velocity of A with respect to B

If bodies A and B do not collide, then the minimum separation can be calculated as follows:

- According to the reference frame of B, the minimum separation will be perpendicular to \(B_0\) along the direction of \(\vec{v}_{A/B}\).

\(B_0=d\;sin\phi\)

\(tan\;\phi=\dfrac{|(v_{A/B})_y|}{|(v_{A/B})_x|}\)

A \(16\,m\)

B \(20\,m\)

C \(10\,m\)

D \(15\,m\)

- A man is running with a velocity \(v_m\) along x-axis.
- The rain is falling with velocity \(\vec{v}_r\) making an angle \(\theta\) with the vertical direction as shown in figure.

\(\vec{v}_r=-(v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j\)

- Let \(\vec{v}_{r/m}\) = velocity of rain with respect to man

\(\vec{v}_{r/m}=\vec{v}_r-\vec{v}_m\)

\(=-(v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j-v_m \hat i\)

\(=-(v_m+v_r\;sin\theta)\hat i -(v_r\;cos\theta)\hat j\)

\(|\vec{v}_{r/m}|=\sqrt{(v_m+v_rsin\theta)^2+(v_r\,cos\theta)^2}\)

\(=\sqrt{v_m^2+v_r^2\;sin^2\theta+v_r^2\;cos^2\theta+2\,v_mv_r\sin\theta}\)

\(|\vec{v}_{r/m}|=\sqrt{v_m^2+v_r^2\;+2\,v_mv_r\sin\theta}\)

A \(3\sqrt{7}\;ms^{-1}\)

B \(4\sqrt{5}\;ms^{-1}\)

C \(5\sqrt{6}\;ms^{-1}\)

D \(3\sqrt{5}\;ms^{-1}\)