Informative line

Problems Involving Two Dimensional Motion

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

Relative Velocity

  • 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

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

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

 

Relative Motion in One Dimension

  • 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 m2 as observed by m1 

\(\vec {\text v}_{2}\rightarrow\) Velocity of m2

\(\vec {\text v}_{1}\rightarrow\) Velocity of m1

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

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

Illustration Questions

Consider two cars A and B, moving with a speed of 40 m/s towards east and 30 m/s towards north respectively. Determine the direction of velocity of A with respect to the direction of velocity of B.

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

×

Velocity,

\(\vec{v}_A=40 \,\hat i\;m/s\)

\(\vec{v}_B=30\, \hat j\;m/s\)

image

Take B, as reference frame,

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

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

\(=40 \hat i -30 \hat j\)

image

Here

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

(Velocity of A as seen by B)

image image

Here, \(tan\,\theta=\dfrac{30}{40}\)

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

So A will be moving \(tan^{-1}\left(\dfrac{3}{4}\right)\) south of east with respect to B

image

Consider two cars A and B, moving with a speed of 40 m/s towards east and 30 m/s towards north respectively. Determine the direction of velocity of A with respect to the direction of velocity of B.

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

Option A is Correct

The Man and the Rain Problem

  • 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.

Illustration Questions

A man is moving with a velocity \((\vec v_m)\) of \(4\,m/s\) towards x-axis. If rain is falling vertically downwards with a velocity \((\vec v_r)\) of \(3\,m/s\), then at which angle should he keep his umbrella to avoid the rain?  

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

×

Take man as reference frame,

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

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

\(=(-3 \hat j -4 \hat i) \;m/s\)

 

image

Here

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

(Velocity of rain as seen by man)

image image

\(tan \,\theta=\dfrac{|\vec{v}_m|}{|\vec{v}_r|}=\dfrac{4}{3}\)

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

So, he should keep his umbrella at an angle \(\theta=tan^{-1}\left(\dfrac{4}{3}\right)\) with the vertical to avoid the rain.

image

A man is moving with a velocity \((\vec v_m)\) of \(4\,m/s\) towards x-axis. If rain is falling vertically downwards with a velocity \((\vec v_r)\) of \(3\,m/s\), then at which angle should he keep his umbrella to avoid the rain?  

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

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

Option D is Correct

Calculation of Angle of Rain Falling 

  • 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

Calculation of \(\theta\)

  • 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 \(tan\,\theta=\dfrac{|(v_{r/m})_x|}{|(v_{r/m})_y|}\)

Illustration Questions

A man is running horizontally with a velocity of 8 m/s towards right. Rain is falling with a velocity of 5 m/s making an angle 37º with the vertical direction. Calculate the angle with direction of velocity of rain with respect to man. \(\Big[Take\, sin\,37º =\dfrac{3}{5} \,\,\,cos \,37^o=\dfrac{4}{5}\Big]\)

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

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

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

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

×

\(\vec{v}_m=8 \hat i\; m/s\)

\(\vec{v}_r\) making an angle 37º with the vertical direction .

\(\vec v_r=(-5\,sin\,37^o) \hat i -(5 \,cos \,37^o) \hat j\)

\(\Big(-5×\dfrac {3}{5}\Big)\hat i-\Big(5×\dfrac{4}{5}\Big)\hat j\)

So, \(\vec{v}_r=(-3 \hat i -4 \hat j) \;m/s\)

image

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

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

\(=(-3 \hat i -4 \hat j) -8 \hat i\)

\(=(-11 \hat i -4 \hat j) \;m/s\)

image image

Here,  \(tan\,\theta=\dfrac{|(v_{r/m})_x|}{|(v_{r/m})_y|}\)

\(tan\,\theta=\dfrac{|-11|}{|-4|}\)

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

image

A man is running horizontally with a velocity of 8 m/s towards right. Rain is falling with a velocity of 5 m/s making an angle 37º with the vertical direction. Calculate the angle with direction of velocity of rain with respect to man. \(\Big[Take\, sin\,37º =\dfrac{3}{5} \,\,\,cos \,37^o=\dfrac{4}{5}\Big]\)

image
A

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

.

B

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

C

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

D

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

Option C is Correct

Condition for Collision 

  • 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.

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

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

 

Illustration Questions

Particle A is projected with a velocity of \(v_A=5\,m/s\), making an angle \(\theta\) with the horizontal direction. Second particle B is projected vertically upward with a velocity \(v_B=3\,m/s\), as shown in figure. At which angle \(\theta\), A should be projected so that A and B will collide?

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

×

Velocity of particle A, \(v_A=5\,m/s\)

Velocity of particle B, \(v_B=3\,m/s\)

image

B as a reference frame

image image

For collision, direction of velocity A must be towards B.

\(5\sin\theta-3=0\)

\(\sin\theta=\dfrac{3}{5}\)

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

image

Particle A is projected with a velocity of \(v_A=5\,m/s\), making an angle \(\theta\) with the horizontal direction. Second particle B is projected vertically upward with a velocity \(v_B=3\,m/s\), as shown in figure. At which angle \(\theta\), A should be projected so that A and B will collide?

image
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

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

Option B is Correct

Velocity of Man such that Rain Appears Vertically downwards to him

  • 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, \((v_{r/m})_x=0\)

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

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

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

 

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

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

\(\Rightarrow v_m=-a\)

\(v_m=-a\hat i\)

 

Illustration Questions

A man is running on a horizontal road. Rain is falling with a velocity of 5 m/s making an angle \(\theta=37^o\) with the vertical, but man observes that rain is falling vertically downwards. Determine the velocity of man. \(\Big[Take \,sin 37º = \dfrac{3}{5}\Big]\)

A \(-5\;m/s\) toward left

B \(-4\;m/s\) toward left

C \(3\;m/s\) toward left

D \(-2\;m/s\) toward left

×

Let \(\vec{v}_m= v \hat i\) m/s

\(\vec{v}_r\) is making an angle \(\theta=37^o\) with the vertical direction

So,  \(\vec{v}_r=(-3 \hat i -4 \hat j) \;m/s\)

image

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

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

\(=(-3 \hat i -4 \hat j) -v\hat i\)

\(\vec{v}_{r/m}=[(-3-v)\hat i-4 \hat j]\,m/s\)

\(=[(v_{r/m})_x \hat i+(v_{r/m})_y \hat j]\)

image

Here, 

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

\(=\vec{v}_r-\vec{v}\)

image image

Man observes that rain is falling vertically downward

So, \((v_{r/m})_x=0\)

\(-3-v=0\)

\(v=-3\,m/s\)

So, man is running with a velocity 3 m/s towards left.

image

A man is running on a horizontal road. Rain is falling with a velocity of 5 m/s making an angle \(\theta=37^o\) with the vertical, but man observes that rain is falling vertically downwards. Determine the velocity of man. \(\Big[Take \,sin 37º = \dfrac{3}{5}\Big]\)

image
A

\(-5\;m/s\) toward left

.

B

\(-4\;m/s\) toward left

C

\(3\;m/s\) toward left

D

\(-2\;m/s\) toward left

Option C is Correct

Relative Trajectory under Constant Acceleration

  • 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.

Illustration Questions

Two particles A and B are projected with initial velocities \(3\,m/s\) and \(5\,m/s\), as shown in figure. What will be the trajectory of particle A as observed by B?

A Parabola

B Straight Line

C Circular Path

D Triangular Path

×

As both particles are projected under gravity, the relative acceleration will be zero.

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

\(=g-g=0\)

where \(a_{A/B}\) = acceleration of A with respect to B

Under gravity, relative trajectory of particles is straight line.

Two particles A and B are projected with initial velocities \(3\,m/s\) and \(5\,m/s\), as shown in figure. What will be the trajectory of particle A as observed by B?

image
A

Parabola

.

B

Straight Line

C

Circular Path

D

Triangular Path

Option B is Correct

Shortest Distance Between Two Moving Bodies 

  • 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 \(\vec{v}_{A/B}\).

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

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

Illustration Questions

Two cars A and B are moving with velocities \(v_A=20\,m/s\) and \(v_B=15\,m/s\) respectively, as shown in figure. Calculate the minimum distance between the cars if they do not collide.

A \(16\,m\)

B \(20\,m\)

C \(10\,m\)

D \(15\,m\)

×

According to ground frame

Velocity of car A = 20 m/s

Velocity of car B = 15 m/s

Car B is taken as reference frame and path of car A is seen from car B.

image

Minimum distance between cars will be the perpendicular distance from car B on the path of car A as observed by car B.

image

If these two cars do not collide,

BD = Minimum separation

image

Here 

\(tan\,\theta=\dfrac{|\vec{v}_B|}{|\vec{v}_A|}\)

\(tan\,\theta=\dfrac{15}{20}\)

\(tan\theta=\dfrac{3}{4}\)

\(\therefore sin\,\theta=\dfrac{3}{5}\;\;,\;\;cos\,\theta=\dfrac{4}{5}\)

Distance, \(OC=AO \;tan\,\theta\)

\(=500×\dfrac{3}{4} =375\,m\)

Distance, \(CB=OB-OC\)

\(=400-375=25 \,m\)

Distance, \(BD=CB\;cos\theta\)

\(=25×\dfrac{4}{5} =20\,m\) 

Two cars A and B are moving with velocities \(v_A=20\,m/s\) and \(v_B=15\,m/s\) respectively, as shown in figure. Calculate the minimum distance between the cars if they do not collide.

image
A

\(16\,m\)

.

B

\(20\,m\)

C

\(10\,m\)

D

\(15\,m\)

Option B is Correct

Speed of Rain with Respect to Man Running Horizontally

  • 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

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

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

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

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

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

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

Illustration Questions

A man is running with a velocity \((\vec v_m)\) of \(5\,m/s\) towards right and the rain is falling with a velocity \((v_r)\) of \(5\,m/s\) making an angle \(\theta=37^o\) with the vertical direction. Calculate the speed of rain with respect to man.

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

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

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

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

×

\(\vec{v}_m=5\hat i\;m/s\)

\(\vec{v}_r\) is making angle 37º with the vertical direction 

\(\vec{v}_r = -5\,sin\;37^o\hat i - 5\,cos\;37^o\hat j\)

\(=(-3\hat i - 4\hat j)\,m/s\)

image

Taking man as reference frame, 

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

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

\(=(-3\hat i - 4\hat j)-5\hat i\)

\(=(-8\hat i - 4\hat j)\;m/s\)

image

Here

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

(Velocity of rain as seen by man)

image image

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

\(|\vec{v}_{r/m}|=\sqrt{(8)^2+(4)^2}\)

\(=\sqrt{64+16}\)

\(=\sqrt{80}\)

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

image

A man is running with a velocity \((\vec v_m)\) of \(5\,m/s\) towards right and the rain is falling with a velocity \((v_r)\) of \(5\,m/s\) making an angle \(\theta=37^o\) with the vertical direction. Calculate the speed of rain with respect to man.

image
A

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

.

B

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

C

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

D

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

Option B is Correct

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