NCERT Solutions for Class 11 Science Physics Chapter 4 - Motion In A Plane

Answer:

$$\begin{gathered} \overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}} = \left|\mathrm{A}\right|\left|\mathrm{B}\right|\mathrm{cos\theta } \\ \Rightarrow ( \hat{\mathrm{i}}+\hat{\mathrm{j}} ).( \hat{\mathrm{i}}-\hat{\mathrm{j}} ) = \sqrt{1+1} \times \sqrt{1+1} \mathrm{cos\theta } \\ \Rightarrow 1-0+0-1 = 2 \mathrm{cos\theta } \\ \Rightarrow \mathrm{cos\theta } =\frac{0}{2} = 0 \\ \Rightarrow \mathrm{\theta }=90° \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

A scalar quantity does not depend on the direction.
So a scalar quantity has the same value for observers with different orientations of the axes.
Hence, the correct answer is option (d).

Answer:

On resolving the components of vectors is XY plane, we can see that a, b and p are along positive axes while q is along negative axes.

Hence the correct answer is option (b).

Answer:

If vector r is along positive X-axis then component of a vector r along X-axis will be,
 $r_x=r \times \cos \theta$ as shown in the figure.

r_x will be maximum when cos$\theta$ will be maximum, i.e $\theta$ = 0$°$.or r vector is along positive X-axis.
Hence, the correct answer is option (b).

Answer:

$$\begin{gathered} \mathrm{Horizontal} \mathrm{Range}, R=\frac{u^2\sin 2\theta }{g} \\ \Rightarrow 50=\frac{u^2\sin (2\times 15°)}{9.8} \\ \Rightarrow u^2=50\times 9.8\times 2 \\ \mathrm{For}\mathit{ }\theta =45°,\mathit{ } \\ R\text{'}= \frac{u^2\sin 90°}{g}=\frac{50\times 9.8\times 2\times 1}{9.8} \\ =100 \mathrm{m} \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

Area is treated as vector. Direction of area vector is outward normal to the plane of the area.
Impulse is also vector quantity.
Hence the correct answer is option (b).

Answer:

In uniform circular motion average velocity can be zero as displacement can be zero and acceleration need not to be zero. Since speed is positive constant then distance covered per unit time is constant.
Hence, the correct answer is option (d).

Answer:

If the path of particle is straight then acceleration will be in the plane either parallel or opposite to velocity.
If the path of particle is curvilinear then acceleration will be in the plane of motion and directed along the center of curvature. 
Hence, the correct answer is option (c)

Answer:

Question is wrong as option (a), (b) and (d) are false and option (c) is true.
$$\begin{gathered} \mathrm{If} \mathrm{A}+\mathrm{B}+\mathrm{C}=0 \mathrm{then} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{are} \mathrm{coplanar} \mathrm{vectors}. \\ \mathrm{We} \mathrm{can} \mathrm{write} \\ \mathrm{B}\times (\mathrm{A}+\mathrm{B}+\mathrm{C}) =\mathrm{B}\times 0=0 \\ \Rightarrow \mathrm{B}\times \mathrm{A} + \mathrm{B}\times \mathrm{B} + \mathrm{B}\times \mathrm{C} =0 \\ \Rightarrow \mathrm{B}\times \mathrm{A} + \mathrm{B}\times \mathrm{C} =0 \\ \Rightarrow \mathrm{B}\times \mathrm{A} =- \mathrm{B}\times \mathrm{C} \\ \Rightarrow \mathrm{A}\times \mathrm{B} = \mathrm{B}\times \mathrm{C} \\ \therefore (\mathrm{A}\times \mathrm{B})\times \mathrm{C} = (\mathrm{B}\times \mathrm{C})\times \mathrm{C} \\ \mathrm{Hence} \mathrm{if} \mathrm{B}\parallel \mathrm{C}, \mathrm{then} \mathrm{B}\times \mathrm{C}=0 \mathrm{But} \mathrm{if} \mathrm{A}\parallel \mathrm{B} \mathrm{then} \mathrm{also} (\mathrm{A}\times \mathrm{B})\times \mathrm{C} \mathrm{is} \mathrm{zero}. \end{gathered}$$

 

Answer:

If A and B are anti-parallel provided |B| = 2|A| or B = 0 then |A+B| =|A|
Hence the correct answer is option (b).

Answer:

$$\begin{gathered} \mathrm{Maximum} \mathrm{height} \mathrm{reached} \mathrm{by} \mathrm{first} \mathrm{particle}, {\mathrm{H}}_1=\frac{{{\mathrm{v}}_0}^2{\mathrm{Sin}}^2{\mathrm{\theta }}_1}{2\mathrm{g}} \\ \mathrm{Maximum} \mathrm{height} \mathrm{reached} \mathrm{by} \mathrm{second} \mathrm{particle}, {\mathrm{H}}_2=\frac{{{\mathrm{v}}_0}^2{\mathrm{Sin}}^2{\mathrm{\theta }}_2}{2\mathrm{g}} \\ \mathrm{Since} {\mathrm{H}}_1>{\mathrm{H}}_{2 }, \mathrm{then} \frac{{{\mathrm{v}}_0}^2{\mathrm{Sin}}^2{\mathrm{\theta }}_1}{2\mathrm{g}}>\frac{{{\mathrm{v}}_0}^2{\mathrm{Sin}}^2{\mathrm{\theta }}_2}{2\mathrm{g}} \\ \mathrm{or} {\mathrm{Sin}}^2{\mathrm{\theta }}_1 >{\mathrm{Sin}}^2{\mathrm{\theta }}_2 \\ \mathrm{or} {\mathrm{Sin}}^2{\mathrm{\theta }}_1-{\mathrm{Sin}}^2{\mathrm{\theta }}_2>0 \\ \mathrm{or} ({\mathrm{Sin\theta }}_1-{\mathrm{Sin\theta }}_2)({\mathrm{Sin\theta }}_1+{\mathrm{Sin\theta }}_2) >0 \\ \mathrm{Since} {\mathrm{\theta }}_1 \mathrm{and} {\mathrm{\theta }}_2 \mathrm{both} \mathrm{are} \mathrm{acute} \mathrm{then} {\mathrm{Sin\theta }}_1-{\mathrm{Sin\theta }}_2 \mathrm{has} \mathrm{to} \mathrm{be} \mathrm{greater} \mathrm{than} \mathrm{zero}. \\ \Rightarrow {\mathrm{Sin\theta }}_1>{\mathrm{Sin\theta }}_2 \\ \mathrm{or} {\mathrm{\theta }}_1>{\mathrm{\theta }}_2 \\ \mathrm{So} \mathrm{option} \left(\mathrm{a}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Time} \mathrm{of} \mathrm{flights}, {\mathrm{T}}_1=\frac{2{\mathrm{v}}_0{\mathrm{Sin\theta }}_1}{\mathrm{g}} \mathrm{and} {\mathrm{T}}_2=\frac{2{\mathrm{v}}_0{\mathrm{Sin\theta }}_2}{\mathrm{g}} \\ \mathrm{Since} {\mathrm{Sin\theta }}_1>{\mathrm{Sin\theta }}_2 \mathrm{then} {\mathrm{T}}_1>{\mathrm{T}}_2. \\ \mathrm{So} \mathrm{option} \left(\mathrm{b}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Horizontal} \mathrm{Ranges}, {\mathrm{R}}_1=\frac{{{\mathrm{v}}_0}^2\mathrm{Sin}2{\mathrm{\theta }}_1}{\mathrm{g}} \mathrm{and} {\mathrm{R}}_2=\frac{{{\mathrm{v}}_0}^2\mathrm{Sin}2{\mathrm{\theta }}_2}{\mathrm{g}} \\ \mathrm{Since} {\mathrm{Sin\theta }}_1>{\mathrm{Sin\theta }}_2 \mathrm{then} {\mathrm{R}}_1>{\mathrm{R}}_2 \mathrm{but} \mathrm{if} {\mathrm{\theta }}_1+{\mathrm{\theta }}_2=90° \mathrm{then} {\mathrm{R}}_1={\mathrm{R}}_2. \\ \mathrm{So} \mathrm{option} \left(\mathrm{c}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Total} \mathrm{Energy} = \mathrm{Initial} \mathrm{Energy} = \frac{1}{2}{{\mathrm{mv}}_0}^2. \\ \mathrm{So} {\mathrm{U}}_1={\mathrm{U}}_2. \\ \mathrm{So} \mathrm{option} \left(\mathrm{d}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Hence} \mathrm{the} \mathrm{correct} \mathrm{answers} \mathrm{are} \mathrm{option} \left(\mathrm{a}\right) \mathrm{and} \left(\mathrm{b}\right). \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{track} \mathrm{is} \mathrm{frictionless} \mathrm{so} \mathrm{the} \mathrm{total} \mathrm{energy} \mathrm{is} \mathrm{conserved} \mathrm{everywhere}. \\ \mathrm{So} \mathrm{total} \mathrm{energy} \mathrm{at} \mathrm{P} = \mathrm{total} \mathrm{energy} \mathrm{at} \mathrm{A}. \\ \mathrm{The} \mathrm{potential} \mathrm{energy} \mathrm{at} \mathrm{A} \mathrm{is} \mathrm{converted} \mathrm{to} \mathrm{KE} \mathrm{and} \mathrm{PE} \mathrm{at} \mathrm{P}, \\ \mathrm{Hence} \mathrm{KE}\hspace{0.17em}\mathrm{at} \mathrm{B}>\mathrm{KE} \mathrm{at} \mathrm{P}. \\ \mathrm{And} \left(\mathrm{PE}\right) \mathrm{at} \mathrm{P} < \left(\mathrm{PE}\right) \mathrm{at} \mathrm{A} \\ \mathrm{As}, \mathrm{Height} \mathrm{of} \mathrm{P} < \mathrm{Height} \mathrm{of} \mathrm{A} \\ \mathrm{Hence}, \mathrm{path} \mathrm{length} \mathrm{AB} > \mathrm{path} \mathrm{length} \mathrm{BP} \\ \mathrm{Hence}, \mathrm{time} \mathrm{of} \mathrm{travel} \mathrm{from} \mathrm{A} \mathrm{to} \mathrm{B} \ne \mathrm{Time} \mathrm{of} \mathrm{travel} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{P}. \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

$$\begin{gathered} \mathrm{If} \mathrm{acceleraion} \mathrm{is} \mathrm{non} \mathrm{uniform} \mathrm{then} \mathrm{option} \left(\mathrm{a}\right) \mathrm{and} \left(\mathrm{c}\right) \mathrm{are} \mathrm{incorrect}. \\ \mathrm{Average} \mathrm{velocity} \mathrm{is} \mathrm{Total} \mathrm{displacement} \mathrm{upon} \mathrm{total} \mathrm{time}. \mathrm{So} \mathrm{option} \left(\mathrm{b}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Option} \left(\mathrm{d}\right) \mathrm{is} \mathrm{also} \mathrm{correct} \mathrm{for} \mathrm{uniform} \mathrm{acceleration}. \\ \mathrm{Hence} \mathrm{the} \mathrm{correct} \mathrm{answer} \mathrm{is} \mathrm{option} \left(\mathrm{a}\right) \mathrm{and} \left(\mathrm{c}\right). \end{gathered}$$

Answer:

In Uniform circular Motion, magnitude of particle velocity (speed) remains constant. Particle velocity remains directed perpendicular to radius vector. Direction of acceleration keeps changing as particle moves. Angular momentum is conserved.
Hence the correct answer is option (a), (b) and (c).

Answer:

$$\begin{gathered} \mathrm{If} \left|\mathbf{A}+\mathbf{B}\right|=\left|\mathbf{A}-\mathbf{B}\right| \mathrm{then} \sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABCos\theta }}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{ABCos\theta }} \\ \mathrm{or} 4\mathrm{ABCos\theta }=0 \\ \Rightarrow \mathrm{Either} \left|\mathbf{A}\right|=0 \mathrm{or} \left|\mathbf{B}\right|=0 \mathrm{or} \mathbf{A}\mathbf{ }\mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathbf{B}\mathbf{.} \\ \mathrm{Hence} \mathrm{the} \mathrm{correct} \mathrm{answer} \mathrm{is} \mathrm{option} \left(\mathrm{b}\right) \mathrm{and} \left(\mathrm{d}\right). \end{gathered}$$

Answer:

Acceleration at R is centripetal acceleration given by v²/r where v is tangential speed and r is radius.
So, acceleration =100/1000 = 0.1 m/s² directed towards the centre.

Answer:

Answer:

(a) Just before it hits the ground. Since mechanical energy is conserved, the ball gains kinetic energy as it's potential energy reduces as it falls from height. So, the kinetic energy will be maximum after potential energy gets totally converted to kinetic energy. 
(b) At the highest point reached since all the kinetic energy gets converted to potential energy.
(c) a = g = constant since gravity is the only external force acting on the ball and acceleration due to gravity is constant.

Answer:

(a) a = g = constant, since gravity is the only external force acting on the ball and acceleration due to gravity is constant.
This acceleration due to gravity always acts towards the ground.
b) As acceleration acts in opposite to the direction of motion during upward motion, so velocity of the ball goes on decreasing and becomes zero at the highest point. 

Answer:

Since B × C is perpendicular to plane of B and C, thus direction of A × (B × C) will lie in the plane of B and C.

Answer:

To the observer standing on the footpath or ground the ball appears to be a projectile and will follow a parabolic path. 

The symbolic diagram is as shown below:



 

Answer:

The boy standing on ground throws the ball at an angle of 60° with horizontal at a speed of 10 m/s.

The horizontal component of the velocity is given by $10\cos \left({60}^{\mathrm{o}}\right)= 5 {\mathrm{ms}}^{-1}$

The vertical component of the velocity is given by $10\sin \left({60}^{\mathrm{o}}\right)= 5\sqrt{3} {\mathrm{ms}}^{-1}$

Speed of the car = 18 km/h = 5 m/s
It is clear that the horizontal component of car and the ball are same. 
So the boy in the car will observe the ball going in the upward direction only .
It can be explained in diagram shown below:

Answer:

If air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.
Due to air resistance, particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure:

​When we are neglecting air resistance path was symmetric parabola. When air resistance is considered path is asymmetric parabola.

Answer:

The plane is travelling in the horizontal direction so the bomb dropped from the plane must have  some horizontal velocity which is equal to the plane's horizontal velocity

Horizontal velocity of the bomb initially = 720 km/h=  200 m/s

Let the target is at an angle θ with horizontal.

Height of he plane =  1.5 km = 1500 m

Let t be the time taken by the bomb to hit the target 

Horizontal distance travelled by the bomb  in time t is given by  200t
 
Vertical distance travelled by the bomb is given by

 $$\begin{gathered} h\mathit{=}\mathit{ }ut\mathit{ }+ \frac{1}{2}g{t}^2 \\ 1500 = \frac{12}{2}\times (9.8) \times t^2 \\ t = \sqrt{\frac{1500}{49}} = 17.49 \mathrm{s} \end{gathered}$$

The horizontal distance travelled by the bomb is 200t = $200\times 17.49 =3498 \mathrm{m} =3.5 \mathrm{km}$

Now to calculate the angle with respect to the horizontal direction:

​ $$\begin{gathered} \tan \theta \mathit{ }\mathit{=}\frac{h}{x}\mathit{ }=\frac{1500}{3498} \\ \theta = {\tan }^{-1}(0.4288) ={23}^{\mathrm{o}}{21}^{\text{'}} \end{gathered}$$


 

Answer:

(a) 
radius of the earth = 6400 km = 6400000 m

Time period of  Rotation of the Earth about its own axis is given by  $T = 24 \times 3600 \mathrm{s} = 86400 \mathrm{s}$

centripetal acceleration $a_{\mathrm{c}} =\mathit{ }{\omega }^{2}R = {\left(\frac{2\mathrm{\pi }}{T}\right)}^{2}R= \frac{4{\mathrm{\pi }}^{\mathit{2}}R}{T^2}= \frac{4{\mathrm{\pi }}^2\times 64\times {10}^5}{{86400}^2} =0.034 {\mathrm{ms}}^{-2}$
$\frac{g }{a_c}= \frac{9.8}{0.034}\approx 288$

 

Answer:

As shown in the diagram below in which vectors A and B are corrected by head and tail.
Resultant vector C = A + B

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{from} \left(\mathrm{iv}\right), \mathrm{it} \mathrm{is} \mathrm{clear} \mathrm{that} \mathit{c}\mathit{ }\mathit{=}\mathit{ }\mathit{a}\mathit{ }\mathit{+}\mathit{ }\mathit{b}\mathit{ } \\ \left(\mathrm{b}\right) \mathrm{from} \left(\mathrm{iii}\right), \mathit{c}\mathit{ }\mathit{+}\mathit{ }\mathit{b}\mathit{ }=\mathit{ }\mathit{a}\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\Rightarrow \mathit{a}\mathit{-}\mathit{ }\mathit{c}\mathit{ }\mathbf{=}\mathit{ }\mathit{b} \\ \left(\mathrm{c}\right) \mathrm{from} \left(\mathrm{i}\right), \mathit{b}\mathit{ }=\mathit{ }\mathit{a}\mathit{ }\mathit{+}\mathit{ }\mathit{c}\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\Rightarrow \mathit{ }\mathit{b}\mathit{-}\mathit{a}\mathit{ }=\mathit{ }\mathit{c}\mathit{ } \\ \left(\mathrm{d}\right) \mathrm{from} \left(\mathrm{ii}\right), \mathit{-}\mathit{c}\mathit{ }=\mathit{ }\mathit{a}\mathit{ }\mathit{+}\mathit{ }\mathit{b} \\ \mathit{ }\Rightarrow \mathit{ }\mathit{a}\mathit{ }\mathit{+}\mathit{ }\mathit{b}\mathit{ }\mathit{+}\mathit{ }\mathit{c}\mathit{ }= 0 \end{gathered}$$

​(Note: a, b , c, A and B are vectors)

Answer:

(a) A.B  = A B cos (90^o ) = 0 

So (a) matches with (ii)

(b)  $\mathit{A}\mathbf{.}\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left(0^o\right) = +8$

  So (b) matches with (i) 

(c) $\mathit{A}\mathbf{.}\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left({60}^o\right) = +4$

  So (c) matches with (iv).


(d) $\mathit{A}\mathbf{.}\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left({180}^o\right) = -8$

So (d) matches with (iii) 

Answer:

(a) $\mathit{A}\mathbf{\times }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \sin \left(0^o\right) = 0$ 

So (a) matches with (iv)

(b)  $\mathit{A}\mathbf{\times }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left({90}^o\right) = +8$

  So (b) matches with (iii) 

(c) $\mathit{A}\mathbf{\times }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left({30}^o\right) = +4$

So (c) matches with (i) 

(d) $\mathit{A}\mathbf{\times }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }2\times 4\times \cos \left({45}^o\right) = 4\sqrt{2}$

 So (d) matches with (ii) 
 

Answer:

As  the hill is 500 m high, 
So, the minimum vertical component of the velocity of packet is,
$u_y=\sqrt{2gh}=\sqrt{2\times 10\times 500}=100 \mathrm{m}/\mathrm{s}$
Also the maximum velocity with which the packet can be thrown is v = 125 m/s.
So, the required horizontal component of the velocity is,
$u_x=\sqrt{v^2-{\left(u_y\right)}^2}=\sqrt{{125}^2-{100}^2}=75 \mathrm{m}/\mathrm{s}$
Time to reach the maximum height,
$t=\frac{u_y}{g}=\frac{100}{10}=10 \mathrm{s}$
Horizontal distance covered by the  packet is,
$$\begin{gathered} =u_x\times t=75\times 10=750 \mathrm{m} \\ \mathrm{Remaining} \mathrm{distance} (800 \mathrm{m}-750 \mathrm{m}=50 \mathrm{m}), \mathrm{that} \mathrm{should} \mathrm{be} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{cannon} \mathrm{with} \mathrm{speed} 2 \mathrm{m}/\mathrm{s}, \\ \mathrm{So} \mathrm{required} \mathrm{time} = \frac{50}{2}=25 \mathrm{s} \end{gathered}$$
Hence, the total time taken by the packet finally to the ground is
$$\begin{gathered} =(25 +10+10) \mathrm{s} \\ =45 \mathrm{s} \end{gathered}$$




 

Answer:

Let us consider for horizontal motion:
$$\begin{gathered} R+∆x=v_0\cos \left(\theta \right)\times t \\ \Rightarrow t=\frac{R+∆x}{v_0\cos \left(\theta \right)} \\ \mathrm{Now} \mathrm{for} \mathrm{vertical} \mathrm{motion}, \\ -h=v_{0}sin\left(\theta \right)t-\frac{1}{2}g{t}^2 \\ \mathrm{Now} \mathrm{by} \mathrm{putting} \mathrm{the} \mathrm{vaue} \mathrm{of} t, and value of \theta ={45}^o \mathrm{we} \mathrm{get}, \end{gathered}$$
(As maximum horizontal range $\left(R=\frac{v_0^2}{g}\right)$  can be achieved only if angle is 45^o)
$h=∆x\left(1+\frac{∆x}{R}\right)$


 

Answer:

Consider new Cartesian coordinates in which X-axis is along inclined plane OP and Y-axis along OY as shown in the diagram. 
Consider the motion of the projectile from OAP.

$$\begin{gathered} h=u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ 0=V_0\sin \beta -\frac{1}{2}g\cos \alpha t^2 \\ \left(\mathrm{In} \mathrm{this} \mathrm{co}-\mathrm{ordinate} \mathrm{system}, \mathrm{when} \mathrm{projectile} \mathrm{reches} \mathrm{to} \mathrm{P}, \mathrm{then} \mathrm{vertical} \mathrm{displacement} \mathrm{is} 0.\right) \\ \mathrm{So}, \mathrm{time} \mathrm{of} \mathrm{flight}: \\ t=\sqrt{\frac{2V_0\sin \beta }{g\cos \alpha }} ...\left(1\right) \end{gathered}$$

(a) Range on the inclined plane surface,
 $$\begin{gathered} \mathrm{Range} \left(\mathrm{OP}\right) \mathrm{on} \mathrm{X}-\mathrm{axis}=L \\ L=u_{x}t+\frac{1}{2}{a}_x{t}^2 \\ \Rightarrow L=V_o\cos \beta \times \frac{2V_o\sin \beta }{g\cos \alpha }-\frac{1}{2}\left(g\sin \alpha \right){\left(\frac{2V_o\sin \beta }{g\cos \alpha }\right)}^2=\frac{2{V_o}^2\sin \left(\beta \right)\cos \left(\alpha +\beta \right)}{g{\cos }^2\left(\alpha \right)} \end{gathered}$$

(b) Time of flight,
$t=\frac{2V_o\sin \beta }{g\cos \alpha } \left[ \mathrm{From} \mathrm{equation} \left(1\right)\right]$

 (c)Expression for range, as calculated in part (a) is:

$L==\frac{2{v_o}^2\sin \left(\beta \right)\cos \left(\alpha +\beta \right)}{g{\cos }^2\left(\alpha \right)}$
As, $\alpha$ is angle of inclination, which is fixed, so:
for maximum range, 
$\beta =\frac{\pi }{4}-\frac{\alpha }{2}$
 

Answer:

After rebound its velocity will be v_o, but at an angle $\theta$ with vertical axis perpendicular to the inclined plane.
So, time of flight be,
$$\begin{gathered} T=\frac{2v_o\cos \left(\theta \right)}{g\cos \left(\theta \right)}=\frac{2v_o}{g} \\ \mathrm{Horizontal} \mathrm{range}, \\ R=u_{x}T+\frac{1}{2}{a}_x{T}^2 \\ =\frac{2v_{\mathrm{o}}\times v_{\mathrm{o}}\sin \left(\theta \right)}{\mathrm{g}}+\frac{1}{2}g\sin \left(\theta \right){\left(\frac{2v_{\mathrm{o}}}{g}\right)}^2 \\ \Rightarrow R=L =\frac{4{\left(v_{\mathrm{o}}\right)}^2\sin \left(\theta \right)}{g} \end{gathered}$$

Answer:

Let unit vector along  north direction be  $\hat{i}$  and vertically downward  be $-\hat{j}$.
and velocity of rain, $\overrightarrow{v_r}=a\hat{i} +b\hat{j}$
Velocity of girl, $\overrightarrow{v_g}=5\hat{i}$
Velocity of rain with respect to girl ,
$\overrightarrow{v_{rg}}=\overrightarrow{v_r} -\overrightarrow{v_g} =(a-5)\hat{i}+b\hat{j}$
As rain falls vertically downwards, so
$$\begin{gathered} a-5=0 \\ \Rightarrow a=5 \mathrm{m}/\mathrm{s} \end{gathered}$$
$\mathrm{As}, \tan {\left(45\right)}^{\mathrm{o}}=\frac{a-10}{b}=\frac{5-10}{b}\Rightarrow b=-5 \mathrm{m}/\mathrm{s}$ 
When rain falls 45^o to the vertical, then v_g= 10 m/s and 
$$\begin{gathered} \overrightarrow{v_r} =a\hat{i}+b\hat{j}=5\left( \hat{i}-\hat{j} \right) \\ \left|\overrightarrow{v_r}\right|=5\sqrt{2} \mathrm{m}/\mathrm{s} \end{gathered}$$


 

Answer:

(a) Resultant velocity of  swimmer = $\sqrt{4^2+3^2}=5 \mathrm{m}/\mathrm{s} \mathrm{at} \mathrm{an} \mathrm{angle} {\tan }^{-1}\left(\frac{3}{4}\right)={37}^o \mathrm{of} \mathrm{North}.$
(b) (i) If he wants to reach the point directly opposite to him so,
           He should swim at an angle ${\tan }^{-1}\left(\frac{3}{\sqrt{7}}\right) \mathrm{of} \mathrm{North}$ 
     (ii) The result of the speed of the swimmer, $\sqrt{4^2-3^2}=\sqrt{7} \mathrm{m}/\mathrm{s}$
(c) In case (a) time taken by the swimmer to cross the river =$\frac{d}{4}$
      In case (b) time taken by the swimmer to cross the river = $\frac{d}{\sqrt{7}}$

So, the swimmer will cross the river in shortest time in case (a)
 

Answer:

(a) Let initial horizonal and vertical component of velocity along x-axis and y-axis respectively as,
$$\begin{gathered} u_x=u+v_o\cos \left(\theta \right) \\ {u}_y=v_o\sin \left(\theta \right) \\ \mathrm{So}, \tan \left(\theta \right)=\frac{u_{\mathrm{y}}}{u_{\mathrm{x}}}=\frac{v_{\mathrm{o}}\sin \left(\theta \right)}{u+v_{\mathrm{o}}\cos \left(\theta \right)} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{v_{\mathrm{o}}\sin \left(\theta \right)}{u+v_{\mathrm{o}}\cos \left(\theta \right)}\right) \end{gathered}$$

(b) Time of flight,
$T=\frac{2u_y}{g}=\frac{2v_o\sin \left(\theta \right)}{g}$

(c) Horizontal range along x-axis,
$R=u_x\times T=\left(u+v_o\cos \theta \right)\frac{2v_o\sin \theta }{g}$

(d) Horizontal range to be maximum if,
$$\begin{gathered} \frac{dR}{d\theta }=0 \\ \mathrm{Solving} \mathrm{we} \mathrm{get}, \\ {\theta }_{max}=co{s}^{-1}\left(\frac{-u+\sqrt{u^2+{8v_o}^2}}{4v_o}\right) \end{gathered}$$
(e) For u = v_0,
$$\begin{gathered} \mathrm{Solving} \mathrm{we} \mathrm{get}, \\ {\theta }_{max}=co{s}^{-1}\left(\frac{-v_o+\sqrt{u^2+{8v_o}^2}}{4v_o}\right)=co{s}^{-1}\left(\frac{1}{2}\right)={60}^o \end{gathered}$$
For u << v₀
 $$\begin{gathered} \mathrm{Solving} \mathrm{we} \mathrm{get}, \\ {\theta }_{max}=co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\mathrm{\pi }}{4} \end{gathered}$$
For u >> v₀
$$\begin{gathered} \mathrm{Solving} \mathrm{we} \mathrm{get}, \\ {\theta }_{max}=\frac{\mathrm{\pi }}{2} \end{gathered}$$
(f)    for u = 0
$$\begin{gathered} \mathrm{Solving} \mathrm{we} \mathrm{get}, \\ {\theta }_{max}>{45}^o \end{gathered}$$




 

Answer:

 Ans.

(a) $\hat{r}$ = $\cos \theta \hat{i}+\sin \theta \hat{j}$  
        $\hat{\theta }$ = $-\sin \theta \hat{i} + \cos \theta \hat{j}$ 

So , $\hat{i}$ = $\cos \theta \hat{r} - \sin \theta \hat{\theta }$
       $\hat{j}$ = $\sin \theta \hat{r} +\hspace{0.17em}\cos \theta \hat{\theta }$

(b) | $\hat{r}$ | = $\sqrt{({\cos }^2\theta + {\sin }^2\theta )}$ = 1
      
      | $\hat{\theta }$ | = $\sqrt{{\sin }^2\theta + {\cos }^2\theta }$ = 1 

$\hat{r} . \hat{\theta } = (\hspace{0.17em}\cos \theta )(-\sin \theta ) + \left(\sin \theta \right)\left(\cos \theta \right) = 0$

Hence , $\hat{r} \mathrm{and} \hat{\theta }$  are perpendicular to each other . 

(c)

$$\begin{gathered} \frac{d}{dt}\left(\hat{r}\right) = \frac{d}{dt}(\cos \theta \hat{i}+\sin \theta \hat{j}) =\frac{d}{d\theta }(\cos \theta \hat{i}+\sin \theta \hat{j})\frac{d\theta }{dt} \\ =( -\sin \theta \hat{i}+\cos \theta \widehat{j }) \omega = \omega \hat{\theta } \\ \frac{d}{dt}\left(\hat{\theta }\right) = \frac{d}{dt}(-\sin \theta \hat{i}+\cos \theta \widehat{j }) = \frac{d}{d\theta }( -\sin \theta \hat{i}+\cos \theta \widehat{j })\frac{d\theta }{dt} \\ = (-\cos \theta \hat{i}+-\sin \theta \hat{j}) \omega =- \omega \hat{r} \end{gathered}$$

(d) $\overrightarrow{r}=a\theta \hat{r}$ 

$\theta$ = dimensionless
$\hat{r}$ = dimensionless

So, dimensions  of  $\overrightarrow{r}$ = dimensions of a = [L] 


(e)
 $$\begin{gathered} velocity = \stackrel{\rightharpoonup }{v}=\frac{d}{dt}\left(\overrightarrow{r}\right)=\frac{d}{dt}\left(a\theta \hat{r}\right) = a \frac{d}{dt}\left(\theta \hat{r}\right) = a \left[\frac{d\theta }{dt}\hat{r}+\theta \frac{d\left(\hat{r}\right)}{dt} \right]= a ( \omega \hat{r}+\theta \omega \hat{\theta }) \\ =a\omega (\hat{r}+\theta \hat{\theta }) \\ acceleration = \overrightarrow{a}= \frac{d}{dt}\left(\overrightarrow{v}\right) = \frac{d}{dt}\left[a\omega (\hat{r}+\theta \hat{\theta }) \right]=a\omega \frac{d}{dt}\left(\hat{r}+\theta \hat{\theta }\right) \\ =a\omega \left[\frac{d\left(\hat{r}\right)}{dt} + \frac{d\left(\theta \hat{\theta }\right)}{dt}\right] =a\omega \left( \omega \hat{\theta } +\frac{d\theta }{dt}\hat{\theta }+\theta \frac{d\hat{\theta }}{dt}\right) = a\omega \left[ 2\omega \hat{\theta }+\theta (-\omega \hat{r}) \right]= a{\omega }^2(2\hat{\theta }-\theta \hat{r}) \end{gathered}$$



      






 

Answer:

Time taken by the man via sand on the path APQC, be
$T_{sand}=\frac{2\times 25\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}=50\sqrt{2}\left(1+\frac{1}{v}\right)$
The shortest path outside the sand will BE ARC. Time to cover this path will be,
$$\begin{gathered} T_{outside}=\frac{\mathrm{AR}+\mathrm{RC}}{1}=\frac{2\sqrt{{75}^2+{25}^2}}{1}=50\sqrt{10} \mathrm{S} \\ \mathrm{For}, \\ {T}_{sand}<T_{outside} \\ 50\sqrt{2}\left(1+\frac{1}{\mathrm{v}}\right)<50\sqrt{10} \\ \Rightarrow v>\frac{1}{\sqrt{5}-1} \\ \Rightarrow v>0.81 \mathrm{m}/\mathrm{s} \end{gathered}$$