Solutions-Class-11-Science-Physics-Chapter-3-Motion in a Plane-Maharashtra Board

(C) Projectile motion

(A) Speed

(B) Uniform motion in a horizontal circle

(D) Velocity-time graph is nonlinear

(C) \(\vec{V_A}\) − \(\vec{V_B}\)

Scalars : speed, work done, power, energy, electric charge.

Vectors : velocity, displacement, force, acceleration, angular velocity.

Average velocity : The average velocity of a particle is the displacement of the particle divided by the time elapsed during which the displacement occurs.

If \(\vec{x_1}\) and \(\vec{x_2}\) are the position vectors of the particle at instants t1 and t2 respectively, then

\(\vec{v_{av}}=\frac{displacement}{elapsed\,time} = \frac{Δ\vec{x}}{Δt}\) = \(\frac{\vec{x_2}-\vec{x_1}}{t_2-t_1}\)

Instantaneous velocity : The velocity at a given instant is called the instantaneous velocity and is defined as the limiting value of the average velocity as the time interval approaches zero.

\(\vec{v}=\lim\limits_{Δt\to{0}}\frac{Δ\vec{x}}{Δt}= \frac{d\vec{x}}{dt}\)

When a particle moves with uniform velocity (i.e., a velocity which remains constant with time both in magnitude and direction), the acceleration is zero and the average velocity equals the instantaneous velocity.

When a body starts with zero velocity at a certain height from the ground and falls under the influence of the gravity of the Earth, ignoring air resistance and buoyancy of air, it is said to be in free fall..

Given x =f(t), the instantaneous velocity is 

\(v=\frac{dx}{dt}=\frac{d}{dt}f(t)\)

and the instantaneous acceleration is

\(a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=\frac{d^2}{dt^2}f(t)\)

Let \(\vec{u}\) be the initial velocity of a particle at an initial time t_i = 0, \(\vec{a}\) its uniform (constant) acceleration, \(\vec{v}\) its final velocity at a final time t_f = t, \(\vec{r_0}\) is the position vector at time t_i, \(\vec{r}\) the position vector at time t_f, and \(\vec{s}\) the displacement of the particle in the elapsed time Δt = t_f — t_i = t. Then,

\(\vec{a}\) = \(\frac{\vec{v}-\vec{u}}{t}\)

∴ \(\vec{v}\) = \(\vec{u}+\vec{a}t\) =  …..(1)

\(\vec{s}\) = \((\vec{v_{av}})t\) = \(\frac{1}{2}(\vec{u}+\vec{v})t\)

= \(\frac{1}{2}(\vec{u}+\vec{u}+\vec{a}t)t\)

∴ \(\vec{s}\) = \(\vec{u}t+\frac{1}{2}\vec{a}t^2\)      …..(2)

The dot product of Eq.(1) with itself gives

\(\vec{v}·\vec{v}\) = \((\vec{u}+\vec{a}t)·(\vec{u}+\vec{a}t)\)

                              = \(\vec{u}·\vec{u}+2\vec{u}·\vec{a}t+\vec{a}·\vec{a}t^2\)

∴ v² = u² + \(2\vec{a}·(\vec{u}t+\frac{1}{2}\vec{a}t^2)\)

Substituting from Eq. (2),

v² = u² + 2\(\vec{a}·\vec{s}\)   ……..(3)

Equations (1), (2) and (3) are the kinematical equations (equations of motion) in vector form.

For a motion in an x-y plane (i.e., a two-dimensional motion), the component equations are

v_x_= u_x + a_xt  ……(1a)

v_y_= u_y + a_yt  …..(1b)

s_x = x — x₀ = u_xt + \(\frac{1}{2}\)a_xt² ...(2a)

s_y = y — y₀ = u_yt + \(\frac{1}{2}\)a_yt²...(2b)

v²_x = u²_x + a_xs_x  ….(3a)

v²_y = u²_y + a_ys_y  …..(3b)

We can see that Eqs. (1a), (2a) and (3a) involve only the x-components of displacement, velocity and acceleration, while Eqs. (1b), (2b) and (3b) involve the y-components of these quantities.

Thus, a two-dimensional motion can be described by two exclusive components along two perpendicular directions.

Consider a particle moving in a straight line with an initial velocity u and constant acceleration a. The v—t graph is a straight line with slope a and y-intercept u. With u and a both positive, the graph is as shown in Fig.

The slope of the line gives the acceleration of the particle,

a = \(\frac{dv}{dt}=\frac{v-u}{t-0}=\frac{v-u}{t}\)

∴ v = u + at

This is the first equation of motion (or the first kinematical equation).

The area under the ‘straight line for a time interval Δt = t − 0, as shown by the shaded region,

area of rect. OACD + area of Δ ABC

= (OA x AC) + \(\frac{1}{2}\)(AC x BC)

= ut + \(\frac{1}{2}\)t(v − u)

= ut +\(\frac{1}{2}\)at²  ( '.' v−u = at)

= s, distance covered in the elapsed time t.

∴ s = ut +\(\frac{1}{2}\)at²

This is the second equation of motion (or the second kinematical equation).

[Alternatively, the shaded area = area of the trapezium OABD = \(\frac{1}{2}\)(OA + BD) x OD

= \(\frac{1}{2}\)(v + u)t

= v_av t      { '.' v_av = \(\frac{1}{2}\)(v + u)}

= distance covered in the elapsed time t]

∴ s = \(\frac{1}{2}(u+v)(\frac{v-u}{a})\)

substituting for t from the first kinematical equation. '

∴ s = \(\frac{v^2-u^2}{2a}\)

∴ v² = u² + 2as

This is the third equation of motion (or the third kinematical equation).

If the projectile is launched with an initial velocity  and a launch angle θ from the origin of the vertical x—y frame of reference containing  the x and y components of the initial velocity are

u_x = u cos θ, u_y = u sin θ …… (1)

and those of the acceleration are

a_x = 0, a_y = −g  ….(2)

(i) The vertical component of the instantaneous velocity is

v_y = u_y + a_yt = u sin θ — gt ………(3)

The time of ascent (t_a) is the time taken by the projectile to reach the peak of its trajectory where v_y = 0. Hence, substituting t = t_a and v_y = 0 in the above equation (3),

 t_a = \(\frac{u\,sin\,θ}{g}\)

(ii) The time of flight (T) of a projectile is the time it takes to cover its entire trajectory or to cover its hofizontal range

The time of descent (t_d) is the time taken by the projectile to return to its initial launch height, usually the ground level, from the peak of its trajectory. It can be shown that t_d = t_a.

∴ T = t_a + t_d = 2t_a = \(\frac{2u\,sin\,θ}{g}\)

(iii) The range or horizontal range R is the horizontal distance travelled by the projectile during its flight.

∴ R = u_xT = \(\frac{2u_xu_y}{g}\)

       = \(\frac{2(u\,sin\,θ)(u\,cos\,θ)}{g}\)

      = \(\frac{u^2(sin\,θ\,cos\,θ)}{g}\)

  ∴ R = \(\frac{u^2sin\,2θ}{g}\)  ….(4)

(iv) The vertical component of the displacement of the projectile at an instant t is

y = u_yt − ½ gt²

At t= t_a, the projectile is at its peak, i.e., at maximum height, y = H.

∴ H = u_yt_a − \(\frac{1}{2}\)gt²_a

= \(u_y(\frac{u_y}{g})-\frac{1}{2}g(\frac{u_y}{g})^2\)

=  \(\frac{u_y^2}{2g}=\frac{u^2sin^2θ}{2g}\) 

Suppose a projectile is launched obliquely with an initial velocity , inclined at an angle θ to the horizontal, called the launch angle or the angle of projection, and that after launching no force, except that due to gravity, acts on it.

Then, the projectile's trajectory or path is confined to a vertical plane containing . Its acceleration, due only to gravity, is constant near the Earth's surface and always vertically downward. In the horizontal direction it is not acted upon by any force. Hence, along that direction, it has no acceleration, i.e., it moves with a constant velocity. We orient the x-axis horizontal and the y-axis vertical such that  lies in the xy plane and take the launch point as the origin of the axes; the motion then lies in the xy plane. (For reference, see Fig.)

The x and y components of the kinematic variables are

(i) the initial velocity :

u_x = u cos θ, u_y = u sin θ …… (1)

(ii) the acceleration :

a_x = 0, a_y = −g  ….(2)

(iii) the displacement :

 x = u_xt + \(\frac{1}{2}\)at² = (u cos θ)t …….(3)

y = u_yt + \(\frac{1}{2}\)a_yt²

   = (u sin θ)t − \(\frac{1}{2}\)gt² ……(4)

From above Eq. (3), t = \(\frac{x}{u\,cos\,θ}\)

Substituting the value of t in Eq. (4),

 y = \((u\,sin\,θ)\frac{x}{u\,cos\,θ}-\frac{1}{2}g(\frac{x}{u\,cos\,θ})^2\)

∴ y = x tan θ \(-(\frac{g}{2u^2\,cos^2\,θ})x^2\)

Since θ, u and g are the constants of the motion, the above equation is of the form y = bx — cx², in which b and c are constants. This being the equation of a parabola, the trajectory of the projectile is a parabola.

Conical pendulum : A conical pendulum is a simple pendulum whose bob revolves in a horizontal circle with constant speed with the centre of its circular path below the point of suspension such that the string makes a constant angle θ with the vertical.

Expression :

Consider a conical pendulum of string length L with its bob of mass m performing UCM along a horizontal circle of radius r (Fig-7.).

The centre O’ of the circular path is vertically below the point of suspension O, and the string makes a constant angle θ with the vertical.

At every instant of its motion, the bob is acted upon by its weight m\(\vec{g}\)  and the tension\(\vec{F}\)  in the string. If the constant linear speed of the bob is v, the necessary horizontal centripetal force is

F_C = mv²/r

\(\vec{F}\) is the resultant of the tension in the string and the weight. Resolve \(\vec{F}\) into components F cos θ vertically opposite to the weight of the bob and F sin θ horizontal. F cos θ balances the weight. F sin θ is the necessary centripetal force.

∴ F sin θ = mv²/r  and

   F cos θ = mg

Dividing Eq. (1) by Eq. (2),

tan θ = v²/rg  ……(3)

The period T of the conical pendulum is the time taken by its bob to complete one revolution in a horizontal circle with constant speed v.

Then,

v = 2πr/T    ….(4)

∴ tan θ = \(\frac{4π^2r^2}{T^2rg}\)=\(\frac{4π^2r}{T^2g}\)

∴ T² = \(\frac{4π^2r}{g\,tan\,θ}\)= \(\frac{4π^2L\,sin\,θ}{g\,tan\,θ}\)   ..(‘.’ r = L sin θ)

     = \(\frac{4π^2L\,cos\,θ}{g}\)

 ∴ T = 2π\(\sqrt{\frac{L\,cos\,θ}{g}}\)=2π\(\sqrt{\frac{h}{g}}\)    … (5)

where h = L cos θ is the axial height of the cone.

Equation (5) is the required expression.

(i) Angular velocity : The time rate of angular displacement of a particle performing circular motion is called the angular velocity.

If the particle has an angular displacement δθ in a short time interval δt its angular velocity

\(\vec{ω} =\lim_{δt→0}\frac{\vec{δθ}}{dt}=\frac{\vec{dθ}}{dt}\)

 is a vector in the direction of \(\vec{δθ}\), given by the right hand rule or right-handed screw rule.

(ii) Consider a particle performing uniform circular motion with a constant angular velocity \(\vec{ω}\) along a circle in the xy plane. Let r be the radius of the circle and its centre be the origin of the coordinate axes. If the angular position of the particle at any instant t is θ as shown in the figure, then θ = ωt.

If x and y are the coordinates of the particle, its position vector or the radius vector at time t is  \(\vec{r}=x\hat{i}+y\hat{j}\)

where \(\hat{i}\) and \(\hat{j}\) are unit vectors along the x and y axes respectively.

From Δ POM,

x =r cos θ = r cos ωt and

y = r sin θ = r sin ωt

 ∴ \(\vec{r}\) = r cos ωt \(\hat{i}\) + r cos ωt \(\hat{j}\)

The linear velocity of the particle is

\(\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}\)(r cos ωt \(\hat{i}\) + r cos ωt \(\hat{j}\))

∴\(\vec{r}\) = −ωr sin ωt \(\hat{i}\) + ωr cos ωt \(\hat{j}\)

Since r and ω are constant in uniform circular motion  is along the tangent to the path in the sense of motion of the particle.

The linear acceleration of the particle is

\(\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}\)(−ωr sin ωt \(\hat{i}\) + ωr cos ωt \(\hat{j}\))

    = −ω²r cos ωt \(\hat{i}\) − ω²r sin ωt \(\hat{j}\)

    = −ω²(r cos ωt \(\hat{i}\) + r sin ωt \(\hat{j}\))

∴\(\vec{a}\)=−ω²\(\vec{r}\)

The acceleration of a particle performing UCM has the magnitude ω²r. The minus sign shows that its direction is opposite to that of \(\vec{r}\).

It means that at every instant the acceleration is directed along the radius towards the centre of the circle. This acceleration is called radial acceleration or centripetal acceleration.

A net real force must act on the particle to produce this acceleration. This force, which at every instant must point towards the centre of the circle, is called the centripetal force.

If m is the mass of the particle, the centripetal force,

\(\vec{F}\)= m\(\vec{a}\)  = −mω²\(\vec{r}\)

Given : s = 500 m, u = 0, constant acceleration, t = 30 s

Since acceleration is constant,

s = v_avt = \((\frac{u+v}{2})=\frac{v}{2}t\)   (‘.’ u = 0)

The velocity of the aeroplane at take off,

v = 2s/t = 2(500)/30 = 100/3 m/s

   = \(\frac{100/1000\,km}{3/3600\,hr}=\frac{360}{3}\) = 120 km/h

Given : u = 120 km/h = (120×10³ )/3600  m/s = 100/3 m/s, s = 100 m, v=0

(i) v² = u² + 2as = 0,  as v = 0

∴ 2as = −u²

∴ a = −u²/2s

 ∴ a = \(-\frac{(100/3)^2}{2×100}\) = \(-\frac{100×100}{9×2×100}\)

      = \(-\frac{50}{9}\) = − 5.556 m/s²

∴ The average retardation of the car

= −a = 5.556 m/s²

(ii) v = u + at = 0, as v = 0

∴  u = −at, ∴ t = —u/a

∴t = \(-\frac{(100/3)}{(-50/9)}\) = \(\frac{300}{50}\) = 6 s

This is the time taken by the car to come to rest

Given : v₁ = 50 km/h, t₁ = 30 min =30/60 h = 0.5 h,

v₂ = 30 krn/h, t₂ = 15 min = 15/60 h = 0.25 h,

v₃ = 70 km/h, t₃ = 45 min = 45/60 h = 0.75 h

The average speed of the car,

v =  \(\frac{s_1+s_2+s_3}{t_1+t_2+t_3}\) = \(\frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}\)

   = \(\frac{50×0.5+30×0.25+70×0.75}{0.5+0.25+0.75}\)

   = \(\frac{25+7.5+52.5}{1.5}=\frac{85}{1.5}\)

    = 56.66 km/h

(a) The initial speed (at t = 0) of the car = speed at O = 0 m/s.

(b) The maximum speed attained by the car = speed at A = 20 m/s.

(c) The car has zero acceleration (or constant speed) between t = 3 s and t = 6 s as shown by the part AB.

(d) In the 2 seconds time interval (t = 6 s to t= 8 s) in which the car decelerates, the part BC of the graph is a straight line showing that there is uniform (constant) retardation.

[Note : The car had varying speed in the regions OA and BC.]

(e) The distance travelled by the car in the first 6 seconds

= area under the curve OAB

= area of Δ OAE + area of rect. ABDE

=  OE x AE + AB x AE

=  x 3 x 20 + 3 x 20 = 30 + 60 = 90 m

Given: R_max = 80 m

R = \(\frac{u^2\,sin\,2θ}{g}\) and H = \(\frac{u^2\,sin\,2θ}{2g}\)

The maximum range is attained for launch angle θ = 45°, so that sin 2θ  = 1 and the maximum range of the projectile is

R_max = u²/g

The corresponding maximum (peak) height is

H = \(\frac{u^2(sin\,45)^2}{2g}\) = \(\frac{u^2(\frac{1}{\sqrt{2}})^2}{2g}\) = u²/4g

∴ H = R_max/4 = 80/4 = 20 m

Suppose a particle is projected with an initial speed u = v₀ and angle of projection θ from the bottom of an inclined surface whose angle of inclination with the horizontal is Φ, Fig.

Let the launch point be the origin of the vertical x-y frame of reference. The particle strikes the inclined surface at a point A. We have to find an expression for the range OA.

The equation of the trajectory of profile is given by

y = x tan θ − \((\frac{g}{2u^2\,cos^2θ})x^2\)  …..(1)

In Δ OAB

R = OA = \(\frac{OB}{cosΦ}\) = \(\frac{x}{cosΦ}\)

∴ x = R cos Φ    …..(2)

And R = OA = \(\frac{AB}{sinΦ}\) = \(\frac{y}{sinΦ}\)

∴ y = R sin Φ    …..(3)

From eq. (1), (2) and (3)

R sin Φ  = R cos Φ tan θ − \((\frac{g}{2v_0^2\,cos^2θ})\)R²cos²Φ   ….(‘.’ u=v₀)

∴ sin Φ  =  cos Φ tan θ − \((\frac{g}{2v_0^2\,cos^2θ})\)R cos²Φ 

∴ \(\frac{gR\,cos^2Φ}{2v_0^2\,cos^2θ}\) = ( cos Φ tan θ - sin Φ )

Hence

R = \((\frac{2v_0^2\,cos^2θ}{g\,cos^2Φ})\)( cos Φ tan θ - sin Φ )

   = \((\frac{2v_0^2\,cos\,θ}{g\,cos^2Φ})(\frac{cos\,θ\,cos\,Φ\,sin\,θ}{cos\,θ} - cos\,θ\,sin\,Φ)\)

   = \((\frac{2v_0^2\,cos\,θ}{g\,cos^2Φ})\)( sin θ cos Φ − cos θ sin Φ )

   = \((\frac{2v_0^2\,cos\,θ\,sin(θ-Φ)}{g\,cos^2Φ})\)  ….[‘.’ sin θ cos ∅ − cos θ sin ∅  = sin(θ − ∅)]

Given :

 (a) v-t graph

(b) Distances from the graph :

The distance between stations A and B = the shaded area ADEBA

= \(\frac{(240-0)+(230-10)}{2}×20\) = (240 + 220) x 10 = 4600 m = 4.6 km

The distance between stations B and C = the shaded area BFGCB

= \(\frac{(440-260)+(430-270)}{2}×20\)  = (180 + 160) x 10 = 3400 m = 3 4 km

Given : Speed of the train relative to the Earth, v_TE = 10 m/s due east;

speed of the waiter relative to the train, v_WT = 1.2 m/s due east;

speed of the fly relative to the waiter, v_FW = 2 m/s due north

Using an xy coordinate system with the positive direction of x due east and that of y due north,

 \(\vec{v}_{TE}=10\hat{i}\) m/s,  \(\vec{v}_{WT}=1.2\hat{i}\) m/s,  \(\vec{v}_{FW}=2\hat{j}\) m/s

The velocity of the fly relative to the Earth is

 \(\vec{v}_{FE}=\vec{v}_{FW}+\vec{v}_{WT}+\vec{v}_{TE}\)

= 2\(\hat{j}\) + 1.2\(\hat{i}\)  + 10\(\hat{i}\)

= (11.2\(\hat{i}\) + 2\(\hat{j}\)) m/s

|\(\vec{v}_{FE}\)| = \(\sqrt{(11.2)^2+(2)^2}\)

          = \(\sqrt{125.44+4}\)

          = 11.38 m/s

 tan θ = 2/11.2 = 0.1786

∴ θ = tan^-1 0.1786 = 10°7'

The velocity of the fly relative to the Earth is (11.2\(\hat{i}\) + 2\(\hat{j}\)) m/s, i.e., 11.38 m/s in the direction of 10°7' north of east.

Given: v = 50 m/s, T= 40s

The magnitude of the acceleration of the car = a

= v ω

=  v\((\frac{2π}{T})\)               ...( \(ω=\frac{2π}{T}\) 

= 50\((\frac{2×3.142}{40})\) = \((\frac{50}{20})\)(3.142)

= 2.5 x 3.142 = 7.855 m/s²

Given : v = 15 m/s, r =2m

The centripetal acceleration of the particle

= a = v²/r = (15)²/2

= 225/2

= 112.5 m/s²

Given:

θ = 30°, R₁ = 40m, R₂ =50 m, g = 10 m/s²

R = \(\frac{u^2\,sin\,2θ}{g}\)

∴ u² = \(\frac{Rg}{sin60°}\) = \(\frac{10R}{\sqrt{3}/2}\)

∴ u² = \(\frac{20R}{\sqrt{3}}\) 

For R = R₁,

u₁² = \(\frac{20×40}{\sqrt{3}}\) = = \(\frac{800}{\sqrt{3}}\) = 461.9 (m/s)²

∴ u₁ = 21.49 m/s

As u² = \(\frac{20R}{\sqrt{3}}\) 

\(\frac{u_2^2}{u_1^2}=\frac{R_2}{R_1}=\frac{50}{40}\) = 1.25

∴ u₂² = 1.25 u₁² = (1.25)(461.9) = 577.3 (m/s)²

∴ u₂ = 24.03 m/s

 21.49 ≤ u ≤ 24.03 m/s

This is the required range of the initial (i.e launch) velocity.