Motion in Plane

NCERT Exercise

Part 3

Question 21: A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0i + 2.0 j) m s^-2. (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

Answer: Velocity of particle, v = 10.0 j m/s

Acceleration, a = (8.0i + 2.0 j) m s^-2

We know, `a=(dv)/(dt)` = 8.0i + 2.0 j

Or, dv = (8.0i + 2.0 j)dt

Integrating on both sides, we get:

v(t) = 8.0i + 2.0 j + u

Integrating the equations with following conditions: at t = 0, r = 0 and at t = t, r = r

`r=ut+1/2xx8.0t^2i+1/2xx2.0t^2j`

= ut + 4.0 t² i + t² j

= (10.0 j)t + 4.0 t² i + t² j

Or, Xi + yj = 4,0 t² i + (10t + t²) j

It is observed that motion of the particle is in x-y plane, so on equating the coefficients of i and j we get:

X = 4t²

Or, `t=(x/4)^(1/2)`

And y = 10t + t²

When y = 16 m

Then, `t=((16)/4)^(1/2)=2` s

So, y = 10 × 2 + 2² = 24 m

Velocity of the particle can be calculated as follows:

v(t) = 8.0t i + 2.0t j + u

At t = 2 second

v(2) = 8.0 × 2 i + 2.0 × 2 j + 10 j

= 16 i + 14 j

So, `v=saqrt(16^2+14^2)`

`=sqrt(256+196)=21.26` m/s

Question 22: i and j are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors i + j and i - j? What are the components of a vector A = 2 i + 3 j along the directions of i + j and i - j? (You may use graphical method)

Answer: Let us take a vector a = i + j

a_x i + a_y j = i + j

So, a_x = a_y = 1

So, |a|`=sqrt(a_x^2+a_y^2) = sqrt2`

So, magnitude of i + j is `sqrt2`

Let us assume that vector a makes an angle θ with x-axis.

So, tan θ `=(a_x)/(a_y)=1`

So, θ = 45°

Let us assume another vector b = i - j

Or, b_xi - b_yj = i - j

Since b_x = b_y = 1

So, |b| `=sqrt2`

So, magnitude of i - j is `sqrt2`

If vector b makes angle θ with x-axis

Then, tan θ `=(b_y)/(b_x)=-1`

So, θ = -45°

Now, component of A in the direction of i - j can be calculated as follows:

[(2i + 3j)(i - j)] ÷ `sqrt2`

`=-1/(sqrt2)` unit

Question 23: For any arbitrary motion in space, which of the following relations are true:

v_average = (1/2)[v (t₁) + v (t₂)]
v_average = [r(t₂) = r(t₁)]/(t₂ - t₁)
v(t) = v(0) + at
r(t) = r(0) + v(0)t + (1/2)at²
a_average = [v(t₂) – v(t₁)]/(t₂ - t₁)

The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)

Answer: (b) and (e) are true while a, c and d hold true only for uniform acceleration

Question 24: Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that

(a) is conserved in a process

Answer: False, because energy is not conserved in case of inelastic collision

(b) can never take negative values

Answer: False, because potential energy may have negative value in gravitational field.

Answer: False, because mass has dimension

(d) does not vary from one point to another in space

Answer: False, because speed varies from point to point in space

(e) has the same value for observations with different orientations of axes.

Answer: True, because scalar has no direction of its own

Question 25: An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

Answer: This figure shows two positions of aircraft at P and Q and we get ∠POQ = 30° and height OR = 3400 m

In Δ PRQ

tan 15° `=(PR)/(OR)=(PR)/(3400)`

Or, PR = 3400 × 0.268 = 911.2

PR is half the distance and hence is covered in 5 sec

So, speed `=(911.2)/5=182.24` m/s

Additional Exercise

Question 26: A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

Answer: A vector may not have a location in space, but a position vector does have a location in space. Velocity and acceleration are examples of vectors which show variation in position with time. Two equal vectors may not have identical physical effects. For example; if equal and opposite forces are acting on an object they cannot have equal physical effect on the object.

Question 27: A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Answer: Finite rotation does not obey the law of vector addition. So in spite of having both magnitude and direction, finite rotation is not a vector.

Question 28: Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.

Answer: We cannot associate vector with the length of a wire in a loop because direction is not definite in a loop. Vector can be associated with a plane area and direction of vector is normal to the plane. In case of sphere, we can associate a vector with the area but not with the volume of the sphere.

Question 29: A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Answer: Given, R = 3 km, θ = 30°

We need to find velocity by using following formula

`R=(u_0^2xx\si\n2θ)/g`

Or, `3=(u_0^2xx\si\n60°)/g`

Or, `(u_0^2)/g=3/(si/n60°)=(3xx2)/(sqrt3)=2sqrt3`

We know that maximum range is possible when angle of projection is 45°

`R_m=(u_0^2xx\si\n2xx45°)/g`

`=(u_0^2xx\si\n90°)/g=(u_0^2)/g=2sqrt3`

`=2xx1.732=3.464` km

This is less than 5 km so, he cannot hit the target.

Question 30: A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly over an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 m s^-2)

Answer: Given, height = 1.5 km = 1500 m, sped of aircraft = 720 km/h `=720xx5/(18)=200` m/s, speed of bullet = 600 m/s

In the given triangle sin θ `(200)/(600)=1/3`

So, θ = 19.47°

Now, height h can be calculated as follows:

`v^2-u^2=2as`

Or, `-(600\co\s\θ)^2=-2xx10xx\h`

Or, `h=(600xx600(1-si\n^2θ))/(20)`

`=30xx600(1-1/9)`

`=8/9xx30xx600=16000` m

= 16 km

Question 31: A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Answer: Given v = 27 km/h `=27xx5/(18)=7.5` m/s, r = 80 m and tangential acceleration a_T = -0.50 m/s²?

Centripetal acceleration can be calculated as follows:

`a_c=(v^2)/r`

`=(7.5^2)/(80)=0.70` m s^-2

As shown in given figure, two accelerations are acting in mutually perpendicular directions

So, resultant acceleration can be calculated as follows:

`a=sqrt(a_T^2+a_c^2)`

`=sqrt(0.5^2+0.7^2)=0.86` ms^-2

Direction of net acceleration can be found as follows:

tan θ `= (a_c)/(a_T)`

`=(0.7)/(0.5)=1.4`

So, θ = 54.56 °

Question 32: (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

`θ=ta\n^(-1)(v_(0y)-g\t)/(v_(0x))`

Answer: Let us assume that v_0x and v_0y are initial components of velocity of projectile along x- and y-axes. If v_x and v_y are components of velocity at point P

Time taken by projectile to reach P = t

Using the first equation of motion along the two axes, we get:

`v_y=v_(0y)=g\t`

And `v_x=v_(0x)`

tan θ `=(v_y)/(v_x)=(v_(0y)-g\t)/(v_(0x))`

Or, `θ=ta\n^(-1)(v_(0y)-g\t)/(v_(0x))`

(b) Show that the projection angle θ₀ for a projectile launched from the origin is given by

`θ(t)=ta\n^(-1)((4h_m)/R)`

Where the symbols have their usual meaning.

Answer: We know that maximum height is given by following equation:

`h_m=(u_0^2si\n^2θ)/(2g)`

Range is given by following equation:

`R=(u_0^2si\n^2\2θ)/g`

So, `(h_m)/R=(si\n^2θ)/(2si\n^2θ)`

`=(si\n\θ\xx\si\n\θ)/(2xx\2xx\si\n\θ\co\s\θ)`

`=(si\n\θ)/(4co\s\θ)=(ta\n\θ)/4`

Or, `ta\n\θ=(4h_m)/R`

Or, `θ=ta\n^(-1)((4h_m)/R)`