Class 11 Physics

# 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

ax i + ay j = i + j

So, ax = ay = 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, bxi - byj = i - j

Since bx = by = 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:

1. vaverage = (1/2)[v (t1) + v (t2)]
2. vaverage = [r(t2) = r(t1)]/(t2 - t1)
3. v(t) = v(0) + at
4. r(t) = r(0) + v(0)t + (1/2)at2
5. aaverage = [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.

(c) must be dimensionless

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

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 aT = -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 v0x and v0y are initial components of velocity of projectile along x- and y-axes. If vx and vy 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)