Motion in Plane
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
= 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
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
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`
Question 23: For any arbitrary motion in space, which of the following relations are true:
- vaverage = (1/2)[v (t1) + v (t2)]
- vaverage = [r(t2) = r(t1)]/(t2 - t1)
- v(t) = v(0) + at
- r(t) = r(0) + v(0)t + (1/2)at2
- 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
We know that maximum range is possible when angle of projection is 45°
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:
= 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:
`=(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:
Direction of net acceleration can be found as follows:
tan θ `= (a_c)/(a_T)`
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
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:
tan θ `=(v_y)/(v_x)=(v_(0y)-g\t)/(v_(0x))`
(b) Show that the projection angle θ0 for a projectile launched from the origin is given by
Where the symbols have their usual meaning.
Answer: We know that maximum height is given by following equation:
Range is given by following equation: