Kinematics

Q: A body is thrown upward from ground covers equal distances in 4^th and 7^th sec . With what initial velocity the body was projected ?

(a) 20 m/s

(b) 50 m/s

(c) 30 m/s

(d) 10 m/s

Ans: (b)

Sol: D₄ = – D₇

$\displaystyle D_4 = u – \frac{g}{2}(2 \times 4 – 1) $

$\displaystyle D_4 = u – 35 $ …(i)

$\displaystyle D_7 = u – \frac{g}{2}(2 \times 7 – 1) $

$\displaystyle D_7 = u – 65 $ …(ii)

Since , D₄ = – D₇

$\displaystyle u – 35 = – (u – 65)$

2 u = 100

u = 50 m/s

Problem :   A bullet is moving horizontally with certain velocity. It pierces two paper discs rotating co-axially with angular speed w separated by a distance l . If the hole made by the bullet on 2^nd disc is shifted through an angle θ with respect to that in the first disc, find the velocity of the bullet. (change of velocity in the bullet is neglected.)

Solution:

Let the bullet take a time t to travel from one disc to the other with a velocity v.

⇒  The distance between the discs  l = v.t

⇒ $ \displaystyle t = \frac{l}{v} $  ……(a)

When the bullet is about to hit the second disc at Q, the hole made by the bullet on the first disc at P will rotate through an angle θ during the time t = l/v

⇒ θ = ω t

$ \displaystyle t = \frac{\theta}{\omega} $ …..(b)

Using (a) & (b)

$ \displaystyle t = \frac{l}{v} = \frac{\theta}{\omega} $

⇒ $ \displaystyle v = \frac{\omega l}{\theta} $

Problem :  Find the radius of a rotating wheel if the linear velocity v₁ of a point on the rim is 2.5 times greater than the linear velocity v₂ of a point 5 cm closer to wheel axle.

Solution :

Let the radius of the disc = r (in cm.)

$ \displaystyle \frac{v_2}{v_1} = \frac{(r-5)\omega}{r \omega} $

v₁ is 2.5 times greater than v₂.

$ \displaystyle \frac{v_2}{2.5 v_2} = \frac{(r-5)}{r} $

⇒ r = 2.5 r – 12.5

⇒ 1.5 r = 12.5

⇒ r = 12.5/1.5 = 8.33 cm.

Problem :    A particle when fired at an angle θ = 60° along the direction of the breadth of a rectangular building of dimension 9m × 8m × 4m so as to sweep the edges.  Find the range of the projectile.

Solution :

Since the projectile touches A & B, both of these points lie on the path of the projectile. After putting the coordinate of A in the trajectory equation of projectile we obtain ,

$ \displaystyle y = h = xtan\theta – \frac{g x^2}{2v_0^2 cos^2 \theta} $   ……(i)

As we know that

$ \displaystyle R = \frac{v_0^2 sin2\theta}{g} $

$ \displaystyle 2(x+h) = \frac{v_0^2 sin2\theta}{g} $  ……(ii)

Using (i) & (ii)

$ \displaystyle h = (\frac{R}{2}-h)tan\theta -\frac{g (\frac{R}{2}-h)^2 }{\frac{2gR}{sin2\theta} cos^2\theta} $

⇒ R² – 4Rh cot θ – 4h² = 0

$ \displaystyle R = \frac{4hcot\theta \pm \sqrt{16h^2 cot^2\theta + 16h^2}}{2} $

⇒ R = 2h (cot θ + cosec θ)

= 2h(1+ cosθ)/sinθ

= 2h cot (θ/2)

putting θ = 60⁰ & h = 4m

R = 2 × 4 cot 30⁰ = 8 √3  m

Problem  :   A particle is projected up from the bottom of an inclined plane of inclination a with velocity v₀.  If it returns to the point of projection after an elastic impact with the plane, find the total time of motion of the particle.

Solution: To return to the point of projection after one elastic collision the particle must meet the plane at right angle of the particle along x axis.

v_x – v_xo  = (-g sin α)t₀

⇒  0 – v₀ cos θ = (-g sin α)t₀

$ \displaystyle t_0 = \frac{v_0 cos\theta}{gsin\alpha} $ ……(i)

Motion of the particle along y axis

$ \displaystyle \vec{v_y} – \vec{v_{y0}} = ( g cos\alpha) t_0 $

$ \displaystyle v_0 sin\theta + v_0sin\theta = ( g cos\alpha) t_0 $

$ \displaystyle t_0 = \frac{2v_0 sin\theta}{gcos\alpha} $ …(ii)

Equating (i) & (ii)

$ \displaystyle \frac{v_0 cos\theta}{gsin\alpha} = \frac{2v_0 sin\theta}{gcos\alpha} $

⇒   $ \displaystyle cot\theta = 2 tan\alpha $    …(iii)

$ \displaystyle cos\theta = \frac{2tan\alpha}{\sqrt{1 + 4tan^2 \alpha}} $

$ \displaystyle sin\theta = \frac{1}{\sqrt{1 + 4tan^2 \alpha}} $

∴ Time of flight  for to & fro motion of the particle

$ \displaystyle T = 2 t_0 = \frac{2 v_0 cos\theta}{gsin\alpha} $ (from (i))

$ \displaystyle T = 2 t_0 = \frac{2 v_0 (\frac{2tan\alpha}{\sqrt{1 + 4tan^2 \alpha}})}{gsin\alpha} $

$ \displaystyle = \frac{4v_0}{g\sqrt{1 + 3 sin^2\alpha}} $

Problem  :   Shown in the figure is a ring sliding freely along an inextensible string that passes through the smooth pegs (1) & (2) at the same level. One end of the string is fixed. The other end P is pulled with a constant velocity v, find the velocity of the ring.

Solution:

 Pythagoras theorem yields

x² + y² = l²,

Differentiating both sides w.r.t. ‘t’, we obtain

$ \displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2l \frac{dl}{dt} $

Since the ring slides smoothly, it always lie at the mid position maintaining equal distance from the pegs; Therefore, x remains constant

$ \displaystyle \frac{dy}{dt} = v_{ring} $

$ \displaystyle \frac{dl}{dt} = v = $rate of pulling of the string

$ \displaystyle \frac{dy}{dt} = v_{ring} = \frac{l}{y}v$

Putting  $ \displaystyle \frac{l}{y}=sec\theta $

v_ring = v sec θ 

Problem  :    An elevator of height h is ascending with an acceleration a. After a time t a coin is dropped from the top of the  elevator. When will the coin strike the bottom of the elevator ?

Solution: Let after a time t from the instant of release, the coin strikes the floor of the elevator.
The coin was moving with the elevator with a velocity, say v. Just after losing contact it begins to ascend with same velocity v (obeying the law of inertia of motion).
Since the net displacement S1 of the coin during time t is downward (w.r.t the initial position or point of release)

$ \displaystyle S_1 = -vt + \frac{1}{2}gt^2 $ …(i)

The net displacement S2 of the ascending elevator during time t is upwards (w.r.t. the initial position)

$ \displaystyle S_2 = vt + \frac{1}{2}at^2 $ …(ii)

Now, S₁ + S₂ = h (iii)

Putting S1 and S2 from (i) & (ii) respectively in (iii) we obtain

$ \displaystyle -vt + \frac{1}{2}gt^2 + vt + \frac{1}{2}at^2 = h $

$ \displaystyle \frac{1}{2}(g+a)t^2 = h $

$ \displaystyle t = \sqrt{\frac{2h}{g+a}} $

Problem :  A particle moves along X-axis obeying the equation x = t (t – 1) (t – 2) where x is in metre &  t in second.  Find the (a) initial velocity and acceleration of the particle (b) velocity & acceleration of the particle when its displacement is zero (c) displacement & acceleration of the particle when its velocity is zero.

Solution :   

(a)  x = t (t –1) (t – 2)                     . . . (i)

      ⇒  x = t³ – 3t² + 2t

⇒ v = dx/dt  = 3t² – 6t + 2               . . . (ii)

⇒ At time t = 0 , v = 2 m/sec.  &

a = dv/dt = 6t – 6 = 6(t –1)        . . . (iii)

⇒ at time t = 0,  a = -6m/sec²  .

(b) Displacement of the particle is zero at time t given by

x = t (t – 1) (t−2) = 0

⇒    t = 0 , t = 1 & t = 2

putting the values of t in eq. (ii) & (iii) we obtain,

velocity v = +2 , -1 & +2 m/sec.  respectively.

Acceleration a = −6 , 0 & +6 m/sec² respectively.

(c) velocity v = 0

⇒ v = 3t² – 6t + 2 = 0

⇒ t = 1 + (1/√3)    & t = 1 − (1/√3)

By putting these values of t in eq. (i) we obtain,

x = (−2/3√3) & (2/3√3) m respectively & the corresponding  acceleration can be obtained

by putting the values of t in  equation (iii) given by

a = 2√3 m/sec²  & – 2√3 m/s² respectively

Q. A particle moving with a constant acceleration describes in the last second of its motion 36% of the whole distance. It is starts from rest, how long is the particle in motion and through what distance does it moves if it describes 6 cm in the first sec.?

(a) 5s; 150 cm

(b) 10s; 150 cm

(c) 15s; 100 cm

(d) 20s; 200 cm

Ans:(a)

Q. A particle moving with uniform retardation covers distances 18 m, 14 m and 10 m in successive seconds. It comes to rest after travelling a further distance of

(a) 50m

(b) 8m

(c) 12m

(d) 42m

Ans: (b)