Kinematics

A particle is projected at angle 60º with speed 10 √3, from the point ‘ A ‘ as shown in the fig….

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Q: A particle is projected at angle 60º with speed 10 √3, from the point ‘ A ‘ as shown in the fig. At the same time the wedge is made to move with speed 10 √3 towards right as shown in the figure. Then the time after which particle will strike with wedge is
(g = 10 m/sec2) :

Numerical

(a) 2 sec

(b) 2 √3 sec

(c) 4/ √3 sec

(d) none of these

Click to See Solution :
Ans: (a)

Sol: Suppose particle strikes wedge at height ‘ S ‘ after time t.

$ S = 15 t – \frac{1}{2}.10 . t^2 $

$ S = 15 t – 5 t^2 $

During this time , distance travelled by particle in horizontal direction = 5 √3 t.

Also wedge has travelled extra distance

Numerical

$x = \frac{S}{tan 30^o}$

$x = \frac{15 t – 5 t^2}{\frac{1}{\sqrt{3}}}$

Total distance travelled by wedge in time t = 10 √3 t

= 5 √3 t + √3 (15 t – 5 t2)

t = 2 sec

 

A particle moves along the parabolic path y = ax2 in such a way that the y-component of the velocity remains constant ….

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Q: A particle moves along the parabolic path y = ax2 in such a way that the y-component of the velocity remains constant, say c. The x and y coordinates are in meters. Then acceleration of the particle at x = 1 m is

(a) $\displaystyle ac \hat{k}$

(b) $\displaystyle 2 a c^2 \hat{j}$

(c) $\displaystyle – \frac{c^2}{4 a^2} \hat{i}$

(d) $\displaystyle – \frac{c}{2 a} \hat{i}$

Click to See Solution :
Ans: (c)
Sol: y = a x2

$\displaystyle \frac{dy}{dt} = c $ (given )

$\displaystyle \frac{d^2y}{dt^2} = 0 $

$\displaystyle \frac{dy}{dt} = 2 a x\frac{dx}{dt} $

$\displaystyle 2 a (\frac{dx}{dt})^2 + 2ax \frac{d^2 x}{dt^2} = 0 $

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{1}{x} (\frac{dx}{dt})^2 $

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{1}{x} (\frac{c}{2ax})^2 $

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{c^2}{4 a^2 x^3} $

$\displaystyle \frac{d^2 x}{dt^2} = – \frac{c^2}{4 a^2 } $

 

Starting at time t = 0 from the origin with speed 1 m/s a particle follows a two-dimensional trajectory in the x-y plane ….

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Q: Starting at time t = 0 from the origin with speed 1 m/s a particle follows a two-dimensional trajectory in the x-y plane so that its co-ordinates are related by the equation $y = \frac{x^2}{2}$ .The x and y components of its acceleration are denoted by ax and ay respectively . Then

(a) ax = 1 m/s^2 implies that when the particle is at the origin , ay = 1 m/s^2

(b) ax = 0 implies ay = 1 m/s^2 at all times

(c) at t = 0 the particle’s velocity points in the x-direction

(d) ax = 0 implies that at t= 1 s , the angle between the particle’s velocity and the x axis is 45°

Click to See Solution :
Ans: (a,b,c,d)

Sol: The given equation is $y = \frac{x^2}{2}$

$\frac{dy}{dt} = x \frac{dx}{dt}$

vy = x vx

Differentiating w.r.t time

$\frac{dv_y}{dt} = x \frac{dv_x}{dt} + \frac{dx}{dt} v_x $

ay = x ax + vx2

 

The distance x covered by a particle in one dimensional motion varies with time t as x^2 = at^2 + 2 bt + c ….

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Q: The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2 bt + c . If the acceleration of particle depends on x as x-n , where n is an integer , the value of n is …….

Sol: x2 = at2 + 2 bt + c

$ \large x = (at^2 + 2 bt + c)^{1/2} $

Differentiating with respect to time t ,

$ \large \frac{dx}{dt} = \frac{1}{2}(at^2 + 2 bt + c)^{-1/2}(2at + 2b) $

Acceleration , $ \displaystyle A = \frac{d^2x}{dt^2} = \frac{1}{2}[ (at^2 + 2 bt + c)^{-1/2}(2a) + (2at + 2b)(-\frac{1}{2})(at^2 + 2bt + c)^{-3/2}(2at + 2b)] $

$\large Ax = a – (\frac{at+b}{x})^2 $

$\large Ax = \frac{ax^2 -(at+b)^2 }{x^2} $

$\large Ax = \frac{a(at^2 + 2bt + c)-(at+b)^2}{x^2}$

$\large A = \frac{ac-b^2}{x^3} $

$\large A \propto \frac{1}{x^3} $

$\large A \propto x^{-3} $

Hence , n = 3