Q32 of 37 Page 25

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed vo and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.


(Hint: (i) After rebound, particle still has speed Vo to start.


(ii) Work out angle particle speed has with horizontal after it rebounds.


(iii) Rest is similar to if particle is projected up the incline.)


Let’s consider the X-axis parallel to the inclined surface and Y-axis perpendicular to it. v be the speed of the projectile.


Range=R, x component of velocity = vsinθ


y component of velocity = vcosθ


x component of acceleration = gsinθ


y component of acceleration = gcosθ



There is no displacement in the y direction.


By 2nd kinematic equation in y-direction,





Now let’s calculate the distance along the plane the particle hit = R,


By 2nd kinematic equation in x-direction,



Substituting the value of t in the above equation.




More from this chapter

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30

A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R = .


If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at least


h = ∆x


(Hint: This problem can be approached in two different ways:


(i) Refer to the diagram: target T is at horizontal distance x = R + ∆x and below point of projection y = – h.


(ii) From point P in the diagram: Projection at speed vo at an angle θ below horizontal with height h and horizontal range ∆x.)


31

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig. 4.6).

(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).


(b) Time of flight.


(c) β at which range will be maximum.


(Hint: This problem can be solved in two different ways:


(i) Point P at which particle hits the plane can be seen as intersection of its trajectory (parabola) and straight line. Remember particle is projected at an angle (α + β) w.r.t. horizontal.


(ii) We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two different components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)


 33

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

(Hint: Assume north to be î direction and vertically downward to be − ĵ. Let the rain velocity vr be a î + b ĵ. The velocity of rain as observed by the girl is always vr – vgirl. Draw the vector diagram/s for the information given and find a and b. You may draw all vectors in the reference frame of ground based observer.)


 34

A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s (Fig. 4.8).

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?


(b) If he wants to start from point A on south bank and reach opposite point B on north bank,


(a) which direction should he swim?


(b) what will be his resultant speed?


(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?