# NCERT Physics Class 11 Exemplar Ch 4 Motion in A Plane Part 4

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29 A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take.

Ans:

The minimm vertical velocity required for crossing the hill is given by

As canon can haul packets with a speed of 125m/s, so the maximum value of horizontal velocity, will be

The time taken to reach the top of the hill with velocity is given by 1 10s.

In 10s the horizontal distance covered.

So cannon has to be moved through a distance of 50 m on the ground.

So total time taken (shortest) by the packet to reach ground across the hill

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

If a target farther away by distance has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least

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

(i) Refer to the diagram: target T is at horizontal distance and below point of projection .

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

Ans:

31 A particle is projected in air at an angle to a surface which itself is inclined at an angle α to the horizontal (Fig. 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) 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 (parobola) and straight line. Remember particle is projected at an angle 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, along the plane and 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.)

Ans:

32 A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle with speed and rebounds elastically. Find the distance along the plane where if will hit second time.

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

Ans:

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 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 bedirection and vertically downward to be. Let the rain velocity be. The velocity of rain as observed by the girl is always. 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.)

Ans:

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.

(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?

Ans:

(a) 5m/s at to N.

(b) (i) of N, (ii)

(c) in case (i) he reaches the opposite bank in shortest time.