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# Projectile Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Projectile Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Introductory Exercise 4.2

Ques 1: A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point A? (g = 10 m/s2)

Ans:
Sol: Time of flight

Using,  v = u + at
vx = ux = u cos 60Â°

Ques 2: In the above problem what is the component of its velocity perpendicular to the plane when it strikes at A?
Ans: 5 m/s
Sol: Component of velocity perpendicular to plane

Ques 3: Two particles A and B are projected simultaneously from the two towers of height 10 m and 20 m respectively. Particle A is projected with an initial speed of 10âˆš2 m/s at an angle of 45Â° with horizontal, while particle B is projected horizontally with speed 10 m/s. If they collide in air, what is the distance d between the towers?

Ans: 20 m
Sol: Let the particle collide at time t.

x1 = (u cos Î¸) t
and x2 = vt
âˆ´ d = x2 - x1
= (v + u cos Î¸) t

For vertical motion of particle 1:

i.e.,   â€¦(i)
or
For the vertical motion of particle 2:

i.e.,   â€¦(ii)
Comparing Eqs. (i) and (ii),

â‡’ t = 1 s
âˆ´ d = 20 m

Ques 4: Two particles A and B are projected from ground towards each other with speeds 10 m/s and 5âˆš2 m/s at angles 30Â° and 45Â° with horizontal from two points separated by a distance of 15 m. Will they collide or not?

Ans: No
Sol: u = 10 m/s
v = 5âˆš2 m/s

Î¸ = 30Â°
Ï† = 45Â°
d = 15 m
Let the particles meet (or are in the same vertical time t).
âˆ´ d = (u cos Î¸) t + (v cos Ï†) t
â‡’ 15 = (10 cos 30Â° + 5âˆš2 cos 45Â°) t
or
or
= 1.009 s
Now, let us find time of flight of A and B

= 1 s
As TA < t, particle A will touch ground before the expected time t of collision.
âˆ´ Ans: NO.

Ques 5: A particle is projected from the bottom of an inclined plane of inclination 30Â°. At what angle Î± (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.
Ans: 60Â°
Sol: For range to be maximum

Ques 6: A particle is projected from the bottom of an inclined plane of inclination 30Â° with velocity of 40 m/s at an angle of 60Â° with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take g = 10 m/s2.
Ans:
Sol: At point A velocity  of the particle will be parallel to the inclined plane.

âˆ´ Ï† = Î²
vx = ux = u cos Î±
vx = v cos Ï† = v cos Î²
or u cos Î± = v cos Î²
â‡’

Ques 7: Two particles A and B are projected simultaneously in the directions shown in figure with velocities vA = 20 m/s and vB = 10 m/s respectively. They collide in air after 1/2 s.
Find:
(a) the angle Î¸
(b) the distance x.

Ans: (a) 30Â°

Sol: (a) At time t, vertical displacement of A
= Vertical displacement of B

i.e.,  vA sin Î¸ = vB

âˆ´ Î¸ = 30Â°
(b) x = (vA cos Î¸) t

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