DC Pandey Solutions Projectile Motion - 2 - Physics for JEE Main & Advanced

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
v_x = u_x = u cos 60°

Q2: 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 10√2 m/s at an angle of 45° 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 θ) t
and x₂ = vt
∴ d = x₂ - x₁
= (v + u cos θ) t


For vertical motion of particle 1:

i.e.,  …(i)
or
For the vertical motion of particle 2:

i.e.,  …(ii)
Comparing Eqs. (i) and (ii),

⇒ t = 1 s
∴ d = 20 m

Ques 4: Two particles A and B are projected from ground towards each other with speeds 10 m/s and 5√2 m/s at angles 30° and 45° 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 = 5√2 m/s

θ = 30°
φ = 45°
d = 15 m
Let the particles meet (or are in the same vertical time t).
∴ d = (u cos θ) t + (v cos φ) t
⇒ 15 = (10 cos 30° + 5√2 cos 45°) t
or
or
= 1.009 s
Now, let us find time of flight of A and B

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

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

Sol: For range to be maximum

Ques 6: A particle is projected from the bottom of an inclined plane of inclination 30° with velocity of 40 m/s at an angle of 60° 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.


∴ φ = β
v_x = u_x = u cos α
v_x = v cos φ = v cos β
or u cos α = v cos β
⇒

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

Ans: (a) 30°

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


i.e., v_A sin θ = v_B


∴ θ = 30°
(b) x = (v_A cos θ) t