Case 1

Rocket Propulsion

Thrust force on the rocket in upwards direction 

Weight of rocket = mg down ward direction







If gravity (g) neglected and

u = 0 then

Case 2

Density of liquid = P   Orifice area = A







A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward with velocity 100 ms-1 from the ground. The bullet gets embedded in the wood. Then, the maximum height to which the combined system reaches above the top of the building before falling below is (take, g = 10 ms-2)                                                                                                  (2019 JEE Main, 10 Jan I)

(a) 20 m                                        (b) 30 m

(c) 10 m                                        (d) 40 m

Solution: Key Idea As bullet gets embedded in the block of wood so, it represents a collision which is perfectly inelastic and hence only momentum of the system is conserved.

Velocity of bullet is very high compared to velocity of wooden block so, in order to calculate time for collision, we take relative velocity nearly equal to velocity of bullet.

So, time taken for particles to collide is


Speed of block just before collision is

v1 = gt = 10 x 1 = 10 ms-1

Speed of bullet just before collisions is

v2 = u – gt

= 100-10 x 1 = 90 ms-1

Let v = velocity of bullet + block system, then by conservation of linear momentum, we get

 - (0.03 x 10) + (0.02 x 90) = (0.05)v

Þ      v = 30 ms-1

Now, maximum height reached by bullet and block is


Þ   h = 45 m

   Height covered by the system from point of collision = 45 m

Now, distance covered by bullet before collision in 1 sec.


Distance of point of collision from the top of the building = 100 – 95 = 5m

   Maximum height to which the combined system reaches above the top of the building before falling below = 45 – 5 = 40 m          

  Answer (d)

Question. n elastic balls are placed at rest on a smooth horizontal plane which is circular at the ends with radius r as shown in the figure. The masses of the ball are  respectively. What is the minimum velocity which should be imparted to the first ball of mass m such that nth ball completes the vertical circle



Solution: Velocity of second ball


Day – 5 and 6 Question Practice Online

Velocity of third ball will become,



                                            Answer (a)