1) Two bodies of mass in the ratio of 1:2 fall from the heights in the ratio of 3:4.What will be the ratio of their linear momentum on reaching the ground?

Sol v²=u²+2as here in both case u will be zero

momentum(p)=velocity × mass

here we need v

v=√2as (a=g=10m/s²)

in first case v = √2×10×4 =4√5

p = 4√5 × 3

= 12√5

in case 2


= 6√5

p= 6√5 × 2



1:1 ANS


Two point masses M and 3M are placed at a

distance L apart. Another point mass m is placed in

between on the line joining them so that the net

gravitational force acting on it due to masses M and

3M is zero. The magnitude of gravitational force

acting due to mass M on mass m will be

(a) \frac{GMm {(1 + \sqrt{3} )}^{2} }{ {l}^{2} } \\ \\ (b) \frac{3gmm}{ {l}^{2} (1 + \sqrt{ 3} )} \\ \\ (c) \frac{gmm {(1 - \sqrt{3} )}^{2} }{ {l}^{2} } \\ \\ (d) \frac{gmm(1 - \sqrt{3} )}{ {l}^{2} }

So 2)Net force on any mass at the position between two masses is zero

Let say the distance between M and m is “x” so the net force due to M and 3M on mass m is zero

so we have

now by solving above equation

now by the formula of force between two mass

plug in the value of “x” in


2 persons just manage to push a block from left to right direction along a horizontal level road with uniform velocity when the same block is pushed by 3 person in same direction a constant acceleration of 0.2 metre per second square is produced in the block if the 5persons push the block in same direction together then the magnitude of the acceleration of the Block will be

3)When 2 person pulls the box it will move to the right with constant speed

So applied force = friction force

When 3 person apply force on it

Now when 5 person apply the same force then we have

from above equation F = 0.2m

so acceleration will be 0.6 m/s^


If the displacement of a particle is equal to zero then its distance 4) Then it’s distance in not equal to zero
hope it helps you


Velocity time graph of a block of mass 100 gram sliding on horizontal concrete floor under the action of a constant force of 5 N shown below the magnitude of frictional Force acting in the block due to the floor is

Share this post

Leave a Reply

Your email address will not be published. Required fields are marked *