Question

Two objects having mass in 1 : 4 ratio are dropped
from the same height. The relation between their
velocity when they strike the ground is
(1) Both objects will have the same velocity
(2) The velocity of the first object is twice that of
the second one
(3) The velocity of the 2nd object is one fourth of
that of the 1st object
(4) The velocity of the 2nd object is 4 times that of
the 1st one​

Answer 1

Answer:

Same velocity     (option 1)

Explanation:  Using the equation of motion:  v = u + at

           For 1st body, a = g, u = 0

⇒ v₁ = 0 + gt₁    ⇒ v₁ = gt₁

           For 2nd body, a = g, u = 0

⇒ v₂ = 0 + gt₂    ⇒  v₂ = gt₂

Using, S = ut + 1/2 at²,

⇒ S₁ = 0(t) + g(t₁)²/2   ⇒ S₁ =  g(t₁)²/2

⇒ S₂ = 0(t) + g(t₂)²/2   ⇒ S₂ =  g(t₂)²/2

  But since they are dropped form same height, S₁ = S₂:

⇒ g(t₁)²/2 = g(t₂)²/2

⇒ t₁ = t₂    

Hence,  v₁ = gt₁  &  v₂ = gt₂

 v₁ = gt₁ = gt₂ = v₂      [as t₁ = t₂]

Answer 2

Answer:

Given :-

Ratio of two objects = 1:4

To Find :-

Relation between veloci

Solution :-

At first the final Velocity

$\bf \: v = u + at(1)$

$\bf \: v = u + at(2)$

Now,

The distance covered

$\sf \: s \: = ut + \dfrac{1}{2} {at}^{2} (1)$

$\sf \: s \: = ut + \dfrac{1}{2} {at}^{2} (2)$

On comparing

$\sf ut \dfrac{a(t_1)}2 = ut \dfrac{a(t_2)}{2}$

$\sf \dfrac{a(t_1)}{2} = \dfrac{a(t_2)}{2}$

Cancelling a and 2

$\sf \: t _1 = t_2$