Question

A ball of mass M1 and another ball of mass M2 are dropped from equal height if time taken by the balls at T1 and t2 respectively then what is the relation between T1 and t2

Answer 1

Answer:

The relation between T₁ and T₂ is T₂ =T₁

Explanation:

Given that,

A ball of mass = M₁

Another ball of mass = M₂

Two balls are dropped from equal height. if time taken by the balls at T₁ and T₂ respectively.

We need to find the relation between T₁ and T₂

For first ball,

Using equation of motion,

$s= ut+\dfrac{1}{2}gt^2$

Where, s = height

u = initial velocity

g = acceleration due to gravity

t = time

Put the value in equation

$h =0+\dfrac{1}{2}g\timesT_{1}^2$

$h = \dfrac{1}{2}gT_{1}^{2}$.....(I)

For another ball,

Using equation of motion again

$s= ut+\dfrac{1}{2}gt^2$

Put the value in equation

$h =0+\dfrac{1}{2}g\times T_{2}^2$....(II)

From equation (I) and (II)

$\dfrac{1}{2}g\times T_{2}^2=\dfrac{1}{2}g\times T_{1}^{2$

$T_{2}=T_{1}$

Hence, The relation between T₁ and T₂ is T₂ =T₁ .

Answer 2

Question-

A ball of mass M1 and another ball of mass M2 are dropped from equal height if time taken by the balls at T1 and t2 respectively then what is the relation between T1 and t2.

Answer-

One thing we have to keep in mind is this-

"Time is independent of mass".

What this means is that whatever the mass may be, it DOES NOT matter.

If you are asking how I am able to say this, check this equation-

• s = ut + ½at2

In the following equation, initial velocity, time and acceleration is needed to find displacement. There is no mention of mass anywhere.

Since the acceleration will be equal as both are dropped from the same height which also implies displacement is equal, the time will be equal. Initial velocity is 0 for both objects.

T₁ = T₂

NOTE: This type of "trick " is best used in competition exams where you don't need to show the complete steps. But in the written exam, please derive it from the equation like the answer above me has done.