
Two blocks A and B of mass 2M and M are moving along X-direction on a frictionless plane. Their instantaneous positions are given by t and ${t^2}$, respectively. At a particular instant of time T, the kinetic energy of both particles are equal. Then,
A. at T, ${V_A} > {V_B}$
B. at T, ${V_B} > {V_A}$
C. at T, ${V_A} = {V_B}$
D. ${V_A} = {V_B}$ always
Answer
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Hint: We know the expression of kinetic energy in terms of velocity. By equating the kinetic energies of the two blocks, we can find that which block has velocity greater than the other at time T.
Formula used:
The kinetic energy of a particle is given as
$K = \dfrac{1}{2}m{v^2}{\text{ }}...\left( i \right)$
Here K represents the kinetic energy of the particle, m is used to represent the mass of the particle and the particle is moving with velocity v.
Detailed step by step solution:
The kinetic energy of a block is the energy possessed by the block due to its state of motion. The expression for kinetic energy is given in equation (i).
We are given two blocks A and B. The mass of the block A is given as
${M_A} = 2M$
And the mass of block B is given as
${M_B} = M$
It is said that the kinetic energy of the two blocks is equal when time is equal to T. Therefore, we can write the following expression
$\dfrac{1}{2}{M_A}V_A^2 = \dfrac{1}{2}{M_B}V_B^2$
Now we can insert the values of mass of block A and B here and get that
$\eqalign{
& \dfrac{1}{2} \times 2MV_A^2 = \dfrac{1}{2}MV_B^2 \cr
& \Rightarrow 2V_A^2 = V_B^2 \cr
& \Rightarrow \sqrt 2 {V_A} = {V_B} \cr
& \Rightarrow {V_B} > {V_A} \cr} $
Now we need to check the relevance of this equation with time. We are given that positions of the blocks are time dependent, that is,
$\eqalign{
& {x_A} = t \cr
& {x_B} = {t^2} \cr} $
The expressions for velocities can be calculated as
$\eqalign{
& {V_A} = \dfrac{{d{x_A}}}{{dt}} = \dfrac{d}{{dt}}t = 1 \cr
& {V_B} = \dfrac{{d{x_B}}}{{dt}} = \dfrac{d}{{dt}}{t^2} = 2t \cr} $
Therefore the relation ${V_B} > {V_A}$ holds true at time T also. And the correct answer is option B.
Note: It should be noted that velocity of block A is constant while the velocity of block B is variable and changes linearly with time. Therefore for T > 0.5, the velocity of B will always be greater than velocity of A.
Formula used:
The kinetic energy of a particle is given as
$K = \dfrac{1}{2}m{v^2}{\text{ }}...\left( i \right)$
Here K represents the kinetic energy of the particle, m is used to represent the mass of the particle and the particle is moving with velocity v.
Detailed step by step solution:
The kinetic energy of a block is the energy possessed by the block due to its state of motion. The expression for kinetic energy is given in equation (i).
We are given two blocks A and B. The mass of the block A is given as
${M_A} = 2M$
And the mass of block B is given as
${M_B} = M$
It is said that the kinetic energy of the two blocks is equal when time is equal to T. Therefore, we can write the following expression
$\dfrac{1}{2}{M_A}V_A^2 = \dfrac{1}{2}{M_B}V_B^2$
Now we can insert the values of mass of block A and B here and get that
$\eqalign{
& \dfrac{1}{2} \times 2MV_A^2 = \dfrac{1}{2}MV_B^2 \cr
& \Rightarrow 2V_A^2 = V_B^2 \cr
& \Rightarrow \sqrt 2 {V_A} = {V_B} \cr
& \Rightarrow {V_B} > {V_A} \cr} $
Now we need to check the relevance of this equation with time. We are given that positions of the blocks are time dependent, that is,
$\eqalign{
& {x_A} = t \cr
& {x_B} = {t^2} \cr} $
The expressions for velocities can be calculated as
$\eqalign{
& {V_A} = \dfrac{{d{x_A}}}{{dt}} = \dfrac{d}{{dt}}t = 1 \cr
& {V_B} = \dfrac{{d{x_B}}}{{dt}} = \dfrac{d}{{dt}}{t^2} = 2t \cr} $
Therefore the relation ${V_B} > {V_A}$ holds true at time T also. And the correct answer is option B.
Note: It should be noted that velocity of block A is constant while the velocity of block B is variable and changes linearly with time. Therefore for T > 0.5, the velocity of B will always be greater than velocity of A.
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