Question

A particle moving with kinetic energy $3\,J$ makes a head on collision with a stationary particle which has twice its mass. During the impact which of the following is true.
A. The minimum kinetic energy of the system is $1\,J$ .
B. The maximum elastic potential energy of the system is $2\,J$
C. Momentum and total energy are conserved at every instant.
D. The ratio of kinetic energy to potential energy of the system first decreases and then increases.

Answer 1

Hint-Energy and momentum is always conserved in an elastic collision. In the case of a head on collision the kinetic energy will be minimum when both the particles move with common velocity. Since kinetic energy is minimum at this point the potential energy will be maximum there. By applying the conservation of momentum and energy we can find the corresponding values of minimum kinetic energy and potential energy.

Step by step solution:
Let the mass of the first particle be m. It is given that it collides with a particle of twice its mass. Therefore, the mass of the second particle will be $2\,m$. Let the initial velocity of the first particle be u. The second particle is stationary so its initial velocity is zero.
Given that it is a head on elastic collision. In an elastic collision both the kinetic energy and momentum is conserved.

In the case of a head on collision the kinetic energy will be minimum at the time of collision when both the particles move with common velocity for a very short time period. Since kinetic energy is minimum at this point the potential energy will be maximum there. Let us assume that this common is v.
It is given that initial kinetic energy of mass m is $3\,J$.
That is,
$\dfrac{1}{2}m{u^2} = 3\,J$
Now let us use the conservation of momentum. The total momentum before collision will be equal to the total momentum at the time of collision.
Initial momentum of first particle m is $mu$
Initial momentum of the second particle is zero.
At the common velocity v the momentum of first particle is $mv$
The momentum of second particle is $2\,mv$
Only equating both we get,
$mu + 0 = mv + 2mv$
On solving we get,
$v = \dfrac{u}{3}$
Now let us find the value of the minimum kinetic energy at velocity v .
$K{E_{\min }} = \dfrac{1}{2}M{v^2}$
Here M is a total mass since both the mass moves together at this velocity.
$\therefore M = m + 2m = 3m$
We already have
$v = \dfrac{u}{3}$
On substituting these values we get
$K{E_{\min }} = \dfrac{1}{2}3m{\left( {\dfrac{u}{3}} \right)^2}$
$ \Rightarrow K{E_{\min }} = \dfrac{1}{2}m{u^2} \times \dfrac{1}{3}$
We know that $\dfrac{1}{2}m{u^2} = 3\,J$
Therefore
$ \Rightarrow K{E_{\min }} = 3 \times \dfrac{1}{3} = 1\,J$

So option A is true.

Here total mechanical energy is conserved. Initially the total energy was the kinetic energy $3\,J$. Let the initial potential energy be P. At the common velocity the kinetic energy is $1\,J$. So, the change in kinetic energy is $3J - 1J = 2J$. This will be the change in potential energy. Since kinetic energy is minimum this value will be the maximum value of potential energy.

So, option B is true.

Since energy and momentum is always conserved in an elastic collision option C is also true.
The ratio of kinetic energy to potential energy initially is $\dfrac{3}{P}$ .
During collision at velocity v this ratio will become $\dfrac{1}{{P + 2}}$
The value of the first ratio is greater than the second.
Again after collision finally the kinetic energy will become $3\,J$ again and potential energy will be $P$
So the ratio will be $\dfrac{3}{P}$ again.
Hence, we can see that first the ratio is decreasing and then it is increasing. Thus option D is true.

Therefore, all the given options are correct.

Note: Remember that the conservation of energy is applicable only if the collision is elastic. In the case of inelastic collision only momentum is conserved and kinetic energy is not conserved.
Also remember that at the time of collision both the particles move together momentarily so both will have the same velocity. When we calculate kinetic energy at this time of collision the mass taken should be the sum of mass of both the particles since they move together.