Question

can anyone give me the answer of these 2 questions please

Answer 1

Both the questions involve a common concept, and we will see it first.

Situation: A ball of mass m moving with velocity v collides with another identical ball of same mass kept at rest. What will be the final velocities of the balls?

Our data:

Mass of both balls = m

Initial Velocity of ball 1 = $u_1$ = v

Initial Velocity of ball 2 = 0

Final Velocity of ball 1 = $v_1$ (say)

Final Velocity of ball 2 = $v_2$ (say)

From Law of Conservation of Momentum, we have:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\\implies mv+m(0)=mv_1+mv_2\\\\\implies mv=m(v_1+v_2)\\\\\implies v_1+v_2=v\quad ---(1)$

We are also assuming the collision to be elastic. This means that Kinetic Energy is conserved.

$\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\\\\\\\implies mv^2=mv_1^2+mv_2^2\\\\\\\implies v_1^2+v_2^2=v^2\\\\\\\implies v_1^2+v_2^2=(v_1+v_2)^2 \quad\text{[Using (1)]}\\\\\\\implies v_1^2+v_2^2=v_1^2+2v_1v_2+v_2^2\\\\\\\implies 2v_1v_2=0\\\\\\\implies v_1v_2=0$

$\implies v_1=0 \quad OR \quad v_2=0$

But Ball 1 hit a stationary ball (Ball 2). The stationary ball must have started moving. So $v_2 \neq 0$

So, $v_1=0$ and hence $v_2=v$

This means that, When Ball 1 moving with velocity v collided with another identical ball, Ball 2, which was at rest, then the velocities of Ball 1 and Ball 2 got exchanged.

That is, now Ball 1 comes to rest, and Ball 2 starts to move with velocity v.

So, we will use this simple concept: When two identical objects collide, their velocities get exchanged. [This is the general statement, and is valid even when both objects are moving]

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Question 1: A particle of mass 0.1 kg moving with a speed v collides with another identical particle at rest.

As we know, velocities will get exchanged. So, after collision, Particle 1 comes to rest, and the second particle starts moving.

The kinetic energy of this particle is given as 0.2 joules. So we have:

$\frac{1}{2}mv^2=0.2\\\\\\\implies\frac{1}{2}\times 0.1\times v^2=0.2\\\\\\\implies v^2=\frac{0.4}{0.1}\\\\\\\implies v^2=4\\\\\\\implies \boxed{\bold{v=2\,\, m/s}}$

In a perfectly elastic collision, no kinetic energy was lost. So, the speed v of initial particle is transferred to the second particle. Thus, we will get the minimum possible value of v.

In a perfectly inelastic collision, loss of kinetic energy is maximum. So, when speed of first particle is v, the speed of each particle after collision will be less than v. In this case, for the energy to equal 0.2, more initial velocity will be required. So, perfectly inelastic collision will give maximum value of v.

So, the answer we obtained just now is Option (1) Minimum value of v is 2 m/s.

Let us also calculate the Maximum Value of v, for which we require Perfectly Inelastic Collision.

Law of Conservation of Momentum states:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\\implies mv=mv_1+mv_2\\\\\implies v=v_1+v_2\quad ---(1)$

Also, since coefficient of restitution is zero, we have:

$\displaystyle -\left( \frac{v_2-v_1}{u_2-u_1}\right)=e\\ \\ \\ \implies -\left( \frac{v_2-v_1}{u_2-u_1}\right)=0 \\ \\ \\ \implies v_2-v_1=0 \\ \\ \\ \implies v_1=v_2=v' \quad (say)$

Putting it in (1), we have:

$v=v'+v' \\ \\ \\ \implies v' = \frac{v}{2}$

Since final kinetic energy is 0.2 joules, we have:

$\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=0.2 \\\\\\\implies mv'^2+mv'^2=0.4 \\ \\ \\ \implies 2mv'^2=0.4 \\ \\ \\ \implies v'^2=\frac{0.2}{0.1} \\ \\ \\ \implies \left( \frac{v}{2} \right)^2=2 \\ \\ \\ \implies \frac{v^2}{4}=2 \\ \\ \\ \implies v^2=8 \\ \\ \\ \implies \boxed{\bold{v=2\sqrt{2} \, \, m/s}}$

This is the maximum possible value of v.

To Summarize, we have:

$\boxed{\begin{array}{ccl}v_{min}&=&2 \, \, m/s\\ \\v_{max} &=& 2\sqrt{2} \, \, m/s\end{array}}$

Answer stands at Option (1)

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Question 2:

This is an interesting question, which can be solved logically by the Situation we solved initially.

A ball is moving with velocity v. It collides with the Ball 1 among the group of five balls.

By the result of our Situation, we know that the initial Ball will come to rest, and Ball 1 will acquire a velocity v.

But Ball 2 is present immediately next to Ball 1. Ball 1 collides immediately with Ball 2. Ball 1, which momentarily had a velocity, now again comes to rest, and Ball 2 now acquires the velocity v.

The same thing happens for other balls. They momentarily gain velocity v, but immediately collide and again come to rest.

In this way, the energy keeps on transferring from Ball 1 to Ball 2, and so on till Ball 5. There is no obstruction in the path of Ball 5. Thus the Ball 5 continues to move with its velocity v. All other balls are now at rest.

So, overall, Energy is sequentially transferred from the initial ball to Ball 5.

Thus, the Answer is Option (3) Ball 5 will move with velocity v, rest all balls will come to rest