Chapter 11, Problem 92A

(a)

To determine

the fraction of the initial energy is transferred to $m_2$ if $m_1=m_2$ .

Answer to Problem 92A

The first ball stops after the collision. So, the entire initial kinetic energy is transferred to the second ball.

Explanation of Solution

Given:

The mass of the two billiard balls is equal that is, $m_1=m_2$ .

Formula used:

From the law of conservation of momentum, the total momentum before and after collision are equal. That is,

  $m_1{v}_1+m_2{v}_2=m_1{v}_{1,f}+m_2{v}_{2,f}$  ......(1)

Here, $v_{1,f}$ and $v_{2,f}$ are the velocities of two balls after collision.

The total kinetic energy is also conserved. That is,

  $\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2=\frac{1}{2}{m}_1{v}_{1,f}{}^2+\frac{1}{2}{m}_1{v}_{2,f}{}^2$  ......(2)

Calculation:

Ball 1 is moving with the velocity $v_1$ and the ball 2 is at rest. As collision is elastic, both momentum and kinetic energy are conserved.

The second ball is at rest initially, and both have same mass i.e. $m_1=m_2$

  $\begin{array}{l}{m}_1{v}_1+m_2\left(0\right)=m_1{v}_{1,f}+m_2{v}_{2,f}\\ v_1=v_{1,f}+v_{2,f}\end{array}$

  $v_1-v_{1,f}=v_{2,f}$  ......(3)

Again, as the second ball is at rest, so from the equation (2),

  $\begin{array}{l}\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{\left(0\right)}^2=\frac{1}{2}{m}_1{v}_{1,f}{}^2+\frac{1}{2}{m}_1{v}_{2,f}{}^2\\ v_1^2=v_{1,f}{}^2+v_{2,f}{}^2\end{array}$

  $v_1^2-v_{1,f}{}^2=v_{2,f}{}^2$  ......(4)

Divide equation (4) with (1),

  $\begin{array}{l}\frac{v_1^2-v_{1,f}{}^2}{v_1-v_{1,f}}=\frac{v_{2,f}{}^2}{v_{2,f}}\\ v_1+v_{1,f}=v_{2,f}\end{array}$

Use this equation to replace $v_{2,f}$ from equation (1)

  $\begin{array}{l}{v}_1-v_{1,f}=v_1+v_{1,f}\\ 2v_{1,f}=0\\ \boxed{v_{1,f}=0}\end{array}$

Conclusion:

Hence, the first ball stops after the collision. So, the entire initial kinetic energy is transferred to the second ball.

(b)

To determine

the fraction of the initial energy is transferred to $m_2$ if $m_1>>m_2$ .

Answer to Problem 92A

A very less energy is transformed to the second ball.

Explanation of Solution

Given:

The mass of the two billiard balls is equal that is, $m_1=m_2$ .

Formula used:

From the law of conservation of momentum, the total momentum before and after collision are equal. That is,

  $m_1{v}_1+m_2{v}_2=m_1{v}_{1,f}+m_2{v}_{2,f}$  ......(1)

Here, $v_{1,f}$ and $v_{2,f}$ are the velocities of two balls after collision.

The total kinetic energy is also conserved. That is,

  $\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2=\frac{1}{2}{m}_1{v}_{1,f}{}^2+\frac{1}{2}{m}_1{v}_{2,f}{}^2$  ......(2)

Calculation:

If $m_1>>m_2$ , the collision is similar to that between a moving train and bicycle. The motion of the first ball is unaffected and the second ball flies off with the large speed. As the ball moves with almost same speed, a very less energy is transformed to the second ball.

Conclusion:

Hence, a very less energy is transformed to the second ball.

(c)

To determine

whether hydrogen, carbon or iron atoms would be more desirable to use for the given purpose or not.

Answer to Problem 92A

Hydrogen is more desirable to use for this purpose.

Explanation of Solution

Given:

The mass of the two billiard balls is equal that is, $m_1=m_2$ .

Calculation:

When the mass of first ball is much smaller than the second ball, the first ball will rebound after collision. So, the neutrons should not be colliding with the large atoms, but it should collies with atoms of nearly same mass. So, hydrogen is more desirable for this purpose.

Conclusion:

Hence, hydrogen is more desirable for this purpose.