Chapter 7, Problem 28P

(a)

To determine

The mass of the second croquet ball.

Answer to Problem 28P

  $m_2=0.84\text{ kg}$

Explanation of Solution

Given:

Mass of first croquet ball is $m_1=0.280\text{ kg}$

Initially second croquet ball is at rest. That means initial speed of the second croquet ball is zero.

After collision, speed of the second croquet ball is half of the initial speed of the first croquet ball.

Formula used:

Let $u_1$ and $u_2$ be the speed of the first and second croquet ball respectively, before collision. Also let $v_1$ and $v_2$ be the speed of the first and second croquet ball respectively, after collision. Since the collision between the balls are perfectly elastic, momentum is conserved. That is total initial momentum of the balls $\left(p_{initial}\right)$ is equal to the total final momentum of the balls $\left(p_{final}\right)$ .

That is,

  $p_{initial}=p_{final}$

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$\left(1\right)$

Where, $m_2$ is the mass of the second croquet ball.

Also, for a perfectly elastic one dimensional collision,

  $u_1-u_2=-\left(v_1-v_2\right)$$\left(2\right)$

Calculation:

It is given that,

  $u_2=0\text{ m/s}$ and $v_2=\frac{u_1}{2}$

Substituting these in equation $\left(2\right)$ ,

  $u_1-0=-\left(v_1-\frac{u_1}{2}\right)$

  $\Rightarrow u_1=\frac{u_1}{2}-v_1$

  $\Rightarrow v_1=\frac{u_1}{2}-u_1$

  $\Rightarrow v_1=-\frac{u_1}{2}$

Substituting $u_2=0\text{ m/s}$ , $v_1=-\frac{u_1}{2}$ and $v_2=\frac{u_1}{2}$ in equation $\left(1\right)$ ,

  $m_1{u}_1+m_2\left(0\text{ m/s}\right)=m_1\left(-\frac{u_1}{2}\right)+m_2\left(\frac{u_1}{2}\right)$

  $\Rightarrow m_1{u}_1=m_1\left(-\frac{u_1}{2}\right)+m_2\left(\frac{u_1}{2}\right)$

  $\Rightarrow m_1=-\frac{m_1}{2}+\frac{m_2}{2}$

  $\Rightarrow m_1+\frac{m_1}{2}=\frac{m_2}{2}$

  $\Rightarrow \frac{3m_1}{2}=\frac{m_2}{2}$

  $\Rightarrow m_2=3m_1$$\left(3\right)$

Substituting the numerical values in equation $\left(3\right)$ ,

  $m_2=3\left(0.280\text{ kg}\right)=0.84\text{ kg}$

Conclusion:

The mass of the second croquet ball is $0.84\text{ kg}$ .

(b)

To determine

The fraction of the original kinetic energy $\left(\frac{\Delta KE}{KE}\right)$ gets transferred to the second croquet ball.

Answer to Problem 28P

  $0.75$

Explanation of Solution

Given:

Mass of first croquet ball is $m_1=0.280\text{ kg}$

Formula used:

Let $m_2$ be the mass of the second croquet ball. Take $u_1$ and $u_2$ be the speed of the first and second croquet ball respectively, before collision. Also let $v_1$ and $v_2$ be the speed of the first and second croquet ball respectively, after collision.

Total kinetic energy of the balls before collision is,

  $KE=K{E}_{1,beforecollision}+K{E}_{2,beforecollision}=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$$\left(4\right)$

Since, $u_2=0\text{ m/s}$ , equation $\left(4\right)$ becomes,

  $KE=\frac{1}{2}{m}_1{u}_1^2$

Kinetic energy of the second croquet ball after collision is,

  $K{E}_{2,aftercollision}=\frac{1}{2}{m}_2{v}_2^2$$\left(5\right)$

Since $v_2=\frac{u_1}{2}$ and $m_2=3m_1$ equation $\left(5\right)$ becomes,

  $K{E}_{2,aftercollision}=\frac{1}{2}\left(3m_1\right){\left(\frac{u_1}{2}\right)}^2=\frac{3}{8}{m}_1{u}_1^2$

Calculation:

The fraction of the original kinetic energy gets transferred to the second croquet ball can be calculated as,

  $\frac{K{E}_{2,aftercollision}}{KE}=\frac{\left(\frac{3}{8}{m}_1{u}_1^2\right)}{\left(\frac{1}{2}{m}_1{u}_1^2\right)}=\frac{3}{8}\times \frac{2}{1}=0.75$

Conclusion:

The fraction of the original kinetic energy gets transferred to the second croquet ball is $0.75$ .