Chapter 11.2, Problem 23PP

(a)

To determine

To sketch the situation and choose the system.

Explanation of Solution

Given:

Mass of a magnetic target suspended on a string is $m_{mt}=0.73\text{ kg}$

Mass of a magnetic dart, shot horizontally, and strikes the target head-on is $m_{md}=0.025\text{ kg}$

The height above the initial level at which the dart and target together swings like a pendulum is $\text{h = 12}\text{.0 cm = 12}\text{.0}\times {\text{10}}^{-2}\text{ m}$

Suspended magnetic target and the magnetic dart which strikes the target makes the system. The situation is sketched as shown in figure 1.

Figure 1

(b)

To determine

To decide and describe what is conserved in each step of the process.

Answer to Problem 23PP

During the inelastic collision between the magnetic target and the magnetic dart, momentum is conserved and the energy is conserved when the magnetic target - magnetic dart combination swings upward.

Explanation of Solution

Introduction:

Law of conservation of momentum: The law of conservation of momentum states that when collision occurs between the two bodies in an isolated system then total momentum of two bodies before collision is equivalent to the total momentum of that two bodies after the collision.

Law of conservation of energy: It states that total energy of an isolated system is conserved. That is, it remains persistent over a time.

During the inelastic collision between the magnetic target and the magnetic dart, momentum is conserved. Therefore, momentum of the system before collision is equal to the momentum of the system after collision. This can be expressed as,

  $p_{mti}+p_{mdi}=p_{mtf}+p_{mdf}$  $\left(1\right)$

Where,

$p_{mti}$ is the momentum of the magnetic target before collision, $p_{mti}=m_{mt}{v}_{mti}$

$p_{mdi}$ is the momentum of the magnetic dart before collision, $p_{mdi}=m_{md}{v}_{mdi}$

$p_{mtf}$ is the momentum of the magnetic target after collision, $p_{mtf}=m_{mt}{v}_{mtf}$

$p_{mdf}$ is the momentum of the magnetic dart after collision, $p_{mdf}=m_{md}{v}_{mdf}$

Substituting all these in equation $\left(1\right)$ ,

  $m_{mt}{v}_{mti}+m_{md}{v}_{mdi}=m_{mt}{v}_{mtf}+m_{md}{v}_{mdf}$  $\left(2\right)$

Where,

$m_{mt}$ and $m_{md}$ are masses of magnetic target and magnetic dart respectively,

$v_{mti}$ is the velocity of magnetic target before collision, $v_{mti}=0.0\text{ m/s}$

$v_{mdi}$ is the velocity of the magnetic dart before collision

$v_{mtf}$ and $v_{mdf}$ are the velocity of magnetic target and the magnetic dart after collision. Since after collision magnetic target and the dart swings together like a pendulum their velocity must be the same. That can be taken as $v_{mtf}=v_{mdf}=v_f$

Now, the equation $\left(2\right)$ becomes,

  $m_{md}{v}_{mdi}=\left(m_{mt}+m_{md}\right)v_f$  $\left(3\right)$

And, the energy is conserved during the upward swing of magnetic target-magnetic dart combination. Thus, at the topmost of the swing, potential energy of the system is equal to the kinetic energy of the system. This can be represented as,

  $P{E}_{mt}+P{E}_{md}=K{E}_{mt}+K{E}_{md}$  $\left(4\right)$

Where,

$P{E}_{mt}+P{E}_{md}$ is the total potential energy of the magnetic target-magnetic dart combination at the top of the swing, $P{E}_{mt}+P{E}_{md}=\left(m_{mt}+m_{md}\right)g{h}_f$

$K{E}_{mt}+K{E}_{md}$ is the total kinetic energy of the magnetic target-magnetic dart combination at the top of the swing, $K{E}_{mt}+K{E}_{md}=\frac{1}{2}\left(m_{mt}+m_{md}\right)v_f^2$

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

  $\left(m_{mt}+m_{md}\right)g{h}_f=\frac{1}{2}\left(m_{mt}+m_{md}\right)v_f^2$  $\left(5\right)$

Where $h_f$ topmost height of the swing

$g$ is acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

(c)

To determine

The initial velocity of the dart.

Answer to Problem 23PP

  $v_{mdi}\text{= 46}\text{.3 m/s}$

Explanation of Solution

Given:

Mass of a magnetic target suspended on a string is $m_{mt}=0.73\text{ kg}$

Mass of a magnetic dart, shot horizontally, and strikes the target head-on is $m_{md}=0.025\text{ kg}$

The height above the initial level at which the dart and target together swings like a pendulum is $\text{h = 12}\text{.0 cm = 12}\text{.0}\times {\text{10}}^{-2}\text{ m}$

Formula used:

From the conservation of momentum equation $\left(3\right)$ in the part (b), Initial velocity of the magnetic dart can be written as,

  $v_{mdi}=\frac{\left(m_{mt}+m_{md}\right)v_f}{m_{md}}$  $\left(6\right)$

From the conservation of energy equation $\left(5\right)$ in the part (b), final velocity of the magnetic target-dart combination is,

  $v_f=\sqrt{2g{h}_f}$

Substituting for $v_f$ in equation $\left(6\right)$ ,

  $v_{mdi}=\frac{\left(m_{mt}+m_{md}\right)}{m_{md}}\sqrt{2g{h}_f}$  $\left(7\right)$

Calculation:

Initial velocity of the magnetic dart can be calculated by substituting the numerical values in equation $\left(7\right)$ ,

  $v_{mdi}=\frac{\left[\left(0.73\text{ kg}\right)+\left(0.025\text{ kg}\right)\right]}{\left(0.025\text{ kg}\right)}\sqrt{2\left(9.8{\text{ m/s}}^2\right)\left(\text{12}\text{.0}\times {\text{10}}^{-2}\text{ m}\right)}$

  $=\frac{0.755\text{ kg}}{0.025\text{ kg}}\sqrt{2.352{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}}$

  $=30.2\times 1.533\text{ m/s}$

  $\text{= 46}\text{.3 m/s}$

Conclusion:

Initial velocity of the magnetic dart is $\text{46}\text{.3 m/s}$ .