Question

A 0.7 kgblock of ice is sliding by you on a very slippery floor at $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}\mathbf{.}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$]As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for $\mathbf{}\mathbf{0}\mathbf{.}\mathbf{003}\mathbf{\hspace{0.33em}}\mathbf{s}$. The block eventually slides at an angle of $\mathbf{}\mathbf{22}\mathbf{\hspace{0.33em}}\mathbf{degrees}$from its original direction. The overhead view shown in Figure 2.54is approximately to scale. The arrow represents the average force your toe applies briefly to the block of ice. (a) Which of the possible paths shown in the diagram corresponds to the correct overhead view of the block’s path? (b) Which components of the block’s momentum are changed by the impulse applied by your foot? (Check all that apply. The diagram shows a top view, looking down on the xzplane.) (c) What is the unit vectorin the direction of the block’s momentum after the kick? (d) What is the x component of the block’s momentum after the kick? (e) Remember that $\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{\left|}\overrightarrow{\mathbf{p}}\mathbf{\right|}\mathbf{}\hat{\mathbf{p}}$. What is the magnitude of the block’s momentum after the kick? (f) Use your answers to the preceding questions to find the zcomponent of the block’s momentum after the kick (drawing a diagram is helpful). (g) What was the magnitude of the average force you applied to the block?

Short answer

$$\begin{gathered} \mathrm{a})\mathrm{Path}\mathrm{B} \\ \mathrm{b})\mathrm{z}\mathrm{component} \\ \mathrm{c})(0.927,0,0.374\left) \\ \mathrm{d}\right)1.75\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s} \\ \mathrm{e})1.887\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s} \\ \mathrm{f})0.707\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s} \\ \mathrm{g})629\hspace{0.33em}\mathrm{N} \end{gathered}$$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of the ice block is, $0.7\hspace{0.33em}kg$.
• The velocity of the block is,$2.5\mathrm{m}/\mathrm{s}$ .
• The time for which the foot is in contact with the ice block is,$0.003\hspace{0.33em}\mathrm{s}$ .
• The angle at which the block slide is,$22°$ .

Significance of the unit vector, momentum principle and impulse of the block

The unit vector is the division between the non-zero vector and the normal of the non-zero vector.

The momentum of a body is described as the product of the mass and the velocity of that body.

The impulse of a body is described as the product of the force exerted on a body and the time taken by the body.

The equation of the unit vector gives the unit vector of the block and the momentum principle gives x and z components of the block’s momentum along with the magnitude of the block and the impulse gives the average force applied to the block.

(a) Determination of the correct possible path for the block

The possible path B is the correct overhead view of the block’s path. The reason for choosing the path is that, when the block slides down the plane in a straight line, then, it will be helpful for calculating the motion of the block.

Thus, path B is the correct overhead view of the block’s path.

(b) Determination of the component of the block’s momentum that is changed by the applied impulse

As the block is sliding down in the xz -plane and there are no forces acting in the x and y direction, hence, only the z component of the block’s momentum is changed due to the effect of the applied impulse.

Thus, the z component of the block’s momentum is changed by the impulse applied by the foot.

(c) Determination of the unit vector in the direction of momentum of block after the kick

The free body diagram for finding the velocity in the direction is given as follows,

As there is no force acting in the x-direction, hence the velocity in the x-direction is 2.5 m/s and the velocity in the y-direction is 0 and the block is moving in the xz plane.

From the momentum principle, the equation of the momentum of the block at x-direction is expressed as-

$p_x=m{v}_x$

Here, m is the mass of the block and$v_x$the velocity in the x-direction.

For localid="1658557927037" $\mathrm{m}=0.7\mathrm{kg}and{\mathrm{v}}_{\mathrm{x}}=2.5\hspace{0.33em}\mathrm{m}/\mathrm{s}$

$$\begin{gathered} {\mathrm{p}}_{\mathrm{x}}=0.7\hspace{0.33em}\mathrm{kg}\times 2.5\mathrm{m}/\mathrm{s} \\ =1.75\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

From the above figure, the equation of the velocity of the block in the z direction is expressed as,

$\tan 22°=\frac{{\mathrm{v}}_{\mathrm{z}}}{{\mathrm{v}}_{\mathrm{x}}}$

Here, $v_z$is the velocity in z direction and $v_x$is the velocity in x direction.

For$tan22°=0.404$.

$$\begin{gathered} 0.404=\frac{v_z}{2.5m/s} \\ {v}_z=0.404\times 2.5m/s \\ =1.01m/s \end{gathered}$$

From the momentum principle, the equation of the momentum of the block in z-direction is expressed as,

$p_z=m{v}_z$

For $\mathrm{m}=0.7\hspace{0.33em}\mathrm{kg},\mathrm{and}{\mathrm{v}}_{\mathrm{z}}=1.01\hspace{0.33em}\mathrm{m}/\mathrm{s}$,

$$\begin{gathered} p_z=0.7\hspace{0.33em}\mathrm{kg}\times 1.01\hspace{0.33em}\mathrm{m}/\mathrm{s} \\ =0.707\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the non-zero momentum vector of the block is$(1.75,0,0.707)\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.

From the momentum law, the equation of the magnitude of the momentum of the block can be expressed as,

$\left|\overrightarrow{p}\right|=\sqrt{{p_x}^2+{p_y}^2+{p_z}^2}$

Here,$\left|\overrightarrow{p}\right|$is the magnitude of the momentum of the block, $p_x$is the momentum in the x direction, $p_y$is the momentum in the y direction, and $p_z$is the momentum in the z direction.

For ${\mathrm{p}}_{\mathrm{x}}=1.75\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s},{\mathrm{p}}_{\mathrm{y}}=0\mathrm{and}{\mathrm{p}}_{\mathrm{z}}=0.707\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$,

$$\begin{gathered} \left|\overrightarrow{p}\right|=\sqrt{{\left(1.75kg\cdot m/s\right)}^2+0+{\left(0.707kg\cdot m/s\right)}^2} \\ =\sqrt{3.0625+0.499849}kg\cdot m/s \\ =\sqrt{3.562}kg\cdot m/s \\ =1.887kg.m/s \end{gathered}$$

From the rule of the unit vector, the equation of the unit vector can be expressed as,

$\hat{p}=\frac{\overrightarrow{p}}{\left|\overrightarrow{p}\right|}$

Here, $\overrightarrow{p}$is the non-zero momentum vector and $\left|\overrightarrow{p}\right|$the magnitude of the momentum of the block.

For localid="1658559034177" $\mathrm{p}⃗=(1.75,0,0.707)\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}\mathrm{and}\hspace{0.33em}\left|\overrightarrow{\mathrm{p}}\right|=1.887\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$,

$$\begin{gathered} \hat{p}=\frac{\left(1.75,0,0.707\right)kg.m/s}{1.887kg.m/s} \\ =(0.927,0,0.374) \end{gathered}$$

Thus, the unit vector $\hat{p}$in the direction of the block’s momentum after the kick is $(0.927,0,0.374)$.

(d) Determination of the x-component of the block’s momentum

As the data gathered from above part, the x component of the block’s momentum after the kick is$1.75\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.

Thus, the x component of the block’s momentum after the kick isrole="math" localid="1658559539898" $1.75\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$ .

(e) Determination of the magnitude of the block’s momentum after the kick

As the data obtained from part (c), the magnitude of the block’s momentum after the kick is$1.887\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.

Thus, the magnitude of the block’s momentum after the kick isrole="math" localid="1658559635585" $1.887\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$ .

(f) Determination of the z-component of the block’s momentum

As the data gathered from part (c), the z component of the block’s momentum after the kick is$0.707\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.

Thus, the z component of the block’s momentum after the kick is$0.707\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$ .

(g) Determination of the magnitude of the average force applied to the block

From the law of impulse, the equation for the magnitude of the average force applied on the block can be expressed as,

$I=F\cdot t$

Here, F is the magnitude of the average force and t is the time taken.

For$\mathrm{I}=1.887\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{t}=0.003\hspace{0.33em}\mathrm{s}$

$$\begin{gathered} 1.887\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}=\mathrm{F}\times 0.003\hspace{0.33em}\mathrm{s} \\ \mathrm{F}=\frac{1.887\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}}{0.003\hspace{0.33em}\mathrm{s}} \\ =629\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =629\mathrm{N} \end{gathered}$$

Thus, the magnitude of the average force applied on the block is $629\hspace{0.33em}\mathrm{N}$.