91Ó°ÊÓ

The given data can be listed below as,

• The mass of the ice block is, $0.7 kg$.
• 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,$0.003 \mathrm{s}$ .
• The angle at which the block slide is,$22°$ .

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.

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.

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.

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" $\mathrm{m}=0.7\mathrm{kg}and{\mathrm{v}}_{\mathrm{x}}=2.5 \mathrm{m}/\mathrm{s}$

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

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$tan22°=0.404$.

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

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

$p_z=m{v}_z$

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

$$\begin{gathered} p_z=0.7 \mathrm{kg}×1.01 \mathrm{m}/\mathrm{s} \\ =0.707\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the non-zero momentum vector of the block is$(1.75,0,0.707) \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}$.

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

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

Here,$\left|\stackrel{→}{p}\right|$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 ${\mathrm{p}}_{\mathrm{x}}=1.75 \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s},{\mathrm{p}}_{\mathrm{y}}=0\mathrm{and}{\mathrm{p}}_{\mathrm{z}}=0.707 \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}$,

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

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

$\hat{p}=\frac{\stackrel{→}{p}}{\left|\stackrel{→}{p}\right|}$

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

For localid="1658559034177" $\mathrm{p}⃗=(1.75,0,0.707) \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}\mathrm{and} \left|\stackrel{→}{\mathrm{p}}\right|=1.887\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}$,

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

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

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

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

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

Thus, the magnitude of the block’s momentum after the kick isrole="math" localid="1658559635585" $1.887 \mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}$ .

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

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

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

$I=Fâ‹\dots t$

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

For$\mathrm{I}=1.887\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{t}=0.003 \mathrm{s}$

$$\begin{gathered} 1.887\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}=\mathrm{F}×0.003 \mathrm{s} \\ \mathrm{F}=\frac{1.887\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s}}{0.003 \mathrm{s}} \\ =629\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2×\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =629\mathrm{N} \end{gathered}$$

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