Question

In Fig. 4-48a, a sled moves in the negative X direction at constant speed${\mathit{v}}_{\mathbf{s}}$while a ball of ice is shot from the sled with a velocity${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{0}}\mathbf{=}{\mathit{v}}_{\mathbf{0}\mathbf{x}}\hat{\mathbf{i}}\mathbf{+}{\mathbf{v}}_{\mathbf{0}\mathbf{y}}\hat{\mathbf{j}}$relative to the sled. When the ball lands its horizontal displacement$\mathbf{∆}{\mathit{x}}_{\mathbf{b}\mathbf{g}}$relative to the ground (from its launch position to its landing position) is measured. Figure 4-48bgives$\mathbf{∆}{\mathit{x}}_{\mathbf{b}\mathbf{g}}$as a function of${\mathit{v}}_{\mathbf{s}}$. Assume the ball lands at approximately its launch height. What are the values of (a)${\mathit{v}}_{\mathbf{0}\mathbf{x}}$and (b)${\mathit{v}}_{\mathbf{o}\mathbf{y}}$? The ball’s displacement$\mathbf{∆}{\mathit{x}}_{\mathbf{b}\mathbf{s}}$relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is$\mathbf{∆}{\mathit{x}}_{\mathbf{bs}}$when${\mathit{v}}_{\mathbf{s}}$is (c)and (d) 15.0 m/s?

Short answer

a) The magnitude of velocity$V_{0x}\mathrm{is}10\mathrm{m}/\mathrm{s}$

b) The magnitude of velocity$V_{0x}\mathrm{is}19.6\mathrm{m}/\mathrm{s}$

c) The value of $∆x_{bs}\mathrm{at}5\mathrm{m}/\mathrm{s}\mathrm{is}40\mathrm{m}$

d) The value of $∆x_{bs}\mathrm{at}15\mathrm{m}/\mathrm{s}\mathrm{is}40\mathrm{m}$

Step-by-step solution

The given data

The launching velocity of the ice ball relative to sled:$\left(V_{0s}\right)=V_{0x}\hat{i}+V_{0y}\hat{j}$

Understanding the concept of the relative motion

When two frames of reference Pand Qare traveling relative to each other at a constant velocity, the velocity of a particleAas measured by an observer in frame Pusually differs from that measured in frame Q. The relation between two measured velocities is,

${\overrightarrow{\mathbf{V}}}_{\mathbf{A}\mathbf{P}}\mathbf{=}{\overrightarrow{\mathbf{V}}}_{\mathbf{A}\mathbf{Q}}\mathbf{+}{\overrightarrow{\mathbf{V}}}_{\mathbf{Q}\mathbf{P}}$

Here${\overrightarrow{\mathbf{V}}}_{\mathbf{QP}}$ is the velocity of Q with respect to P.

Using the relative motion concept, we can find the magnitude and the direction of the resultant vector.

Formulae:

The second equation of the kinematic motion, $∆y=V_{y}t-\left(\frac{1}{2}\right)g{t}^2$ (i)

The displacement of a body, x=vt (ii)

a) Calculation of the magnitude of the velocity

The velocity of sled relative to the ground$\left(V_{sg}\right)$can be written in the vector notation as,

${\stackrel{\rightharpoonup }{V}}_{sg}=-V_s\hat{i}$

The launching velocity of the ice ball relative to sled,$\left(\stackrel{\rightharpoonup }{V_{os}}\right)=V_{0x}\hat{i}+V_{0y}\hat{j}$

So, the launching velocity of the ice ball relative to sled using the given equation can be written as follows:

$$\begin{gathered} \left(\stackrel{\rightharpoonup }{V_{os}}\right)=\left(V_{0x}-V_s\right)\hat{i}+V_{0y}\hat{j} \\ {V}_{ogy}=\left(V_{0x}-V_s\right)\mathrm{and} \\ {\mathrm{V}}_{\mathrm{ogy}}={\mathrm{V}}_{\mathrm{oy}} \end{gathered}$$

The horizontal displacement of the ball is given as:

$∆X_{bg}=\left(V_{0X}-V_s\right)t.....................\left(a\right)$

The vertical displacement of the ball is given suing equation (i) and the given data as:

$$\begin{gathered} ∆Y_{bg}=V_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ 0=V_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ {V}_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ t=\frac{2V_{o{y}_{}}}{g}..............................\left(1\right) \end{gathered}$$

Putting this value for t in the equation (a) for$X_{og}$we get,

$∆X_{bg}=\frac{2V_{ox}{V}_{oy}}{g}-\frac{2V_{oy}{V}_s}{g}$

From graph, we can get the above value as follows:

$∆X_{bg}=40-4V_s...........................\left(b\right)$

Comparing these two equations (a) and (b) from part (a) calculations for, we can write that,

$$\begin{gathered} \frac{2V_{oy}}{g}=4 \\ {V}_{oy}=\frac{4g}{2} \end{gathered}$$

Substitute the given values.

$$\begin{gathered} V_{oy}=\frac{4\left(9.8\right)}{2} \\ =19.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the velocity vector of the x-component can be calculated as:

$$\begin{gathered} \frac{2V_{0x}{V}_{oy}}{g}=40 \\ {V}_{0x}=\frac{40g}{2V_{0y}} \end{gathered}$$

Substitute the given values.

$$\begin{gathered} V_{0x}=\frac{40\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{2\left(19.6\mathrm{m}/\mathrm{s}\right)} \\ =10\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the velocity is 10 m/s.

b) Calculation of the magnitude of the velocity

So, from part (a) calculations, we can get that the value of the velocity is 19.6 m/s.

c) Calculation of the distance at 5 m/s

The distance$∆X_{bs}$is independent of the speed of the sled.

Thus, the distance value can be given using equations (ii) and (1) as follows:

$∆X_{bs}=10\times \frac{2V_{oy}}{g}$

Substitute the given values.

$$\begin{gathered} ∆X_{bs}=10\times \left(\frac{2\times 19.6\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}\right) \\ =10\times 4 \\ =40\mathrm{m} \end{gathered}$$

Hence, the value of the distance at this given speed is $40\mathrm{m}$.

d) Calculation of the distance at 15 m/s

The distance$∆X_{bs}$ is independent of the speed of the sled.

So, the value of distance at this given speed is 40 m.