Question

Figure 15-34 shows block 1 of mass $\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathit{k}\mathit{g}$sliding to the right over a frictionless elevated surface at a speed of. The block undergoes an elastic collision with stationary block, which is attached to a spring of spring constant1208.5N/m. (Assume that the spring does not affect the collision.) After the collision, block2 oscillates in SHM with a period of 0.140s, and block 1 slides off the opposite end of the elevated surface, landing a distance from the base of that surface after falling height $\mathbf{h}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{90}\mathbf{m}$. What is the value role="math" localid="1655106415375" $\mathit{o}\mathit{f}\mathbf{d}$?

Short answer

The horizontal distancetravelled from the base of the surface from which the block is falling from the surface is 4.0m

Step-by-step solution

The given data

1. Mass of block 1,$m_1=0.2\mathrm{kg}$
2. Speed of block 1 before collision,$v_i=8.0\mathrm{m}/\mathrm{s}$
3. Spring constant,$k=1208.5\mathrm{N}/\mathrm{m}$
4. Period of second block after collision,$T=0.140\mathrm{s}$
5. Height of fall,$h=4.90\mathrm{m}$

Understanding the concept of collision and kinematic equations

Using the formula for the time period, we first find the mass of the block which is attached to the spring. Next, using the formula for elastic collision, we find the velocity of the block which gets rebounded. Using the velocity and kinematic equations, we find the horizontal distance traveled.

Formula:

The time period of oscillation,

$T=2\pi \sqrt{\frac{m}{k}}$(i)

The second equation of motion,

$x-x_0=v_{i}t+\frac{1}{2}a{t}^2$…….(ii)

Law of conservation of momentum,

$v_{1f}=\left|\frac{m_1-m_2}{m_1-m_2}\right|v_{1i}$(iii)

Calculation of horizontal distance

From equation (i) and the given values, we get the mass of system as:

$$\begin{gathered} 0.140=2\pi \sqrt{\frac{m}{1208.5}} \\ 0.{140}^2=4{\pi }^2\times \left(\frac{m}{1208.5}\right) \\ \frac{0.{140}^2\times 1208.5}{4{\pi }^2}=m \\ m=0.6\mathrm{kg} \end{gathered}$$

As the collision is elastic, the rebound velocity for the block will be given as:

$$\begin{gathered} v_{1f}=\left|\frac{m_1-m_2}{m_1+m_2}\right|v_{1i} \\ =\left|\frac{0.2-0.6}{0.2+0.6}\right|\times 8 \\ =\left|\frac{-0.4}{0.8}\right|\times 8 \\ =4.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, as the block is moving along a straight surface, the block will launch at an angle of $0^{°}$with respect to the surface.

This makes the velocity component in y- direction 0m/s and 4.0m/s in x-direction.

Here,$x-x_0=h\mathrm{and}g=9.8m/s$ and. (As the block is falling,the acceleration due to gravity will be in the direction of motion, so we take it as positive.) So, the equation (ii) can be given as:

$$\begin{gathered} h=\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2h}{g}} \\ =\sqrt{\frac{2\times 4.9}{9.8}} \\ =1s \end{gathered}$$

Now using equation (ii) in x-direction$a=0$and $x-x_0=d$, we get

$$\begin{gathered} \mathrm{d}=\mathrm{v},\mathrm{t} \\ =4.0\mathrm{m}/\mathrm{s}\times 1\mathrm{s} \\ =4.0\mathrm{m} \end{gathered}$$

Hence, the value of horizontal distance is 4.0 m .