In Fig.9-67, block 1 of mass ${\mathit{m}}_{\mathbf{1}}$ slides from rest along a frictionless ramp from heighth=2.50 m and then collides with stationary block 2, which has mass${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{m}}_{\mathbf{1}}$. After the collision, block 2 slides into a region where the coefficient of kinetic friction ${\mathit{\mu }}_{\mathbf{k}}$is 0.500 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?

The height, h=2.50 m

The mass of block 2 is,$m_2=2.00m_1$

The coefficient of friction is,${\mu }_k=0.500$

By finding the value of ${\mathit{v}}_{\mathbf{2}}$for elastic and completely inelastic collision, find the value of distanced.

Formulae are as follow:

The velocity $v_2$is,$v_2=\frac{2m_1}{m_1+m_2}{v}_{1i}$

The velocity $v_{1i}$is,$v_{1i}=\sqrt{2gh}$

The distance d is,$d=\frac{v_2^2}{2{\mu }_{k}g}$

where, m is mass,vis velocity, g is an acceleration due to gravity, h is height, d is distance and ${\mu }_k$is coefficient of kinetic friction.

If the collision is perfect elastic, then,

$$\begin{gathered} v_2=\frac{2m_1}{m_1+m_2}{v}_{1i} \\ {v}_2=\frac{2m_1}{m_1+\left(2.00\right)m_1}\times \sqrt{2gh} \\ {v}_2=\frac{2}{3}\times \sqrt{2gh} \end{gathered}$$

Where the speed of block 1 at the bottom of the frictionless ramp is,

$$\begin{gathered} v_{1i}=\sqrt{2gh} \\ =\sqrt{2\times 9.8\times 2.50} \\ =7.00\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus,

$$\begin{gathered} v_2=\frac{2}{3}\times 7.00 \\ {v}_2=4.67\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, for block 2’s rough slide,

$$\begin{gathered} \frac{1}{2}{m}_2{v}_2^2=∆E_{th} \\ \frac{1}{2}{m}_2{v}_2^2=f_{k}d \\ \frac{1}{2}{m}_2{v}_2^2={\mu }_k{m}_{2}gd \end{gathered}$$

Thus, the distancedis given by,

$$\begin{gathered} d=\frac{v_2^2}{2{\mu }_{k}g} \\ d=\frac{4.{67}^2}{2\times 0.500\times 9.8} \\ d=2.22m \end{gathered}$$

Hence, the value of distance d if the collision is elastic is, d =2.22 m

Similarly, in a completely inelastic collision,

$$\begin{gathered} v_2=\frac{m_1}{m_1+\left(2.00\right)m_1}\times 7.00 \\ {v}_2=\frac{1}{3}\times 7.00 \\ {v}_2=2.33\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus,the value of distancedis given by,

$$\begin{gathered} d=\frac{v_2^2}{2{\mu }_{k}g} \\ d=\frac{2.{33}^2}{2\times 0.500\times 9.8} \\ d=0.556\mathrm{m} \end{gathered}$$

Hence,the value of distancedif the collision is completely inelastic is, d =0.556 m

Therefore, by using the equation for velocity for elastic and completely inelastic collisions, this problem can be solved.