Question

Block 1, with mass${\mathit{m}}_{\mathbf{1}}$ and speed 4.0 m/s, slides along anx axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}{\mathit{m}}_{\mathbf{1}}$.The two blocks then slide into a region where the coefficient of kinetic friction is 0.50; there they stop. How far into that region do (a) block 1 and (b) block 2 slide?

Short answer

a) Sliding distance of block is,$d_1=30\mathrm{cm}$

b) Sliding distance of block is,$d_2=3.3m$

Step-by-step solution

Step 1: Given Data

Mass of first block 1 is$m_1$.

Initial velocity of block 1 is$V_{1i}=4.0\mathrm{m}/\mathrm{s}$

Mass of block 2 is$m_2=0.40m_1$.

Initial velocity of second block is$V_{2i}=0\mathrm{m}/\mathrm{s}$i.e. block is at rest.

Coefficient of kinetic friction is $\mu =0.50$.

Determining the concept

From the law of conservation of mechanical energy, it can be said that the total energy of the system must be conserved during, before, and after the collision.According to conservation of energy, energy can neither be created, nor be destroyed.

Formulae are as follow:

For conservation of linear momentum,$m_1{V}_{1i}+m_2{V}_{2i}=m_1{V}_{1f}+m_2{V}_{2f}$

Conservation of kinetic energy,$\frac{1}{2}{m}_1{V}_{1i}^2=\frac{1}{2}{m}_1{V}_{1f}^2+\frac{1}{2}{m}_2{V}_{2f}^2$

Speed in center of mass is,$V_{com}=\frac{\left(m_1{V}_{1i}\right)}{m_1+m_2}{V}_{1i}$

For elastic collision,$V_{1f}=\frac{m_1-m_2}{m_1+m_2}$

From equation 9-68,$V_{2f}=\frac{2m_1}{m_1+m_2}{V}_{1i}$

Where, m₁, m₂ are masses and V is velocity.

(a) Determining to find the sliding distance of block 1

Let,andare masses of blockand block 2 respectively. Initial velocity of blockis zero.

$m_1{V}_{1i}=m_1{V}_{1f}+m_2{V}_{2f}$

Similarly, the total kinetic energy is conserved,

$\frac{1}{2}{m}_1{V}_{1i}^2=\frac{1}{2}{m}_1{V}_{1f}^2+\frac{1}{2}{m}_2{V}_{2f}^2$

Solving equation tois given by equation 9-67,

$$\begin{gathered} V_{1f}=\frac{m_1-m_2}{m_1+m_2}{V}_{1i} \\ {V}_{1f}=\frac{m_1-0.40m_1}{m_1+0.40m_1}\left(4.0\right) \\ {V}_{1f}=\frac{0.60}{1.40}\times 4.0 \\ {V}_{1f}=1.71\mathrm{m}/\mathrm{s} \end{gathered}$$

Solving equation $V_{2f}$is given by equation 9-68,

$$\begin{gathered} V_{2f}=\frac{2m_1}{m_1+m_2}{V}_{1i} \\ {V}_{2f}=\frac{2m_1}{m_1+0.40m_2}\left(4\right) \\ {V}_{2f}=\frac{2}{1.40}\times 4 \\ {V}_{2f}=\frac{2}{1.40}\times 4 \\ {V}_{2f}=5.71\mathrm{m}/\mathrm{s} \end{gathered}$$

During the sliding of block, the kinetic energy of block is converted to the thermal form. Thus,$K_{1f}=\frac{1}{2}{m}_1{V}_{1f}^2$is the kinetic energy converted to thermal form, i.e.$\Delta E_{ht}={\mu }_k{m}_{1}g{d}_1$

$$\begin{gathered} \frac{1}{2}{m}_1{V}_{1f}^2={\mu }_k{m}_{1}g{d}_1 \\ {d}_1=\frac{1}{2}\times \frac{V_{1f}^2}{{\mu }_{k}g} \\ {d}_1=\frac{1}{2}\times \frac{{\left(1.71\right)}^2}{0.50\times 9.80} \\ {d}_1=0.2983\mathrm{m} \\ {d}_1=0.30\mathrm{m} \\ {d}_1=30\mathrm{cm} \end{gathered}$$

Hence, sliding distance of block is,$d_1=30\mathrm{cm}$

(b) Determining to find the sliding distance of block 2

During the sliding of blockthe kinetic energy of the blockis converted tothermal energy. Therefore,$K_{2f}=\frac{1}{2}{m}_2{V}_{2f}^2$is the kinetic energy converted to the thermal form, i.e.

$$\begin{gathered} ∆E_{th}={\mu }_k{m}_{2}g{d}_2 \\ \frac{1}{2}{m}_2{V}_{2f}^2={\mu }_k{m}_{2}g{d}_1 \\ {d}_2=\frac{1}{2}\times \frac{V_{2f}^2}{{\mu }_{k}g} \\ {d}_1=\frac{1}{2}\times \frac{{\left(5.71\right)}^2}{0.50\times 9.80} \\ {d}_1=3.3269\mathrm{m} \\ {d}_1=3.3\mathrm{m} \end{gathered}$$

Hence,sliding distance of block is,$d_2=3.3\mathrm{m}$

Therefore, by using the concept of conservation linear momentum, the sliding distance of block and block and speed of center of mass can be found.