Speed amplifier.In Fig. 9-75, block 1 of mass m₁ slides along an x axis on a frictionless floor with a speed of ${\mathit{v}}_{\mathbf{1}\mathbf{i}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$.Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass ${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}{\mathbf{m}}_{\mathbf{1}}$. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass ${\mathbf{m}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}{\mathbf{m}}_{\mathbf{2}}$. (a) What then is the speed of block 3? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 is greater than, less than, or the same as the initial values for block 1?

Masses of the blocks are, $m_1,m_2=0.5m_1\mathrm{and}{\mathrm{m}}_3=0.5{\mathrm{m}}_2$.

Speed of $m_1$is, $v_{1i}=4.00\frac{m}{\mathrm{s}}$.

Using the law of conservation of momentum, we can measure the velocity of blocks after elastic collision in 1 dimension. Then comparing the velocity of block 3 with block 1 we can write the answer for part b). Using the formulae for K.E and momentum, we can write the answer for parts c) and d).

Collision of block 1 with block 2:

The velocity of blocks after elastic collision in 1 dimension,

$v_2=\frac{2m_1}{m_1+m_2}{v}_{1i}$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} v_2=\frac{2m_1}{m_1+0.5m_1}\left(4\right) \\ =\frac{2}{1.5}\left(4\right) \\ =\frac{16}{3}\mathrm{m}/\mathrm{s} \end{gathered}$$

Collision of block 2 with block 3:

The velocity of blocks after elastic collision in 1 dimension,

role="math" localid="1661250556459" $$\begin{gathered} v_3=\frac{2m_2}{m_2+0.5m_2}\left(\frac{16}{3}\right) \\ =\frac{2}{1.5}\left(\frac{16}{3}\right) \\ =\frac{64}{9} \\ =7.11\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of block 3 is$7.11\frac{\mathrm{m}}{\mathrm{s}}$ .

From part a), we can conclude that speed of the block 3 is greater than the initial speed of block 1.

Thus, speed of the block 3 is greater than the initial speed of block 1.

The kinetic energy of block 3 is,

$K.E_3=\frac{1}{2}{m}_3{v}_3^2$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} K.E_3=\frac{1}{2}\left(0.25m_1\right){\left(\left(\frac{2}{1.5}\right)\left(\frac{2}{1.5}\right)v_{1i}\right)}^2 \\ =\frac{1}{2}\left(0.79\right)m_1{v}_{1i}^2 \\ =0.79K.E_{1i} \end{gathered}$$

Therefore, the kinetic energy of block 3 is less than the kinetic energy of block 1.

The final momentum of block 3 is,

$P_3=m_3{v}_3$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} P_3=\left(0.25\right)\left(\frac{2}{1.5}\right)\left(\frac{2}{1.5}\right)v_{1i} \\ =0.44m_1{v}_{1i} \\ =0.44P_1 \end{gathered}$$

Therefore, the momentum of block 3 is less than the momentum of block 1.