Question

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?

Short answer

The speed of block 3 is$7.11\frac{\mathrm{m}}{\mathrm{s}}$ .

b) Speed of the block 3 is greater than the initial speed of block 1.

c) The kinetic energy of block 3 is less than the initial kinetic energy of block 1.

d) The momentum of block 3 is less than the initial momentum of block 1.

Step-by-step solution

Listing the given quantities

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, $v_{1i}=4.00\frac{m}{\mathrm{s}}$.

Understanding the concept of the law of conservation of momentum

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).

(a) Calculations of the speed of block 3  

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$7.11\frac{\mathrm{m}}{\mathrm{s}}$ .

(b) Comparison of the speed of block 3 with block 1

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.

(c) Calculations of the kinetic energy of block 3

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.

(d) Calculations of momentum of block 3

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.