91Ó°ÊÓ

$$\begin{gathered} \mathrm{Masses}\mathrm{of}\mathrm{the}\mathrm{blocks}\mathrm{are},{\mathrm{m}}_1,{\mathrm{m}}_2=2{\mathrm{m}}_1,{\mathrm{m}}_3=2{\mathrm{m}}_2. \\ \mathrm{Speed}\mathrm{of}{\mathrm{m}}_1\mathrm{is},\mathrm{v}=4.00\frac{\mathrm{m}}{\mathrm{s}}. \end{gathered}$$

Using the law of conservation of momentum, we can find 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 formula 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 1 and 2 after elastic collision in 1 dimension can be written as,

$v_{1f}=\frac{m_1-m_2}{m_1-m_2}{v}_{1i}$

Here $v_{1i}$is the initial velocity of block 1.

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

role="math" localid="1661326318150" $$\begin{gathered} v_{1f}=\frac{m_1-2m_2}{m_1-2m_2}{v}_{1i} \\ {v}_{1f}=-\frac{1}{3}{v}_{1i} \\ \mathrm{And} \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{2{\mathrm{m}}_1}{{\mathrm{m}}_1+2{\mathrm{m}}_1}{\mathrm{v}}_{1\mathrm{i}} \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{2}{3}{\mathrm{v}}_{1\mathrm{i}} \end{gathered}$$

(1)

Collision of block 2 with block 3:

The velocity of blocks 2 and 3 after elastic collision in 1 dimension,

$$\begin{gathered} v_{2ff}=\frac{m_2-m_3}{m_2-m_2}{v}_{1i} \\ {v}_{3ff}=\frac{2{\mathrm{m}}_2}{{\mathrm{m}}_2+{\mathrm{m}}_2}{v}_{2f} \end{gathered}$$

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

$$\begin{gathered} v_{2ff}=\frac{m_2-2m_2}{m_2-2m_2}\left(\frac{2}{3}{v}_{1i}\right) \\ =-\frac{2}{9}{v}_{2f} \\ \mathrm{And} \\ {\mathrm{v}}_{3\mathrm{ff}}=\frac{2{\mathrm{m}}_2}{{\mathrm{m}}_2+2{\mathrm{m}}_2}\left(\frac{2}{3}{\mathrm{v}}_{1\mathrm{i}}\right) \\ =\frac{4}{9}{\mathrm{v}}_{1\mathrm{f}} \end{gathered}$$

(3)

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

$$\begin{gathered} V_{3ff}=-\frac{4}{9}\left(4\right) \\ =1.78\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the speed of block 3 is $1.78\frac{\mathrm{m}}{\mathrm{s}}$.

From part a),

From Equations 1 and 3, we can conclude thatthe speed ofblock 3 is less than the speed of block 1.

Thus, the speed of block 3 is less than the speed of block 1.

The kinetic energy of block 3 can be calculated as,

$$\begin{gathered} K.E{.}_2=\frac{1}{2}{m}_3{v}_{3ff}^2 \\ =\frac{1}{2}\left(4m_1\right){\left(\frac{4}{9}{v}_{1i}\right)}^2 \end{gathered}$$

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

$$\begin{gathered} K.E_3=\frac{64}{81}{m}_1{v}_{1i}^2 \\ =\frac{64}{81}K.E_{1i} \end{gathered}$$

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

The final momentum of block 3 is,

$P=m_3{v}_{3ff}$

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

$$\begin{gathered} P=4m_1\frac{4}{9}{v}_{1i} \\ =\frac{16}{9}{m}_1{v}_{1i} \end{gathered}$$

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