91Ó°ÊÓ

Masses of blocks 1 and 2 arerespectively.

Block 2 is stationary.

Block 1 undergoes elastic collision with block 2.

The plot of x versus t for block 1 is given

Using the law of conservation of momentum and the concept of relativity, we can write an expression for the final velocity of block 1 (v1). Then inserting the given conditions in it, we can find the answers for a) b) and c).

Formula:

According to the conservation of the momentum of a body, ${\mathrm{p}}_{\mathrm{i}}={\mathrm{p}}_{\mathrm{f}}$ (1)

Using equation 9-75 from the book, we can write, the final velocity of block 1 as:

${\mathrm{v}}_{1\mathrm{f}}=\frac{({\mathrm{m}}_1-{\mathrm{m}}_2)}{({\mathrm{m}}_1+{\mathrm{m}}_2)}{\mathrm{v}}_{1\mathrm{i}}+\frac{2{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_{2\mathrm{i}}$

Since mass 2 is at rest,

${\mathrm{v}}_{2\mathrm{i}}=0,$

Hence, we can write, the above equation as follows:

${\mathrm{v}}_{1\mathrm{f}}=\frac{({\mathrm{m}}_1-{\mathrm{m}}_2)}{({\mathrm{m}}_1+{\mathrm{m}}_2)}{\mathrm{v}}_{1\mathrm{i}}.............\left(2\right)$

If,${\mathrm{m}}_1<{\mathrm{m}}_2,({\mathrm{m}}_1-{\mathrm{m}}_2)$will be negative. Hence, the final velocity ${\mathrm{v}}_{1\mathrm{f}}$will be negative.

This implies that after collision block 1 will move in the negative direction.

Therefore, in region C on the graph, the plot will be continued after the collision if${\mathrm{m}}_1<{\mathrm{m}}_2$

If,${\mathrm{m}}_1>{\mathrm{m}}_2,({\mathrm{m}}_1-{\mathrm{m}}_2)$will be positive. Hence, the final velocity ${\mathrm{v}}_{1\mathrm{f}}$will be positive.

This implies that after collision block 1 will move in a positive direction.

Therefore, in region B on the graph, the plot will be continued after the collision if$m_1>m_2$

If,$m_1=m_2,(m_1-m_2)$becomes zero. Hence, the final velocity is ${\mathrm{v}}_{1\mathrm{f}}=0$.

This implies that after collision block 1 will stop at that position.

Therefore, along line 3 on the graph, the plot will be continued after the collision if $m_1=m_2$