Question

Block 1 with mass ${\mathit{m}}_{\mathbf{1}}$ slides along an x-axis across a frictionless floor and then undergoes an elastic collision with a stationary block 2 with massFigure 9-33 shows a plot of position xversus time tof block 1 until the collision occurs at position${\mathit{x}}_{\mathbf{c}}$and time ${\mathit{t}}_{\mathbf{c}}$. In which of the lettered regions on the graph will the plot be continued (after the collision) if (a)${\mathit{m}}_{\mathbf{1}}\mathbf{<}{\mathit{m}}_{\mathbf{2}}$and (b)${\mathit{m}}_{\mathbf{1}}\mathbf{>}{\mathit{m}}_{\mathbf{2}}$(c) Along which of the numbered dashed lines will the plot be continued if${\mathit{m}}_{\mathbf{1}}\mathbf{=}{\mathit{m}}_{\mathbf{2}}$?

Short answer

1. On the graph, in region C, the plot will be continued after the collision if$m_1<m_2$
2. On the graph, in region B, the plot will be continued after the collision if$m_1>m_2$
3. Along the dashed line 3 the plot will be continued after the collision if$m_1=m_2$

Step-by-step solution

The given data

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

Understanding the concept of the conservation of the momentum

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)

a) Calculation of the region after the collision if m1 < m2

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${\mathrm{m}}_1<{\mathrm{m}}_2$

b) Calculation of the region after the collision if m1 > m2

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$m_1>m_2$

c) Calculation of the region after the collision if m1 = m2

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 $m_1=m_2$