Question

The diagram shows the velocity-time graph for two masses $R$ and $S$ that collided head on elastically. Which of the following statements is true?

• A. $R$ and $S$ moved in the same direction after the collision.
• B. The mass of $R$ was greater than mass of $S$
• C. Both options $\left(a\right)$ and $\left(b\right)$ are correct.
• D. Mass of $S$ is greater than mass of $R$.

Answer 1

The correct option is C Both options $\left(a\right)$ and $\left(b\right)$ are correct.

From the velocity ($v$) vs time ($t$) graph, mass $R$ is having initial velocity $u_1=0.8\text{m/s}$ and mass $S$ is having initial velocity $v_1=0\text{m/s}$ just before collision.

$\Rightarrow$After the collision velocity of mass $R$ becomes $u_2=0.2\text{m/s}$ and velocity of mass $S$ is $v_2=1\text{m/s}$
$\because$From the graph, the velocity of masses just before and after the collision are positive, hence masses are moving in the same direction before and after the collision.

$\Rightarrow$Applying conservation linear momentum on the system of $R\&S$ since net $F_{\text{ext}}=0$ on system.
$m_R{u}_1+m_S{v}_1=m_R{u}_2+m_S{v}_2$
$\Rightarrow m_R\times 0.8+m_S\times 0=m_R\times 0.2+m_S\times 1$
Or, $0.8m_R-0.2m_R=m_S$
Or, $\frac{m_R}{m_S}=\frac{5}{3}$
$\therefore m_R\ge m_S$
Hence, mass of $R$ is greater than $S$