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 a and b are correct-D- Mass of S is greater than mass of R-

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Perfectly Elastic Collision

The diagram s...

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?

R

and

S

moved in the same direction after the collision.

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The mass of

R

was greater than mass of

S

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Both options

(a)

and

(b)

are correct.

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Mass of

S

is greater than mass of

R

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Solution

The correct option is C Both options $(a)$ and $(b)$ are correct.
From the velocity ( $v$ ) vs time ( $t$ ) graph, mass $R$ is having initial velocity $u_{1} = 0.8 m/s$ and mass $S$ is having initial velocity $v_{1} = 0 m/s$ just before collision.

$\Rightarrow$ After the collision velocity of mass $R$ becomes $u_{2} = 0.2 m/s$ and velocity of mass $S$ is $v_{2} = 1 m/s$
$∵$ 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 $F_{ext} = 0$ 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, $0.8 m_{R} - 0.2 m_{R} = m_{S}$
Or, $\frac{m_{R}}{m_{S}} = \frac{5}{3}$
$∴ m_{R} \geq m_{S}$
Hence, mass of $R$ is greater than $S$

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