Question

An object of mass $500g$ moving with a speed of $10m{s}^{-1}$collides with another object of mass $250g$ moving with a speed of $2m{s}^{-1}$ in the opposite direction. After collision both the objects moves in the same direction with common velocity. Write the proper order of steps to find out their common velocity after collision.

• A. B D A C E
• B. B D C E A
• C. A D B C E
• D. C D E B A

Answer 1

The correct option is A

B D A C E

Step 1: Write down the given values of $m_1$ and $m_2$ in S.I. system.

$$\begin{gathered} m_1=500g=\frac{1}{2}kg \\ {m}_2=250g=\frac{1}{4}kg \end{gathered}$$
Step 2: Assign proper signs to the initial velocities $u_1$ and $u_2$.

$$\begin{gathered} u_1=10m{s}^{-1} \\ {u}_2=-2m{s}^{-1} \end{gathered}$$

Step 3: Calculating the total momentum of two bodies before collision.

Total momentum before collision is given by,

$$\begin{gathered} m_1{u}_1+m_2{u}_2 \\ \frac{1}{2}\times 10+\frac{1}{4}\times (-2) \\ \frac{9}{2}kgm{s}^{-1} \end{gathered}$$

Step 4: Calculating the total momentum of two bodies after collision.

Total momentum after collision is given by, $(m_1+m_2)v$

Where $v$ is their common velocity.

Putting the values in the above equation, we get

$$\begin{gathered} (m_1+m_2)v \\ \left(\frac{1}{2}+\frac{1}{4}\right)v \\ \frac{3}{4}v \end{gathered}$$

Step 5: Calculating the value of $v$ by using law of conservation of momentum

The law of conservation of momentum states that the total momentum of a system is always conserved for an isolated system.

This means that total momentum before collision is equal to the total momentum after collision

Therefore, $m_1{u}_1+m_2{u}_2=(m_1+m_2)v$

Putting the values, we get

$$\begin{gathered} \frac{9}{2}=\frac{3}{4}v \\ v=\frac{36}{6}m{s}^{-1} \\ v=6m{s}^{-1} \end{gathered}$$.

Therefore, from the above solution it is clear that the sequence of steps will be B D A C and then E.

Hence, option (A) is the correct answer.