eed of $3~\mathrm{m/s}$, in the same direction as the first.

After the collision, the first particle moves at $4~\mathrm{m/s}$, still in the same direction.

The type of collision is elastic, and both particles move along a straight line (head-on).

We need to find the new speed of the second particle ($v_2$) after the collision.

Step 2: Use Elastic Collision Formula for Head-On Collisions

In an elastic head-on collision, the difference in the starting speeds equals the difference in the speeds after the collision, but in the opposite order:

$u_1 - u_2 = v_2 - v_1$

Step 3: Put in the Numbers

Plug in the values we have:

$5~\mathrm{m/s} - 3~\mathrm{m/s} = v_2 - 4~\mathrm{m/s}$

$2~\mathrm{m/s} = v_2 - 4~\mathrm{m/s}$

Step 4: Solve for $v_2$

Add $4~\mathrm{m/s}$ to both sides to find $v_2$:

$v_2 = 2~\mathrm{m/s} + 4~\mathrm{m/s}$

$v_2 = 6~\mathrm{m/s}$

Step 5: Check the Direction

Because the answer is positive, this means the second particle moves in the same direction as before, just faster.

Step 6: Final Answer

The second particle's speed after the collision is $6~\mathrm{m/s}$, in the same direction as before.

The matching answer choice is:

Option A: $6~\mathrm{m/s}$ in the same direction.

So, the correct answer is $\boxed{6~\mathrm{m/s}~\text{in the same direction.}}$

Center Of Mass

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