We are dealing with a problem involving collisions between spheres, and we are asked to show that after the impacts, the first two spheres will come to rest. The key given condition is that the coefficient of restitution, $e = 1.0$, which means the collisions are perfectly elastic.

Step 1: Apply Conservation of Momentum and Coefficient of Restitution

Let's start by considering the given masses and velocities of the spheres:

• $m_1 = 2 \, \text{kg}$, $u_1 = 12 \, \text{m/s}$ (velocity before collision)
• $m_2 = 6 \, \text{kg}$, $u_2 = 4 \, \text{m/s}$
• $m_3 = 12 \, \text{kg}$, $u_3 = 2 \, \text{m/s}$

We will first consider the collision between the first and second spheres, and then between the second and third spheres.

1.1: Collision between the first and second spheres

For a perfectly elastic collision, the following two conditions must hold:

1. Conservation of linear momentum:
   $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ where $v_1$ and $v_2$ are the velocities of the first and second spheres after the collision, respectively.

2. The coefficient of restitution $e = 1.0$:
   $e = \frac{v_2 - v_1}{u_1 - u_2} = 1.0$ This simplifies to: $v_2 - v_1 = u_1 - u_2$

Now, let's solve these two equations.

From the restitution equation: $v_2 - v_1 = 12 - 4 = 8 \quad \Rightarrow \quad v_2 = v_1 + 8$

Substitute this into the momentum conservation equation: $2 \times 12 + 6 \times 4 = 2 v_1 + 6 (v_1 + 8)$ $24 + 24 = 2 v_1 + 6 v_1 + 48$ $48 = 8 v_1 + 48$ $0 = 8 v_1 \quad \Rightarrow \quad v_1 = 0$ Substituting $v_1 = 0$ into the equation $v_2 = v_1 + 8$: $v_2 = 0 + 8 = 8 \, \text{m/s}$ So, after the first collision, the first sphere comes to rest, and the second sphere moves with a velocity of $8 \, \text{m/s}$.

1.2: Collision between the second and third spheres

Now, we consider the collision between the second sphere (which now has a velocity of $8 \, \text{m/s}$) and the third sphere.

Momentum conservation: $m_2 v_2 + m_3 u_3 = m_2 v_2' + m_3 v_3'$ where $v_2'$ and $v_3'$ are the velocities of the second and third spheres after the collision, respectively.

Restitution equation: $v_3' - v_2' = v_2 - u_3 = 8 - 2 = 6$ This gives: $v_3' = v_2' + 6$

Substitute this into the momentum conservation equation: $6 \times 8 + 12 \times 2 = 6 v_2' + 12 (v_2' + 6)$ $48 + 24 = 6 v_2' + 12 v_2' + 72$ $72 = 18 v_2' + 72$ $0 = 18 v_2' \quad \Rightarrow \quad v_2' = 0$ Substitute $v_2' = 0$ into $v_3' = v_2' + 6$: $v_3' = 0 + 6 = 6 \, \text{m/s}$ Thus, after the second collision, the second sphere comes to rest, and the third sphere moves with a velocity of $6 \, \text{m/s}$.

Conclusion:

• After the first collision, the first sphere comes to rest, and the second sphere moves with a velocity of $8 \, \text{m/s}$.
• After the second collision, the second sphere comes to rest, and the third sphere moves with a velocity of $6 \, \text{m/s}$.

Therefore, after both collisions, the first two spheres are brought to rest, as required.

Would you like further details or have any questions?

Further Exploration:

1. What would happen if the coefficient of restitution was less than 1?
2. How would the results change if the spheres had different masses?
3. What if all three collisions were simultaneous?
4. How does energy conservation apply to these collisions?
5. What are some real-world applications of perfectly elastic collisions?

Tip:

Always check both the momentum conservation and the restitution equation carefully to ensure accuracy in collision problems.