The problem provided involves a collision between two pieces of clay, where:

• $m_1 = 0.3 \, \text{kg}$
• $m_2 = 0.6 \, \text{kg}$
• Both pieces of clay are moving with the same speed $v = 5.7 \, \text{m/s}$.
• The angle between the velocities of the two masses is $\theta = 35^\circ$.
• After the collision, the two pieces of clay stick together and form a single lump with a combined mass of $0.9 \, \text{kg}$.

We are tasked with calculating:

1. The speed of the combined lump of clay after the collision.
2. How much kinetic energy was lost in the collision.

1. Speed of the combined lump of clay after collision

This is a completely inelastic collision, so we can apply the principle of conservation of momentum. Since momentum is a vector quantity, we need to consider the x and y components of momentum separately.

• For $m_1$, moving horizontally along the x-axis: $p_{1x} = m_1 \cdot v_1 = 0.3 \, \text{kg} \times 5.7 \, \text{m/s}$

• For $m_2$, moving at an angle $\theta$ from the x-axis: $p_{2x} = m_2 \cdot v_2 \cdot \cos(\theta) = 0.6 \, \text{kg} \times 5.7 \, \text{m/s} \times \cos(35^\circ)$ $p_{2y} = m_2 \cdot v_2 \cdot \sin(\theta) = 0.6 \, \text{kg} \times 5.7 \, \text{m/s} \times \sin(35^\circ)$

Now, we sum up the momentum components along each axis:

• Total momentum in the x-direction: $p_x = p_{1x} + p_{2x}$
• Total momentum in the y-direction: $p_y = p_{2y}$

The magnitude of the total momentum is: $p_{\text{total}} = \sqrt{p_x^2 + p_y^2}$

After the collision, the combined mass is $m_1 + m_2 = 0.9 \, \text{kg}$, so the speed of the combined lump $v_f$ is: $v_f = \frac{p_{\text{total}}}{m_1 + m_2}$

2. Kinetic Energy Lost

The initial kinetic energy $KE_{\text{initial}}$ is the sum of the kinetic energies of the two masses before the collision: $KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$

The final kinetic energy $KE_{\text{final}}$ after the collision is: $KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v_f^2$

The kinetic energy lost is: $\Delta KE = KE_{\text{initial}} - KE_{\text{final}}$

Let me calculate the numerical answers now.### Final Results:

1. The speed of the combined lump of clay after the collision is approximately $5.47 \, \text{m/s}$.
2. The kinetic energy lost during the collision is approximately $1.18 \, \text{J}$.

Would you like further details or clarifications?

Here are five questions to deepen your understanding:

1. How would the result change if the angle $\theta$ was larger?
2. What if the masses $m_1$ and $m_2$ were equal—how would the velocity after the collision differ?
3. Why does the kinetic energy decrease in a completely inelastic collision?
4. Can momentum be lost in a collision? Why or why not?
5. How would the results change if the collision was elastic instead of inelastic?

Tip: Remember, in an inelastic collision, kinetic energy is not conserved, but momentum is!