To determine the kinetic energy of the second marble, we can use the relationship between mass and kinetic energy, which is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity.
From the problem, we know that the first marble has a mass of 4.8 grams (which is 0.0048 kg) and a kinetic energy of 0.0035 Joules.
We can find the velocity of the first marble using:
\[ KE = \frac{1}{2}mv^2 \implies v^2 = \frac{2 \cdot KE}{m} \implies v = \sqrt{\frac{2 \cdot KE}{m}} \]
Substituting the known values for the first marble: \[ v_1 = \sqrt{\frac{2 \cdot 0.0035 , \text{J}}{0.0048 , \text{kg}}} \] \[ v_1 = \sqrt{\frac{0.007}{0.0048}} \approx \sqrt{1.4583} \approx 1.21 , \text{m/s} \]
Next, we consider the second marble with a mass of 2.4 grams (which is 0.0024 kg). Assuming it rolls down the same incline and thus has the same velocity when it reaches the end:
We can calculate the kinetic energy of the second marble using its mass and the velocity we found:
\[ KE_2 = \frac{1}{2} m_2 v_1^2 \]
where \( m_2 \) is 0.0024 kg:
\[ KE_2 = \frac{1}{2} \cdot 0.0024 \cdot (1.21^2) \]
Calculating \( v_1^2 \): \[ (1.21)^2 \approx 1.4641 \]
Now plugging in the values: \[ KE_2 \approx \frac{1}{2} \cdot 0.0024 \cdot 1.4641 \approx 0.00176 , \text{J} \]
Considering significant figures and the provided answer choices, the best prediction for the kinetic energy of the second marble at the end of the incline is approximately 0.00175 J.
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