Question

first, gavin shoots a 3.0 - gram pea at the model. its speed is measured at 12 meters per second as it leaves the straw. the pea does not break through the paper towel.
for gavins next trial, he decides to use a different pea and shoot it at a different speed. which of the peas illustrated below is most likely to break through the paper towel?

Explanation:

Step1: Recall kinetic - energy formula

The kinetic - energy formula is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the speed. We need to calculate the kinetic energy for each pea to determine which one has the most energy and is most likely to break through the paper towel.

Step2: Calculate kinetic energy for the first pea

For the first pea with $m_1=1.5\ g = 0.0015\ kg$ and $v_1 = 10\ m/s$, $K_1=\frac{1}{2}m_1v_1^{2}=\frac{1}{2}\times0.0015\times(10)^{2}= 0.075\ J$.

Step3: Calculate kinetic energy for the second pea

For the second pea with $m_2 = 4.0\ g=0.004\ kg$ and $v_2 = 0\ m/s$, $K_2=\frac{1}{2}m_2v_2^{2}=\frac{1}{2}\times0.004\times(0)^{2}=0\ J$.

Step4: Calculate kinetic energy for the third pea

For the third pea with $m_3 = 3.2\ g = 0.0032\ kg$ and $v_3=14\ m/s$, $K_3=\frac{1}{2}m_3v_3^{2}=\frac{1}{2}\times0.0032\times(14)^{2}=\frac{1}{2}\times0.0032\times196 = 0.3136\ J$.

Step5: Calculate kinetic energy for the fourth pea

For the fourth pea with $m_4 = 2.8\ g=0.0028\ kg$ and $v_4 = 12\ m/s$, $K_4=\frac{1}{2}m_4v_4^{2}=\frac{1}{2}\times0.0028\times(12)^{2}=\frac{1}{2}\times0.0028\times144 = 0.2016\ J$.

Answer:

The pea with a mass of $3.2$ grams and a speed of $14$ m/s is most likely to break through the paper towel.