Question

how fast must a 2.7 - g ping - pong ball move in order to have the same kinetic energy as a 145 g baseball moving at 32.0 m/s?
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Explanation:

Step1: Write the kinetic energy formula

The kinetic energy formula is \(K = \frac{1}{2}mv^{2}\), where \(m\) is the mass and \(v\) is the velocity.
Let \(m_1 = 145\space g=0.145\space kg\), \(v_1 = 32.0\space m/s\), \(m_2 = 2.7\space g = 0.0027\space kg\), and \(v_2\) be the velocity of the ping - pong ball.
Since \(K_1=K_2\), we have \(\frac{1}{2}m_1v_1^{2}=\frac{1}{2}m_2v_2^{2}\).

Step2: Solve for \(v_2\)

Cancel out \(\frac{1}{2}\) from both sides of the equation \(\frac{1}{2}m_1v_1^{2}=\frac{1}{2}m_2v_2^{2}\), we get \(m_1v_1^{2}=m_2v_2^{2}\).
Then \(v_2^{2}=\frac{m_1v_1^{2}}{m_2}\).
Substitute \(m_1 = 0.145\space kg\), \(v_1 = 32.0\space m/s\), \(m_2 = 0.0027\space kg\) into the formula:
\(v_2^{2}=\frac{0.145\times(32.0)^{2}}{0.0027}\)
First calculate \((32.0)^{2}=1024\), then \(0.145\times1024 = 148.48\)
\(v_2^{2}=\frac{148.48}{0.0027}\approx54992.59\)
Take the square root of both sides: \(v_2=\sqrt{54992.59}\approx234\space m/s\)

Answer:

\(234\space m/s\)