Question

a block slides along a frictionless horizontal surface and collides elastically with a second block that is stationary. the mass of the stationary block is 1.72 times the mass of the sliding block. if the sliding blocks initial velocity is 3.8 m/s (in the +x - direction), find its final velocity in meters per second. express it as a negative number if it bounces backward. provide at least two decimal places

Explanation:

Step1: Set up the equations for elastic collision

For an elastic collision, momentum conservation \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\) and kinetic - energy conservation \(\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2\) apply. Given \(v_{2i} = 0\) (stationary block), \(m_2=1.72m_1\), and \(v_{1i}=3.8\ m/s\).
From momentum conservation: \(m_1v_{1i}=m_1v_{1f}+m_2v_{2f}\), substitute \(m_2 = 1.72m_1\) gives \(v_{1i}=v_{1f}+1.72v_{2f}\), so \(v_{2f}=\frac{v_{1i}-v_{1f}}{1.72}\).
From kinetic - energy conservation: \(m_1v_{1i}^2=m_1v_{1f}^2+m_2v_{2f}^2\), substitute \(m_2 = 1.72m_1\) gives \(v_{1i}^2=v_{1f}^2 + 1.72v_{2f}^2\).

Step2: Substitute \(v_{2f}\) into the kinetic - energy equation

Substitute \(v_{2f}=\frac{v_{1i}-v_{1f}}{1.72}\) into \(v_{1i}^2=v_{1f}^2 + 1.72v_{2f}^2\):

$$
LATEXBLOCK0
$$

Using the quadratic formula \(ax^{2}+bx + c = 0\) (\(a = 2.72\), \(b=-2v_{1i}\), \(c=-0.72v_{1i}^2\)), \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here \(v_{1f}\) is the variable.

$$
LATEXBLOCK1
$$

We take the physical solution. Since the first block can't pass through the second block (because the second block was initially at rest), we use \(v_{1f}=\frac{2v_{1i}-3.44v_{1i}}{5.44}\) (the other root \(v_{1f}=\frac{2v_{1i}+3.44v_{1i}}{5.44}\) would mean the first block continues in the same direction which is unphysical for \(m_2>m_1\) in an elastic collision when \(v_{2i} = 0\)).
Substitute \(v_{1i}=3.8\ m/s\):

$$
LATEXBLOCK2
$$

Answer:

\(-1.01\)