Question

Two gliders are moving on a horizontal frictionless air track. Glider I has mass \(m_{1}=176.3 \mathrm{~g}\) and is moving to the right (positive \(x\) -direction) with a speed of \(2.199 \mathrm{~m} / \mathrm{s} .\) Glider 2 is moving to the left (negative \(x\) -direction) with a speed of \(3.301 \mathrm{~m} / \mathrm{s}\). The gliders undergo a totally elastic collision. The velocity of glider 1 after the collision is \(-4.511 \mathrm{~m} / \mathrm{s} .\) What is the mass of glider \(2 ?\)

Short answer

Answer: The mass of glider 2 is approximately 0.2686 kg or 268.6 g.

Step-by-step solution

Write the conservation of linear momentum equation

Conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as: $$ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} $$ where \(m_1\) and \(v_{1i}\) are the mass and initial velocity of glider 1, \(m_2\) and \(v_{2i}\) are the mass and initial velocity of glider 2, and \(v_{1f}\) and \(v_{2f}\) are the final velocities of glider 1 and glider 2, respectively.

Insert the given values into the momentum equation

We are given \(m_1 = 176.3 \mathrm{~g} = 0.1763\, \mathrm{kg}\), \(v_{1i} = 2.199 \mathrm{~m} / \mathrm{s}\), \(v_{2i} = -3.301 \mathrm{~m} / \mathrm{s}\), and \(v_{1f} = -4.511 \mathrm{~m} / \mathrm{s}\). Insert these values into the momentum equation: $$ 0.1763 (2.199) + m_2(-3.301) = 0.1763(-4.511) + m_2v_{2f} $$

Rearrange the equation to solve for \(v_{2f}\)

Rearrange the formula to isolate \(v_{2f}\) on one side of the equation: $$ m_2v_{2f} = 0.1763 (2.199)- m_2(3.301) - 0.1763 (-4.511) $$ Note that \(v_{1i} - v_{1f} = v_{2f} - v_{2i}\) for elastic collisions. We can rewrite the equation like this: $$ v_{2f} = 0.1763\left(\frac{2.199 + 4.511}{m_2} - 3.301\right) + 3.301 $$

Use the conservation of kinetic energy to create an equation for \(m_2\)

In a totally elastic collision, the total kinetic energy is conserved. Therefore: $$ \frac{1}{2} m_1v_{1i}^2 + \frac{1}{2} m_2v_{2i}^2 = \frac{1}{2} m_1v_{1f}^2 + \frac{1}{2} m_2v_{2f}^2$$ We can plug in the values for \(v_{1i}\), \(v_{2i}\), \(v_{1f}\), and \(v_{2f}\), and then solve for \(m_2\): $$ \frac{1}{2} \cdot 0.1763 \cdot (2.199)^2 + \frac{1}{2} m_2 (-3.301)^2 = \frac{1}{2} \cdot 0.1763 \cdot (-4.511)^2 + \frac{1}{2} m_2\left(0.1763\left(\frac{2.199 + 4.511}{m_2} - 3.301\right) + 3.301\right)^2$$

Solve for \(m_2\) in the energy conservation equation

After simplifying the kinetic energy equation, we get: $$ m_2 = \frac{0.1763((-4.511)^2 - (2.199)^2) + 0.5 \cdot 0.1763(2.199 + 4.511)(2.199 - 4.511)(3.301)}{0.5((-3.301)^2 - (2.199 + 4.511)(2.199 - 4.511)(3.301))} $$ Now, by solving the equation, we can find the mass of glider 2: $$ m_2 = 0.2686\, \mathrm{kg}$$ The mass of glider 2 is approximately \(0.2686\, \mathrm{kg}\) or \(268.6\,\mathrm{g}\).