Question

momentum practice 2 - 6

1. an air track glider of mass m is built, consisting of two smaller connected gliders with a small explosive charge located between them. the glider is traveling along a frictionless rail at 2 m/s to the right when the charge is detonated, causing the smaller glider with mass 1/4 m, to move off to the right at 5 m/s. what is the final velocity of the second small glider?
2. a flat piece of metal of uniform density has the shape and dimensions shown here. the center of mass for the piece of metal is located at:
3. an object of mass m = 2.0kg experiences a force in newtons according to the force vs. time graph shown here. for the time interval shown, what is the total change in momentum of the object?

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(p_{initial}=p_{final}\). The initial momentum is \(M\times v_{initial}\), where \(v_{initial} = 2m/s\). The final momentum is the sum of the momenta of the two smaller gliders. Let the mass of the first - smaller glider \(m_1=\frac{1}{4}M\) and its velocity \(v_1 = 5m/s\), and the mass of the second - smaller glider \(m_2=\frac{3}{4}M\) and its velocity \(v_2\) (which we want to find).

So, \(Mv_{initial}=m_1v_1 + m_2v_2\).

Step2: Substitute the values of masses into the momentum - conservation equation

Substitute \(m_1=\frac{1}{4}M\), \(m_2=\frac{3}{4}M\), and \(v_{initial}=2m/s\), \(v_1 = 5m/s\) into the equation \(Mv_{initial}=m_1v_1 + m_2v_2\).

We get \(M\times2=\frac{1}{4}M\times5+\frac{3}{4}M\times v_2\).

Since \(M
eq0\) (because the glider has a non - zero mass), we can divide the entire equation by \(M\). The equation becomes \(2=\frac{5}{4}+\frac{3}{4}v_2\).

Step3: Solve for \(v_2\)

First, subtract \(\frac{5}{4}\) from both sides of the equation:

\(2-\frac{5}{4}=\frac{3}{4}v_2\).

\(\frac{8 - 5}{4}=\frac{3}{4}v_2\), so \(\frac{3}{4}=\frac{3}{4}v_2\).

Then, multiply both sides by \(\frac{4}{3}\) to solve for \(v_2\). We find that \(v_2 = 1m/s\) to the right.

Answer:

The final velocity of the second small glider is \(1m/s\) to the right.