Collisions Continued

A jet flies at a speed of \(\text{275}\) \(\text{m·s$^{-1}$}\). The pilot fires a missile forward off a mounting at a speed of \(\text{700}\) \(\text{m·s$^{-1}$}\) relative to the ground. The respective masses of the jet and the missile are \(\text{5 000}\) \(\text{kg}\) and \(\text{50}\) \(\text{kg}\). Treating the system as an isolated system, calculate the new speed of the jet immediately after the missile had been fired.

Step 1: Analyse what you are given and draw a sketch

The question explicitly gives a number of values which we identify and convert into SI units if necessary:

• the mass of the jet (\({m}_{1}\) = \(\text{5 000}\) \(\text{kg}\))
• the mass of the rocket (\({m}_{2}\) = \(\text{50}\) \(\text{kg}\))
• the initial velocity of the jet and rocket (\(\vec{v}_{i1}\) = \(\vec{v}_{i2}\) = \(\text{275}\) \(\text{m·s$^{-1}$}\) to the left)
• the final velocity of the rocket (\(\vec{v}_{f2}\) = \(\text{700}\) \(\text{m·s$^{-1}$}\) to the left)

We need to find the final speed of the jet and we can use momentum conservation because we can treat it as an isolated system. We choose the original direction that the jet was flying in as the positive direction, to the left.

After the missile is launched we need to take both into account:

jet and missile

Step 2: Apply the Law of Conservation of momentum

The jet and missile are connected initially and move at the same velocity. We will therefore combine their masses and change the momentum equation as follows:

\begin{align*} \vec{p}_{i}& = \vec{p}_{f} \\ \left({m}_{1}+{m}_{2}\right)\vec{v}_{i}& = {m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2} \\ \left(\text{5 000}+50\right)\left(275 \right)& = \left(\text{5 000}\right)\left(\vec{v}_{f1}\right)+\left(50\right)\left(700 \right) \\ \text{1 388 750} -\text{35 000} & = \left(\text{5 000}\right)\left(\vec{v}_{f1}\right) \\ \vec{v}_{f1}& = \text{270.75} \text{m·s$^{-1}$}~\text{in the original direction}\end{align*}

Step 3: Quote the final answer

The speed of the jet is the magnitude of the final velocity, \(\text{270.75}\) \(\text{m·s$^{-1}$}\).

A bullet of mass \(\text{50}\) \(\text{g}\) travelling horizontally to the right at \(\text{500}\) \(\text{m·s$^{-1}$}\) strikes a stationary wooden block of mass \(\text{2}\) \(\text{kg}\) resting on a smooth horizontal surface. The bullet goes through the block and comes out on the other side at \(\text{200}\) \(\text{m·s$^{-1}$}\). Calculate the speed of the block after the bullet has come out the other side.

Step 1: Analyse what you are given and draw a sketch

The question explicitly gives a number of values which we identify and convert into SI units if necessary:

• the mass of the bullet (\({m}_{1}\) = \(\text{50}\) \(\text{g}\)=\(\text{0.50}\) \(\text{kg}\))
• the mass of the block (\({m}_{2}\) = \(\text{2.00}\) \(\text{kg}\))
• the initial velocity of the bullet (\(\vec{v}_{i1}\) = \(\text{500}\) \(\text{m·s$^{-1}$}\) to the right)
• the final velocity of the bullet (\(\vec{v}_{f1}\) = \(\text{200}\) \(\text{m·s$^{-1}$}\) to the right)

We need to find the velocity of the wood block. We choose the direction in which the bullet was travelling to be the positive direction, to the right.

Step 2: Apply the Law of Conservation of momentum

\begin{align*} \vec{p}_{i}& = {p}_{f} \\ {m}_{1}\vec{v}_{i1}+{m}_{2}\vec{v}_{i2}& = {m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2} \\ \left(\text{0.05}\right)\left(+500 \right)+\left(2\right)\left(0 \right)& = \left(\text{0.05}\right)\left(+200 \right)+\left(2\right)\left(\vec{v}_{f2}\right) \\ 25+0-10& = 2 \vec{v}_{f2} \\ \vec{v}_{f2}& = \text{7.5} \text{m·s$^{-1}$} \text{in the same direction as the bullet}\end{align*}

Step 3: Apply the Law of Conservation of momentum

The block is travelling at \(\text{7.5}\) \(\text{m·s$^{-1}$}\).