A cannon, mass \(\text{500}\) \(\text{kg}\), fires a shell, mass \(\text{1}\) \(\text{kg}\), horizontally to the right at \(\text{500}\) \(\text{m·s$^{-1}$}\). What is the magnitude and direction of the initial recoil velocity of the canon?

We treat the system as an isolated system and conserve momentum. We choose
to the right to be the positive direction. The initial velocity of the system
is zero.

Momentum conservation means that \(\vec{p}_{Ti} = \vec{p}_{Tf}\):

\begin{align*}
\vec{p}_{Ti} &= \vec{p}_{Tf} \\
0 &= m_{cannon}\vec{v}_{cf} + m_{shell}\vec{v}_{sf} \\
0 &= (500)\vec{v}_{cf} + (1)(+500) \\
-(500)\vec{v}_{cf} &= (1)(+500) \\
\vec{v}_{cf} &= \frac{500}{-500} \\
\vec{v}_{cf} &= -1 \\
\vec{v}_{cf} &= \text{1}\text{ m·s$^{-1}$}~\text{towards the left}
\end{align*}

The canon recoils at \(\text{1}\) \(\text{m·s$^{-1}$}\) towards the left.

A trolley of mass \(\text{1}\) \(\text{kg}\) is moving with a speed of \(\text{3}\) \(\text{m·s$^{-1}$}\). A block of wood, mass \(\text{0,5}\) \(\text{kg}\), is dropped vertically into the trolley. Immediately after the collision, the speed of the trolley and block is \(\text{2}\) \(\text{m·s$^{-1}$}\). By way of calculation, show whether momentum is conserved in the collision.

We calculate the momentum of the system before and after the collision.

Before the collision the velocity of the block is 0. The momentum is:

\(\vec{p}_{i} = m_{1}\vec{v}_1 + m_{2}\vec{v}_2 = 0 + (\text{1})(\text{3}) = \text{3}\text{ kg·m·s$^{-1}$}\)

After the collision the momentum is:

\(\vec{p}_{f} = (m_1 + m_2)\vec{v} = (\text{1} + \text{0,5})(\text{2}) = \text{3}\text{ kg·m·s$^{-1}$}\)

Since the momentum before the collision is the same as the momentum after the collision, momentum is conserved.

A child drops a squash ball of mass \(\text{0,05}\) \(\text{kg}\). The ball strikes the ground with a velocity of \(\text{4}\) \(\text{m·s$^{-1}$}\) and rebounds with a velocity of \(\text{3}\) \(\text{m·s$^{-1}$}\). Considering only the squash ball, does the law of conservation of momentum apply to this situation? Explain.

The principle of conservation of linear momentum states:

"The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside."

This means that in an isolated system the total momentum before a
collision or explosion is equal to the total momentum after the
collision or explosion. Taking downwards as positive.

The momentum before the ball hits the floor is:

\begin{align*}
\vec{p}_{down} & = m\vec{v}_i \\
& = (\text{0,05})(+4) \\
& = (\text{0,2}) \\
& = \text{0,2}\text{ kg·m·s$^{-1}$}~\text{downwards}
\end{align*}

The momentum after the ball hits the floor is:

\begin{align*}
\vec{p}_{up} & = m\vec{v}_f \\
& = (\text{0,05})(-3) \\
& = (\text{0,15}) \\
& = \text{0,15}\text{ kg·m·s$^{-1}$}~\text{upwards}
\end{align*}

Since the momentum before the ball hits the floor is not equal to the
momentum after the ball hits the floor the law of conservation of
momentum does not apply to this situation.

We say that the system is not isolated and that there is a force acting on the ball from outside the system.