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.