Dynamics - Impulse and momentum

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Dynamics - Impulse and momentum

Question 1

Find the magnitude of the momentum, $p \ \mathrm{Ns}$, of the following objects. Give your answers to 3 decimal places.

a)

A particle of mass $\var{mass1} \mathrm{g}$ moving at $\var{velocity1} \mathrm{ms^{-1}}$.

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b)

A caravan of mass $\var{tonnes1} \ \mathrm{tonnes}$ moving at $\var{velocity2} \mathrm{ms^{-1}}$.

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Advice

Momentum $p \ \mathrm{Ns} = $ mass $\mathrm{kg} \ \times $ velocity $\mathrm{ms^{-1}}$.

Remember that in the above formula, the mass is measured in $\mathrm{kg}$.

a)

We have

\begin{align}
p & = \frac{\var{mass1}}{1000} \times \var{velocity1}, \\
&= \var{precround(kg1*velocity1,3)} \, \mathrm{Ns}.
\end{align}

The magnitude of the momentum is $\var{precround(kg1*velocity1,3)} \, \mathrm{Ns}.$

b)

We have

\begin{align}
p & = (\var{tonnes1} \times 1000) \times \var{velocity2}, \\
& = \var{precround(tonnes1*1000*velocity2,3)} \, \mathrm{Ns}.
\end{align}

The magnitude of the momentum is $\var{precround(tonnes1*1000*velocity2,3)} \, \mathrm{Ns}.$

Question 2

A body of mass $\var{mass} \mathrm{kg}$ is initially at rest on a smooth horizontal plane. Suppose that a horizontal force of magnitude $\var{force} \mathrm{N}$ acts on the body for $\var{time} \mathrm{s}$.

a)

To 3 significant figures, find the magnitude of the impulse given to the body by the force (in $\mathrm{Ns}$).

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b)

To 3 significant figures, find the final speed of the body in $\mathrm{ms^{-1}}$.

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Advice

a)

The impulse imparted is given by $\text{impulse} = \text{force} \times \text{time}$.

So we have $\var{force} \times \var{time} = \var{impulse} \mathrm{Ns}.$

b)

The Impulse-Momentum Principle states that $\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$.

So we have

\begin{align}
\var{impulse} & = \var{mass}v - \var{mass}u, \\
& = \var{mass}v - 0, \\
v & = \frac{\var{impulse}}{\var{mass}}, \\
& =\var{impulse/mass} \mathrm{ms^{-1}}.
\end{align}

Question 3

Give your answers to the following questions to 3 significant figures.

a)

A football of mass $\var{football_mass} \, \mathrm{kg}$ is at rest before it is kicked.

What is the impulse received by the ball in $\mathrm{Ns}$ if its speed is $\var{football_speed} \, \mathrm{ms^{-1}}$ after it is struck?

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b)

A squash ball of mass $\var{squash_mass} \, \mathrm{kg}$ hits a fixed vertical wall at right angles with speed $\var{squash_speed} \, \mathrm{ms^{-1}}$. The ball rebounds at right angles to the wall.

What is the speed of the ball in $\mathrm{ms^{-1}}$ just after it has hit the wall, given that the magnitude of the impulse exerted by the wall on the ball is $\var{squash_impulse} \, \mathrm{Ns}$?

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Advice

a)

Impulse is change in momentum. The ball begins at rest so initially has a momentum of zero.

Therefore

\begin{align}
I & = \var{football_mass} \times\var{football_speed} - 0, \\
& = \var{siground(football_mass*football_speed,3)} \, \mathrm{Ns}.
\end{align}

The impulse received by the ball is $ \var{siground(football_mass*football_speed,3)} \, \mathrm{Ns}$.

b)

Impulse is change in momentum. Therefore we take the rebound direction as being positive and have that

\begin{align}
I & = mv - mu, \\
\var{squash_impulse} & = \var{squash_mass} v - (\var{squash_mass} \times \var{- squash_speed}), \\
v & = \frac{\var{squash_impulse} + (\var{squash_mass} \times \var{-squash_speed})}{\var{squash_mass}}, \\
& = \var{siground((squash_impulse + squash_mass*-squash_speed)/squash_mass,3)} \, \mathrm{ms^{-1}}.
\end{align}

The speed after the ball rebounds is $\var{siground(squash_second_speed,3)} \, \mathrm{ms^{-1}}$.

Question 4

Two particles $A$ and $B$ are at rest and connected by a light inextensible string which is slack. Particle $A$ is projected directly away from particle $B$ with speed $\var{A_speed} \mathrm{ms^{-1}}$.

When the string goes taut the common speed of the particles is $\var{particle_speed} \mathrm{ms^{-1}}$.

Give your answers to the following questions to 3 significant figures.

a)

If the mass of particle $A$ is $\var{A_mass} \, \mathrm{kg}$, what is the mass of particle $B$ in $\mathrm{kg}$?

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b)

Find the magnitude in $\mathrm{Ns}$ of the impulse transmitted through the string when it goes taut.

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Advice

We can draw a diagram with all the speeds and impulses with arrows.

Here particle $A$ has mass $m_1 = \var{A_mass} \, \mathrm{kg}$ and is projected with a speed $u_1 \, \mathrm{ms^{-1}} = \var{A_speed} \, \mathrm{ms^{-1}}$. Particle $B$ has mass $m_2 \, \mathrm{kg}$ to be determined and is initially at rest: $u_2 = 0 \, \mathrm{ms^{-1}}$. Both particles have common final velocities $v = \var{particle_speed} \, \mathrm{ms^{-1}}$ and exert an impulse $I \, \mathrm{Ns}$ through the string as shown.

a)

To find the mass of particle $B$ we can use the principle of conservation of momentum: the total momentum in a system remains constant.

\begin{align}
m_1 u_1 + m_2 u_2 & = m_1 v_1 + m_2v_2, \\
(\var{A_mass} \times \var{A_speed}) + 0 & = \var{particle_speed} ( \var{A_mass} + m_2 ), \\
\frac{ \var{A_mass} \times \var{A_speed}}{\var{particle_speed}} & = \var{A_mass} + m_2, \\
m_2 & = \frac{ \var{A_mass} \times \var{A_speed}}{\var{particle_speed}} - \var{A_mass}, \\
& = \var{siground( (A_mass*A_speed)/particle_speed - A_mass,3)} \, \mathrm{kg}.
\end{align}

So the mass of particle $B$ is $\var{siground( (A_mass*A_speed)/particle_speed - A_mass,3)} \, \mathrm{kg}$.

b)

To find the impulse we consider one of the particles and apply the Impulse-Momentum Principle. Although it is easier to consider particle $B$ as it is initially at rest we will consider particle $A$ incase the mass of particle $B$ was calculated incorrectly in part a).

We will consider particle $A$ and apply the Impulse-Momentum Principle in the direction of the impulse shown in the diagram, $(\leftarrow)$. This means our velocities will now be negative as they are acting in the opposite direction to the impulse.

\begin{align}
I & = mv - mu_1, \\
& = m (v - u_1), \\
& = \var{A_mass} \times ( - \var{particle_speed} - ( - \var{A_speed})), \\
& = \var{A_mass} \times ( \var{ - particle_speed + A_speed} ), \\
& = \var{siground( A_mass*(A_speed - particle_speed),3)} \, \mathrm{Ns}.
\end{align}

The magnitude of the impulse transmitted through the string is $\var{impulse} \, \mathrm{Ns}$.

Question 5

Two balls $A$ and $B$ of mass $\var{A_mass} \mathrm{kg}$ and $\var{B_mass} \mathrm{kg}$ respectively are moving in the same straight line on a smooth horizontal surface. The balls collide. After the collision both of the balls are moving in the same direction.

After the collision, the $A$ travels at $\var{A_speed} \mathrm{ms^{-1}}$ and $B$ travels at $\var{B_speed} \mathrm{ms^{-1}}$. The magnitude of the impulse of $A$ on $B$ is $\var{impulse} \mathrm{Ns}$.

a)

Find the magnitude of the speed of $A$ before the collision, in $\mathrm{ms^{-1}}$ to 3 s.f.

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b)

Find the magnitude of the speed of $B$ before the collision, in $\mathrm{ms^{-1}}$ to 3 s.f.

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Advice

We can draw a diagram.

The impulse applied to both balls is $I = \var{impulse} \, \mathrm{Ns}$, and the speeds after the collision are $v_1 = \var{A_speed} \, \mathrm{ms^{-1}}$ and $v_2 = \var{B_speed} \, \mathrm{ms^{-1}}$ for the balls $A$ and $B$ respectively.

a)

To calculate $u_1$, the speed of ball $A$ before the collision, we use the equation $I = m_1v_1 - m_1u_1$, where $m_1$ is the mass of ball $A$, which is $\var{A_mass} \, \mathrm{kg}$. We resolve in the direction of the impulse, therefore the signs of $v_1$ and $u_1$ will be reversed.

\begin{align}
I & = m_1v_1 - m_1u_1, \\
\var{impulse} & = \var{A_mass} (v_1 - u_1), \\
\frac{\var{impulse}}{\var{A_mass}} & = - \var{A_speed} - (- u_1), \\
u_1 & = \frac{\var{impulse}}{\var{A_mass}} + \var{A_speed}, \\
& = \var{siground( impulse/A_mass + A_speed,3)} \, \mathrm{ms^{-1}}.
\end{align}

The magnitude of the speed of $A$ before the collision is $\var{siground( impulse/A_mass + A_speed,3)} \, \mathrm{ms^{-1}}$.

b)

To calculate $u_2$, the speed of ball $B$ before the collision we use the equation $I = m_2v_2 - m_2u_2$, where $m_2$ is the mass of ball $B$, which is $\var{B_mass} \, \mathrm{kg}$. We resolve in the direction of the impulse shown in the diagram, which is the same as the direction of the speeds.

\begin{align}
I & = m_2v_2 - m_2u_2, \\
\var{impulse} & = \var{B_mass} ( \var{B_speed} - u_2), \\
\frac{\var{impulse} }{ \var{B_mass}} & = \var{B_speed} - u_2, \\
u_2 & = \var{B_speed} - \frac{\var{impulse} }{ \var{B_mass}}, \\
& = \var{siground( B_speed - (impulse/B_mass),3)} \, \mathrm{ms^{-1}}.
\end{align}

If this is positive it means the direction of ball $B$ we assumed is correct; if it is negative it means the ball was originally travelling in the other direction. However we were asked to find the magnitude of the speed so we take our answer as $\var{B_before_speed} \, \mathrm{ms^{-1}}$.

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Dynamics - Impulse and momentum

Dynamics - Impulse and momentum

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Question 2

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Question 3

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Question 4

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Question 5

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