Unit 6: Momentum

A mouse running along the street.

Linear momentum is defined as mass times the velocity of an object. Therefore, if the truck is at rest, its momentum is equal to zero. By contrast, the momentum of the much lighter mouse is greater than zero because it has speed.

Increase the magnitude or duration of the applied force.

Impulse is the result of exerting a force $F$ on an object for a time $t$, and the amount of impulse delivered to the object is given by \begin{equation} \text{Impulse} = F\times t \end{equation} Therefore, you can increase impulse by exerting a greater force on the object or by extending the time period during which the force is applied.

Both.

As discussed in Question 4, the change in momentum (or impulse) is equal to the product of force and time. Therefore, to impart the greatest momentum, you should exert the strongest force possible for the longest period of time.

No.

Only external forces can produce changes in momentum. When you sit inside an automobile, you are part of the automobile–passenger system. Therefore, if you apply a force on the dashboard in the forward direction, your body simultaneously applies an equal force on the seat in the backward direction. Such internal forces result in a zero net force on the automobile, and, consequently, no impulse is produced.

The speeds are equal.

When two objects of the same mass collide elastically, they exchange speeds. In this case, car B was initially at rest. So, after the collision, car A comes to a stop and transfers all its speed to car B, as shown in Video Q6.20. Note that the total momentum of the two-car system is conserved during the collision.

Their speed after the collision is one-half of car A’s initial speed.

If car A has a mass $m$ and travels at a speed $v$ before the collision, the initial momentum of the two-car system is expressed as \begin{align} \text{Net momentum before collision} &= m v + 0 \nonumber\\[6pt] &= m v \end{align} After the collision, when the two cars travel at a common speed $v^\prime$, the momentum of the system is given by \begin{align} \text{Net momentum after collision} &= m v^\prime + m v^\prime \nonumber\\[6pt] &= 2m v^\prime \end{align} Since no external forces are involved during the collision, the net momentum is conserved and you can write \begin{align} m v &= 2 m v^\prime \nonumber\\[6pt] \Rightarrow\quad v^\prime &= \tfrac{1}{2} v \end{align} So, after the collision, the two cars move at half the initial speed of car A (see Video Q6.21).

During the crash, the dummy’s momentum reduces from \begin{align} \text{Momentum before crash} &= (75\,\text{kg})\, (25\,\text{m/s}) \nonumber\\[6pt] &= 1875\,\text{kg}\cdot\text{m/s} \end{align} to \begin{equation} \text{Momentum after crash} = 0 \end{equation} The change in the dummy’s momentum is equal to the impulse delivered by the backward force ($F$) for the duration of the crash ($t = 0.1\,\text{s}$), where you have \begin{align} \text{Impluse} &= F t \nonumber\\[6pt] \Rightarrow\quad 1875\,\text{kg}\cdot\text{m/s} &= F\, (0.1\,\text{s}) \end{align} So the average force the seat belt exerts on the dummy is calculated as \begin{align} F &= \frac{1875\,\text{kg}\cdot\text{m/s}}{0.1\,\text{s}} \nonumber\\[6pt] &= 18{,}750\,\text{N} \end{align}

Since Jean and her dog move as one body after the catch, the event is considered an inelastic collision and the horizontal momentum of the girl–dog system is conserved. So the total (or net) momentum of Jean and her dog before the catch is equal to their joint momentum afterward. Note that Jean is at rest before she catches the dog. Mathematically, the conservation of momentum of the system is expressed as follows: \begin{align} \text{Momentum before the catch} &= \text{momentum after the catch} \nonumber\\[6pt] m_\text{dog}\, v &= (m_\text{dog} + m_\text{Jean})\, v^\prime \\[6pt] \end{align} By substituting for the masses and for the dog’s initial speed ($v = 3.0\,\text{m/s}$), you calculate the final speed ($v^\prime$) of Jean and her dog: \begin{align} v^\prime &= \left(\frac{m_\text{dog}}{m_\text{dog} + m_\text{Jean}}\right) v \nonumber\\[6pt] &= \left(\frac{15\,\text{kg}}{15\,\text{kg} + 40\,\text{kg}}\right) (3.0\,\text{m/s}) \nonumber\\[6pt] &= 0.82\,\text{m/s} \end{align}

The jumper would experience a strong force that could cause serious injuries.

A bungee jumper gains considerable momentum during a free fall. After the jumper falls a distance equal to the cord’s length, the tension in the cord exerts an upward force on the jumper, who consequently slows down and comes to a stop over a time period $t$. The impulse delivered to the jumper while decelerating must be equal to the momentum gained while falling. If an average force ($F$) acts on the jumper in the upward direction, you can write \begin{align} \text{Impulse} &= F t \nonumber\\[6pt] \Rightarrow\quad F &= \frac{\text{impulse}}{t} \end{align} For the jump to be as safe and comfortable as possible, the tension ($F$) in the cord should be reasonably small, which is achieved by increasing the deceleration time $t$. Because a steel cable is almost nonstretchable, the jumper would come to an almost immediate stop, which would result in a very strong and harmful force on the jumper.

To allow for a longer impact time and reduce forces on passengers during accidents.

A speeding car has a relatively large momentum and requires an equal amount of impulse to bring it to a stop. In an accident, the car would stop almost immediately if its body were built on a rigid frame, thus putting the passengers inside under a harmful and destructive amount of force. A car with a frame designed to crumple, however, takes a relatively longer period to stop as it crumples, thus reducing the impact forces on passengers.

Its momentum is the same.

Momentum is defined as the product of an object’s mass and velocity. Because mass is an intrinsic property of an object, the vehicle has the same mass on the Moon as it has on Earth.

The swarm includes a large number of random momenta that cancel out one another.

Momentum is a vector quantity that has magnitude and direction. When two objects of equal mass and speed move in opposite directions, their net momentum is equal to zero. Therefore, it is possible to have a large number of bees flying in random directions with random speeds such that the vector sum of the momenta of the individual bees is zero. In this case, the center of the swarm system appears to remain in the same place. Any nonzero net momentum in a specific direction causes the swarm to fly in that direction.

Think about the apple and Earth as two bodies that attract each other in the same system.

The force of gravity in the larger apple–Earth system becomes an internal force. If the falling apple is taken as the system, the force of gravity on the apple becomes an external force that results in a change in the apple’s momentum. If you expand the system to include Earth, the force of gravity becomes an internal force within the apple–Earth system. In this case, the downward momentum of the falling apple is equal to the upward momentum gained by Earth. As a result, the net momentum of the larger apple–Earth system is conserved (i.e., remains unchanged). Note that the acceleration of Earth toward the apple is so tiny that it is unnoticeable.

Yes.

The tennis ball and the bowling ball experience equal (but opposite) action–reaction forces during the common time of contact. As a result, the change in momentum of the tennis ball is equal in magnitude to the change in momentum of the bowling ball. Note, however, that the tennis ball experiences a greater velocity change than the heavier bowling ball.

To the left.

Start by calculating the momentum of each cart as follows: \begin{align} \text{Momentum of cart 1} &= (0.5\,\text{kg})\, (1.0\,\text{m/s}) \nonumber\\[6pt] &= 0.5\,\text{kg}\cdot\text{m/s}\;\;\; \text{to the right} \end{align} \begin{align} \text{Momentum of cart 2} &= (0.8\,\text{kg})\, (1.2\,\text{m/s}) \nonumber\\[6pt] &= 0.96\,\text{kg}\cdot\text{m/s}\;\;\; \text{to the left} \end{align} So the net momentum of the two-cart system is equal to the vector sum of momenta values calculated above. Taking the left direction as positive, you have \begin{align} \text{Net momentum} &= (0.96\,\text{kg}\cdot\text{m/s}) - (0.5\,\text{kg}\cdot\text{m/s}) \nonumber\\[6pt] &= 0.46\,\text{kg}\cdot\text{m/s}\;\;\; \text{to the left}\textrm{.} \end{align} Note that the net momentum of the system remains unchanged after the two carts collide.

The law of conservation of momentum.

While in the air, the hero falls straight down with zero horizontal velocity, while the boat moves with a horizontal velocity $v$ on the water below. Therefore, the net horizontal momentum of the hero–boat system before the hero lands is equal to that of the boat. So you can write

\begin{align} \text{Net} &\text{ horizontal momentum before landing} \nonumber\\[6pt] &= 0 + m_\text{boat}\, v \nonumber\\[6pt] &= m_\text{boat}\, v \label{ch06_p_initial} \end{align}

After the hero lands, the boat and the hero continue to move as one body with a final horizontal velocity $v^\prime$. So the net horizontal momentum of the system after landing is given by

\begin{align} \text{Net} &\text{ horizontal momentum after landing} \nonumber\\[6pt] &= (m_\text{hero} + m_\text{boat})\, v^\prime \label{ch06_p_final} \end{align}

Based on the law of conservation of momentum, Equations 6.19 and 6.20 must be equal, or

\begin{align} m_\text{boat}\, v &= (m_\text{hero} + m_\text{boat})\, v^\prime \nonumber\\[6pt] \Rightarrow\quad v^\prime &= \left(\frac{m_\text{boat}}{m_\text{hero} + m_\text{boat}}\right) v \label{ch06_v_prime} \end{align}

From Equation 6.21, you see that $v^\prime \lt v$. In other words, to conserve the net momentum of the system, the final speed of the boat after the hero lands on it must be less than the initial speed before landing.

Note that in a system of particles, internal impulses generated by action–reaction pairs cancel each other.

According to Newton’s third law, forces always appear in pairs. Therefore, any internal force ($F$) created within a system also generates a reaction force ($-F$) that is equal in magnitude and opposite in direction. Since the two forces act within the system over the same time interval ($t$), they generate impulses that cancel each other ($Ft - Ft = 0$). As a result, the net momentum of the system remains unchanged (i.e., is conserved).

Machine gun bullets that bounce from the plate.

Upon impact with a steel plate, a bullet of mass $m$ and speed $v$ experiences a change in momentum equal to the impulse delivered. Since momentum is a vector quantity, the direction of motion becomes important. Assume the bullet initially (i.e., before impact) moves in the negative $x$ direction (i.e., $ v_\text{before} = - v$). If it squashes and sticks to the plate, the bullet experiences a change in momentum, which you can write as

\begin{align} \text{Change in momentum} &= m\, v_\text{after} - m\, v_\text{before} \nonumber\\[6pt] &= m\, (0) - m\, (-v) \nonumber\\[6pt] &= 0 + mv \nonumber\\[6pt] &= m v \label{ch06_mv} \end{align}

In the case where the bullet bounces back in the positive $x$ direction with a speed $ v_\text{after} = v’$, the change in momentum is given by

\begin{align} \text{Change in momentum} &= m\, v_\text{after} - m\, v_\text{before} \nonumber\\[6pt] &= m\, (v’) - m\, (-v) \nonumber\\[6pt] &= m\, (v + v’) \label{ch06_mvv} \end{align}

From Equations 6.22 and 6.23, you see that the impulse delivered when the bullet bounces back is more than the impulse when the bullet sticks to the plate.