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Using Equations as a Recipe for Algebraic Problem-Solving

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Consider the following problem:

A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity ( v ) across the ice.

If it can be assumed that the effect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system . Momentum should be conserved and the post-collision velocity ( v ) can be determined using a momentum table as shown below.

Observe in the table above that the known information about the mass and velocity of the two objects was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball. To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:

Using algebra skills, it can be shown that v = 4 km/hr. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision. (NOTE: The unit km/hr is the unit on the answer since the original velocity as stated in the question had units of km/hr.)  

Now consider a similar problem involving momentum conservation.

A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.

Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity ( v ).

The collision between the ball and the catcher's mitt occurs in an isolated system , total system momentum is conserved . Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt). The table below depicts this principle of momentum conservation.

Observe in the table above that the known information about the mass and velocity of baseball and the catcher's mitt was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expression 0.15 • v and 0.25 • v are used for the after-collision momentum of the baseball and catcher's mitt. To determine v (the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results:

Using algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.

The two collisions above are examples of inelastic collisions. Technically, an inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. In an inelastic collision, the kinetic energy of the colliding objects is transformed into other non-mechanical forms of energy such as heat energy and sound energy. The subject of energy will be treated in a later unit of The Physics Classroom . To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision.  

Now we will consider the analysis of a collision in which the two objects do not stick together . In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together .

A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).

The collision can be analyzed using a momentum table similar to the above situations.

Observe in the table above that the known information about the mass and velocity of the truck and car was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. The after-collision velocity of the car is used (in conjunction with its mass) to determine its momentum after the collision. Finally, the expression 3000•v was used for the after-collision momentum of the truck ( v is the velocity of the truck after the collision). To determine v (the velocity of the truck), the sum of the individual after-collision momentum of the two objects is set equal to the total momentum. The following equation results:

Using algebra skills, it can be shown that v = 5.0 m/s. The truck's velocity immediately after the collision is 5.0 m/s. As predicted, the truck has lost momentum (slowed down) and the car has gained momentum.

The next section of this lesson involves examples of problems that provide a real test of your conceptual understanding of momentum conservation in collisions. Before proceeding with the practice problems , be sure to try a few of the more conceptual questions that follow .

We Would Like to Suggest ...

• Momentum Conservation Times Two

Impulse and Momentum

Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach. Timothy Treadwell, 2001

• Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
• Compute the momentum of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell.
• How fast would a 250 lb man have to run to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
• How fast would a 4000 lb car have to drive to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)

For the first month of its journey, the Mars Orbiter Spacecraft actually orbited the Earth. After a series of orbit raising maneuvers, the tiny main engine gave the spacecraft enough speed in the right direction to escape. For the next nine months the main engine was only used twice and only very briefly (less than a minute of total burn time) to correct the spacecraft's trajectory.

Nine months after leaving Earth orbit, the spacecraft arrived at Mars. To enter orbit around Mars, the spacecraft needed to slow down — a maneuver called orbit insertion. Since the main engine hadn't been used much in nine months, a quick little test was needed.

Press Release: September 22, 2014 Mars Orbiter Spacecraft's Main Liquid Engine Successfully Test Fired The 440 newton Liquid Apogee Motor (LAM) of India's Mars Orbiter Spacecraft, last fired on December 01, 2013, was successfully fired for a duration of 3.968 seconds at 1430 hrs IST today (September 22, 2014). This operation of the spacecraft's main liquid engine was also used for the spacecraft's trajectory correction and changed its velocity by 2.18 metre/second. With this successful test firing, Mars Orbiter Insertion (MOI) operation of the spacecraft is scheduled to be performed on the morning of September 24, 2014 at 07:17:32 hrs IST by firing the LAM along with eight smaller liquid engines for a duration of about 24 minutes. ISRO, 2014

• Using the press release, determine the mass of the Mars Orbiter Spacecraft at the time of this test firing.
• Why was the qualifying phrase "at the time of this test firing" added to the end of the previous question? Why does the mass of the spacecraft vary with time?

After all, the faster we go, the more difficult it is to avoid collisions with small objects and the more damage such a collision will wreak. Even if we are fortunate enough to miss all sizable objects, we can scarcely expect to miss the dust and individual atoms that are scattered throughout space. At two-tenths of the speed of light, dust and atoms might not do significant damage even in a voyage of 40 years, but the faster you go, the worse it is — space begins to become abrasive. When you begin to approach the speed of light, each hydrogen atom becomes a cosmic ray particle and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly the speed of light is a cosmic ray particle, and there is no difference if the ship strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza: "Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for the pitcher.") So 60,000 kilometers per second may be the practical speed limit for space travel. Isaac Asimov, 1987

The density of the interstellar medium is about one hydrogen atom per cubic centimeter. Imagine a 1,000 tonne, 4 by 6 meter, classroom-sized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri (the nearest solar system to our own) 4.243 light years away.

Kinematics:

• How long would it take our hypothetical spacecraft to complete its hypothetical journey?

Impulse-Momentum:

• Determine the momentum of our spacecraft.
• What mass of interstellar medium is swept up during the journey?
• What impulse does the interstellar medium deliver to the spacecraft?
• How does this impulse compare to the momentum of the spacecraft?

Work-Energy:

• Determine the kinetic energy of our spacecraft.
• What is the effective drag force of the interstellar medium during the journey?
• How much work does the interstellar medium do on the spacecraft?
• How does this work compare to the kinetic energy of the spacecraft?
• Write something.
• In older passenger cars, body panels were attached to a single frame around the perimeter, making them very rigid. This is known as body-over-frame construction. In newer cars, different body parts have stress-bearing elements within them and these parts are then welded to each other. This is known as unitized body construction. Repairing "unibody" cars after collision is comparatively difficult as stress (and thus damage) are distributed throughout the different parts. Why then are cars now built this way
• To escape from a horrible fire, two people are forced to jump from the third story of a burning building on to solid concrete. Which person is more likely to sustain serious injuries: the jumper who comes to an abrupt halt when he lands or the jumper who bounces after impact?
• What is the basic idea behind crash safety features in cars like seatbelts, airbags, crumple zones, etc. What quantities in the impulse-momentum theorem ( F ∆ t  =  m ∆ v ) change as a result of these features, how are they changed, and how does this result in increased safety during a crash?
• The phrase "roll with the punches" has its origin in boxing. What does it mean to roll with a punch (or ride a punch)? How does rolling reduce the severity of a punch?
• Is it possible for a motorcycle to have more momentum (that is, a momentum with a larger magnitude) than a train?
• When hit, the velocity of a 0.145 kg baseball changes from +20 m/s to −20 m/s. What is the magnitude of the impulse delivered by the bat to the ball?
• the ground is soft and the ball stops dead
• the ground is hard and the ball bounces straight up at 2.0 m/s
• Draw a free body diagram showing all the forces acting on the model rocket.
• the weight of the rocket
• the net force on the rocket while the engine was running
• the net impulse on the rocket while the engine was running
• the speed of the rocket when the engine stopped
• the height of the rocket above the ground when the engine stopped
• What is the acceleration of the rocket after the engine shut down?
• What maximum height above the ground did the rocket reach?
• What does the area under this curve represent?
• Calculate its cumulative value at 2 s intervals. Compile your results in a table like the one below.

• Sketch a graph of this quantity with respect to time.

statistical

• Use the given data to create a force-time graph.
• Determine the impulse acting on the projectile as a function of time.
• Compute the launch speed of the projectile.

Data adapted from Kampen, Kaczmarczik, and Rath; 2006 .

investigative

• Which plate have you chosen to work with?
• What is the speed of this plate ?
• What is the area of this plate?
• Is the plate you've chosen continental or oceanic?
• What is a typical density of this kind of crust?
• What is a typical thickness of this type of crust?
• Compute the momentum of the tectonic plate you've chosen from the data you've found. State your answer to the nearest order of magnitude (the nearest power of ten). Don't forget the unit.
• the thrust-time data used to generate the graph on the data sheet
• the propellant mass
• the mass after firing (i.e., the mass of the empty rocket)
• impulse provided by the motor
• fractional propellant mass loss (You need to determine the initial and final mass of the rocket. Assume that the loss of mass is directly proportional to the cumulative impulse. When the impulse is zero at the beginning, the mass loss is zero. When the impulse reaches its final value, the mass loss is 100%.)
• mass of the rocket
• speed of the rocket (Don't forget to include the force of gravity in your calculations.)
• acceleration of the rocket
• altitude of the rocket

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Impulse and Momentum – Physics Example Problem

Impulse and momentum are physical concepts that are easily seen from Newton’s Laws of Motion.

Start with this equation of motion for constant acceleration.

v = v 0 + at

where v = velocity v 0 = initial velocity a = acceleration t = time

If you rearrange the equation, you get

v – v 0 = at

Newton’s second law deals with force.

where F = force m = mass a = acceleration

solve this for a and get

Stick this into the velocity equation and get

v – v 0 = (F/m)t

Multiply both sides by m

mv – mv 0 = Ft

The left side of the equation deals with momentum (often denoted by a lower-case p) and the right side is impulse (often denoted by an upper-case letter J).

Mass times velocity is known as momentum and force applied over time is called impulse.

Impulse and Momentum Example Problem

Question: A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) What is the initial momentum of the mass? b) What is the final momentum of the mass? c) What was the force acting on the mass? d) What was the impulse acting on the mass?

Part a) What is the initial momentum?

Momentum is mass times velocity. Since the mass is at rest, the initial velocity is 0 m/s.

momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s

Part b) What is the final momentum?

After the force is finished acting on the mass, the velocity is 3 m/s.

momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s

Part c) What was the force acting on the mass?

From parts a and b, we know mv 0 = 0 kg⋅m/s and mv = 150 kg⋅m/s.

150 kg⋅m/s – 0 kg⋅m/s = Ft 150 kg⋅m/s = Ft

Since the force was in effect over 2 seconds, t = 2 s.

150 kg⋅m/s = F ⋅ 2s F = (15 kg⋅m/s) / 2 s F = 75 kg⋅m/s 2

Unit Fact: kg⋅m/s 2 can be denoted by the derived SI unit Newton (symbol N )

Part d) What was the impulse acting on the mass?

The impulse is the force multiplied by the time passed. It is also equal to the change in momentum over the same time period.

Ft = 75 N ⋅ 2 s Ft = 150 Ns or 150 kg⋅m/s

The impulse was 150 kg⋅m/s.

These problems are relatively simple as long as you keep your units straight. Impulse and momentum should have the same units: mass⋅velocity or force⋅time. Check your units when you check your answer.

Another possible way to cause errors is to confuse your vector directions. Velocity and Force are both vector quantities. In this example, the mass was pushed in the direction of the final velocity. If another force pushed in the opposite direction to slow down the mass, the force would have a negative value compared to the velocity vector.

If you found this helpful, check out other Physics Example Problems .

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Momentum - Problem Solving 1D

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Section Heading

Linear momentum is an important classical concept for describing the (change in) translational motion of an object. It gets defined by \( \overrightarrow{p} = m\cdot \overrightarrow{v} \) .

Linear momentum can be used to rephrase Newton's second law. If we assume mass is constant, we get \[ \overrightarrow{F} = \dfrac{d \overrightarrow{p}}{dt} = m \cdot \overrightarrow{a} .\] When mass is not constant, we can use the first equation to describe for example, rocket motion.

Although velocity and mass themselves can be used to describe the motion of an object and its changes, linear momentum gives us a different approach, which seems to come in very handy when dealing with collisions. It is so helpful that it survived the quantum revolution, where the concept of force was given up. %

%Want to delete the section between percentages signs%

Conservation of momentum is seen by physicists as fundamental.

It is used in solving a variety of problems and especially in collisions. We can write it down in an equation \( \overrightarrow{p_{i}} = \overrightarrow{p_{f}} \) .

A baseball player with a mass of \(80.0\text{ kg}\) throws a \(145\text{ g}\) baseball with a speed of \(160 \text{ km/h}\) towards the batter. What is the recoil speed of the batter? (Not yet finished.) Before the throw, neither the batter nor the baseball has speed. Therefore, \( \overrightarrow{p_{i}} = \overrightarrow{0} \). After the throw, with index 1 for the player and 2 for the ball, we get \( \overrightarrow{p_{f}} = m_{1}\cdot \overrightarrow{v_{1}} + m_{2}\cdot \overrightarrow{v_{2}} \). Using a one dimensional line, pointing in the direction of batter, and using conservation of momentum, we get \[ m_{1}\cdot v_{1} + m_{2}\cdot v_{2} = 0. \ _\square \]

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Momentum and Impulse Example Problems and Solutions AP Physics

Some impulse and momentum practice problems in one dimension are presented and solved. All these problems are helpful for your homework and/or the AP Physics 1 test. 

For better practice, all subjects are broken down into sections.

Linear momentum and its conservation

Problem (1): What is the momentum of a $0.057-\rm kg$ small object moving with a constant speed of $30\,\rm m/s$?

Solution : Linear momentum in physics is defined as the product of mass times the velocity, i.e., $p=mv$. Thus, we have \[p=0.057\times 30=1.71\,\rm \frac{kg\cdot m}{s}\] 

Problem (2): Suppose you are kicking a $410-\rm g$ soccer ball against a wall. The ball, traveling at $25\,\rm m/s$ strikes the wall and rebounds at the same speed.  (a) Find the direction and magnitude of the ball's momentum before and after striking. (b) What is the change in momentum of the ball?

Solution : According to the definition of momentum, it is the product of mass and velocity. Hence, momentum is a definition of a vector quantity in physics . 

(a) To find the direction of the momentum of a moving object, first, choose a reference coordinate. In all momentum and impulse problems, it is better and simpler to take the positive $x$ axis as the reference. 

Here, we assume that the ball initially is kicking toward the positive $x$ axis, so its original momentum just before striking the wall is \begin{align*} \vec{p}_b&=m\vec{v}\\ &=0.410 \times 25\,(\hat{i}) \\ &=10.25\,(\hat{i}) \quad {\rm kg\cdot m/s}\end{align*} where $b$ stands for '' before''. The ball, after rebounding, moves in the negative $x$ direction. So, the ball's momentum just after rebounding is \begin{align*} \vec{p}_a &=m\vec{v}\\ &=0.410 \times 25\,(-\hat{i}) \\ &=10.25\,(-\hat{i}) \quad {\rm kg\cdot m/s}\end{align*}  (b) The change in momentum is the difference between after and before momenta, i.e., $\Delta \vec{p}=\vec{p}_a-\vec{p}_b$. Thus, \begin{align*} \Delta\vec{p}&=(-10.25\,\hat{i})-(10.25\,\hat{i}) \\ &=20.5\,(-\hat{i}) \quad {\rm kg\cdot m/s}\end{align*} where $-\hat{i}$ indicates the direction of the vector of momentum change, which is to the left. 

Problem (3): A $0.025-\rm kg$ tennis ball traveling at a speed of $34\,\rm m/s$ hits a wall at a $45^\circ$ angle, as shown in the figure below.  (a) What is the direction and magnitude of the change in the ball's momentum? (b) Find the direction of the force exerted on the wall by the ball.

Solution : As in all momentum and impulse problems, take the positive $x$ axis as the reference of the coordinate system. Momentum is a vector quantity with components. In this problem, the ball is hit at an angle to the horizontal. 

As shown in the figure below, resolve the velocity vector into its components: \begin{align*} \vec{v}_i&=v\cos\theta\,\hat{i}+v\sin\theta\,(-\hat{j}) \\ &=34\,\left(\cos 45^\circ\,\hat{i}+\sin 45^\circ\,(-\hat{j})\right) \\&=17\sqrt{2}\,(\hat{i}-\hat{j}) \end{align*} where we used the fact that the sine and cosine of $45^\circ$ are $\frac{\sqrt{2}}{2}$. After rebounding from the wall, the ball moves at the same angle as before. Now, the velocity just after hitting the wall is found as below \begin{align*} \vec{v}_f&=v\cos\theta\,(-\hat{i})+v\sin\theta\,(-\hat{j}) \\ &=34\,\left(\cos 45^\circ\,(-\hat{i})+\sin 45^\circ\,(-\hat{j})\right) \\&=17\sqrt{2}\,(-\hat{i}-\hat{j}) \\&=-17\sqrt{2}\,(\hat{i}+\hat{j}) \end{align*} Its corresponding momentum is \[\vec{p}_f=m\vec{v}_f\] The subtraction of final momentum from initial momentum is defined as the change in momentum of a moving object. Thus, we will have \begin{align*} \Delta \vec{p}&=\vec{p}_f-\vec{p}_i \\\\ &=m(\vec{v}_f-\vec{v}_i) \\\\ &=0.025(-34\sqrt{2}\,\hat{i}) \\\\ &=0.85\sqrt{2}\,(-\hat{i}) \end{align*}

(b) According to Newton's second law, the net force applied to an object is equal to the rate of change of its momentum \[\vec{F}=\frac{\Delta\vec{p}}{\Delta t}\] This formula tells us that the direction of the applied force is in the same direction as the change in momentum. In this case, the change in momentum is toward the negative $x$ axis, so the force exerted on the ball due to the wall is also in the same direction.

Problem (4): The velocity of a $2-\rm kg$ object is given as $(3\,\hat{i}-4\,\hat{j}) \,\rm m/s$. Find: (a) The $x$ and $y$ components of its momentum. (b) The magnitude of the direction of its momentum.

Solution : Momentum is defined as the scalar multiplication of mass and velocity vector, i.e., $\vec{p}=m\vec{v}$. As you know, momentum is a vector quantity with some components. This problem is nothing but a vector problem in physics.

(a) The components of momentum are: \begin{align*} p_x&=mv_x \\ &=2(3\,\hat{i}) \\&=6\,\hat{i}\quad \rm kg\cdot m/s \\\\ p_y&=mv_y \\ &=2(-4\,\hat{j}) \\&=-8\,\hat{j}\quad \rm kg\cdot m/s \end{align*} Note that $\hat{i}$ and $\hat{j}$ are the unit vectors along the $x$ and $y$ directions. Recall that unit vectors are defined as ones with a magnitude of unity. 

(b) The magnitude of a given vector with its components is found by following the Pythagorean equation \begin{align*} p&=\sqrt{p_x^2+p_y^2} \\\\&=\sqrt{6^2+8^2}\\\\&=10\quad \rm kg\cdot m/s\end{align*} The direction of a vector is also found as below \begin{align*} \theta &=\tan^{-1}\left(\frac{p_y}{p_x}\right) \\\\ &=\tan^{-1}\left(\frac{-8}{6}\right) \\\\ &=-53^\circ \end{align*} The negative indicates that the angle is below the horizontal.

Problem (5): A $8000-\rm kg$ car moving at a constant speed of $14\,\rm m/s$ strikes a barrier. The two stick together and move for a while at a speed of $5\,\rm m/s$. What is the mass of the barrier? 

Solution : According to the law of conservation of linear momentum, the sum of momenta before any collision equals the sum of momenta after the collision, provided that there is no external force involved in the collision. In this case, the sum of momenta before the collision is \begin{align*} \vec{p}_i &=m_1 \vec{v}_1+m_2\vec{v}_2\\ &=(8000)(14)+m_2(0) \\ &=114000 \quad \rm kg\cdot m/s \end{align*} where we set the velocity of the barrier zero. After the collision, the two stick together and have a common velocity, say $V_f$, so \[ p_f=(m_1+m_2)V_f\] Equating these two momenta and solving for the unknown mass $m_2$, we will have \begin{gather*} p_i=p_f \\ 114000=(8000+m_2)(5) \\ \Rightarrow \quad \boxed{m_2=14800\,\rm kg} \end{gather*} To have a better understanding of this fundamental law in physics, refer to the solved problems of conservation of momentum .

Problem (6): A car of mass $m$ traveling at a speed of $v$ to the right collides with an identical car moving right at half the speed. As a result, the two stick together. Find their combined speed after the collision. 

Solution : Again, there is a collision, and we should apply the principle of conservation of momentum. Before the collision, the sum of momenta is \begin{align*} \vec{p}_i&=m_1 \vec{v}_1+m_2\vec{v}_2 \\\\ &=m(v+\frac v2) \end{align*} Similarly, after the collision, we have \[ \vec{p}_f=(m_1+m_2)\vec{V}_f \] Applying the principle of the conservation of linear momentum, and solving for the common velocity $V_f$, we get \begin{gather*} \vec{p}_i=\vec{p}_f \\\\ \frac 32 mv= (2m)V_f \\\\ \Rightarrow \quad \boxed{V_f= \frac{3}{4}v} \end{gather*} Hence, after the collision, the two cars move at three-fourths of the initial speed of the faster car. 

Impulse and Momentum Problems:

Problem (7): As a result of a force exerted on a ball lasting for about $2\,\rm s$, its momentum increases by $20\,\rm N\cdot s$. What is the average force on the ball? 

Solution : Applying Newton's second law of motion in its momentum form, we get \begin{align*} F_{av}&=\frac{\Delta p}{\Delta t} \\\\ &=\frac{20}{2}\\\\&= 10\,\rm N \end{align*}

Problem (8): Consider a car standing at rest behind the traffic light. After the light turns green, the car accelerates and increases its speed from zero to $6\,\rm m/s$ in $0.6\,\rm s$. What impulse and average force does a $75-\rm kg$ person in the car experience? 

Solution : Impulse and average force are related together by the following formula \[\text{Impulse}=\vec{F}_{av}\Delta t =\Delta \vec{p}\] Thus, if we could find the change in momentum of the passenger, then the impulse and average force on it would also be found. 

The change in the passenger's momentum is \begin{align*} \Delta \vec{p}&=m(\vec{v}_2-\vec{v}_1) \\ &=75(6-0) \\ &=450\,\rm kg \cdot m/s \end{align*} This is also equal to the impulse delivered to the person, so \[\text{impulse}=450\,\rm kg \cdot m/s\] The average force that exerted on the person in the car is also found as below \begin{align*} F_{av}&=\frac{\Delta p}{\Delta t} \\\\ &=\frac{450}{0.6}\\\\ &= 750\,\rm N \end{align*} 

Problem (9): Assume you kick a baseball of mass $0.145-\rm kg$ traveling at a speed of $30\,\rm m/s$. As a result of the collision, it leaves the bat with a speed of $45\,\rm m/s$. If the contact of time between them is $5\times 10^{-3}\,\rm s$,  (a) what is the direction and magnitude of the net force exerted on the ball? (b) how about the force on the bat?  (c) What work do you do on the ball?

Before and after the collision, velocities are given and we are asked about the net force between the bat and baseball. In such problems, we should use the definition of impulse since it relates the average external force exerted during a short time interval of a collision to the change in momentum of the object as below \[\text{Impulse}=\vec{F}\Delta t=\Delta \vec{p}\] (a) According to this definition, the net force on the ball is found as the change in the ball's momentum divided by that short time interval. But there is a subtle point to this problem. 

Assume the baseball originally travels to the right, and we set this direction as the positive direction of our system. After the collision, the ball rebounds in the opposite direction, namely in the negative direction ($-\hat{i}$). With these assumptions, the change in momentum of the ball is written as \begin{align*} \Delta\vec{p} &=m(\vec{v}_2-\vec{v}_1) \\\\ &=(0.145) \left(45\,(-\hat{i})-30\,\hat{i}\right) \\\\ &=10.875(-\hat{i}) \quad \rm kg\cdot m/s \end{align*} Now, using the definition of impulse, we get \begin{align*} \vec{F}&=\frac{\Delta \vec{p}}{\Delta t}\\\\ &= \frac{10.875(-\hat{i})}{5\times 10^{-3}} \\\\ &=2175\,(-\hat{i}) \quad \rm N \end{align*} Impulse is a vector quantity pointing in the same direction as the force. Thus, a net force of $2175\,\rm N$ toward the negative $x$-axis is applied to the ball. 

(b) According to Newton's third law of motion, the force that the bat exerts on the ball is the same force that the ball exerts on the bat but in the opposite direction.

(c) By definition, the work done by a constant force on an object over a displacement $\Delta x$ is \[W=F\Delta x \cos\theta\] where $\theta$ is the angle between $F$ and $\Delta x$ but in this case, the displacement is not known. But there is another method to solve work problems in physics. 

According to the work-kinetic energy theorem, the change in kinetic energy of a moving body is just the work done on it. \[W=\Delta \rm K.E.\] Thus, this part is converted to a work-kinetic energy theorem problem ! \begin{align*} W&=\Delta K \\\\ &=\frac 12 m(v_a^2-v_b^2) \\\\ &=\frac 12 (0.06)\left((-45)^2-30^2\right) \\\\ &=33.75\quad \rm J \end{align*}

Problem (10): A hammer of mass $m=15\,\rm kg$ strikes a nail with a speed of $8\,\rm m/s$ and comes to a stop in a time interval of $5\,\rm ms$.  (a) What impulse is imparted to the nail? (b) What is the average net force exerted on the nail?

Solution : Impulse, which is denoted by $\vec{J}$, is defined as the product of the average net external force applied to an object during a time interval $\Delta t$ and that short time interval, i.e., $\vec{J}=\vec{F}_{av}\Delta t$. Keep in mind that in all momentum and impulse problems, the impulse is also equal to the change in momentum of the object. \[\text{impulse}=\Delta \vec{p}\] Impulse and momentum are vector quantities, meaning they have direction. Hence, in all such problems, you should choose a positive direction and compare all velocities with that. Consider up as positive, in this case. 

(a) The initial and final velocities of the hammer are $v_i=-8\,\rm m/s$ and $v_f=0$, so the change in momentum of the hammer is \begin{align*} \Delta p&=m(v_2-v_1) \\ &=15(0-(-8))=+120\quad \rm kg\cdot m/s \end{align*} The initial velocity is to the down, so we put an extra negative in front of it. Therefore, the impulse imparted to the nail is \[\text{impulse}=+120\quad \rm kg\cdot m/s\]  (b) According to Newton's second law in momentum form, the force exerted on the nail by the hammer is equal to the change in momentum of the nail divided by the contact time between them, \[F_{av}=\frac{\Delta p}{\Delta t}=\frac{120}{5\times 10^{-3}}=24000\,\rm N\] 

Problem (11): A $0.06-\rm kg$ golf ball strikes a wall at an angle of $37^\circ$ with a speed of $28\,\rm m/s$ and rebounds at the same angle and speed.  Determine the magnitude and direction of the impulse given to the ball. (Take $\cos 37^\circ=0.8$ and $\sin 37^\circ=0.6$)

Solution : Take right and up as positive directions. Impulse is the change in momentum of an object. On the other side, momentum is defined as $\vec{p}=m\vec{v}$. 

The velocity vectors before ($b$) and after ($a$) strike make a $37^\circ$ angle with the wall, so resolve them into their components as below \begin{align*} v_{by}&=v_b \cos 37^\circ \\ &=28\times 0.8\\ &=22.4\,\rm m/s \\\\ v_{bx}&=v_b \sin 37^\circ \\&=28\times 0.6 \\&=16.8\,\rm m/s \\\\ v_{ay}&=v_a \cos 37^\circ \\&=28\times 0.8 \\ &=22.4\,\rm m/s \\\\  v_{ax}&=v_b \sin 37^\circ \\&=28\times 0.6 \\&=-16.8\,\rm m/s \end{align*} As you can see in the figure, the component $v_{ax}$ is to the left in the negative $x$ direction, so we put a negative in front of it.

Now, it's time to find the change in momentum of the golf ball. The $y$ components (or vertical ones) are in the same direction, so their subtraction gets zero, but the horizontal ones are in the opposite direction, and their subtraction is written as below \begin{align*} \Delta \vec{p}&=m\Delta\vec{v} \\ &=m(\vec{v}_{after-x}-\vec{v}_{before-x}) \\ &=m(v_{ax}-v_{bx}) \\&=(0.06)(-16.8-16.8) \\&=-2.016 \,\rm kg\cdot m/s \end{align*} The minus sign indicates that the direction of the change in momentum (or impulse) is to the left in the $x$ direction. Therefore, the total impulse imparted to the ball is equal to the change in the ball's momentum, \[\text{Impulse}=\Delta \vec{p}=-2.016\,\rm kg\cdot m/s\] The absolute value of this is the magnitude of the impulse. 

For more problems on resolving vectors into their components, go here .

Problem (12): A ball of mass $0.150\,\rm kg$, initially at rest, is dropped from a height of $1.25\,\rm m$ and rebounds from the floor to a height of $0.96\,\rm m$. Assume the ball was in contact with the floor for $0.2\,\rm s$.  (a) What impulse was delivered to the ball by the floor? (b) What is the average force given to the ball by the floor?

Solution: As before, the impulse in physics equals the change in momentum of a moving body. After the ball strikes the floor, its momentum changes both in direction and magnitude. We should find the value of this change. 

In this problem, the ball's velocities immediately before and after striking the floor are not given. In this case, we must use the law of conservation of mechanical energy. 

Initially, the ball is at rest at a height of $h_1=1.25\,\rm m$ and hits the floor with a speed of, say, $v_b$, so in this part of the motion, the sum of its kinetic ($K=\frac 12 mv^2$) and potential ($U=mgh$) energies (or mechanical energy $E$) is written as below \begin{align*}  E_i&= E_f \\\\ K_i+U_i&=K_f+U_f \\ 0+mgh_1&=\frac 12 mv_b^2 + 0 \\ \Rightarrow \quad v_b&=\sqrt{2gh_1} \end{align*} Just after the ball strikes the floor, its velocity is, say, $v_a$ and it reaches a height of $h_2$. Thus, for this part, the law of conservation of energy tells us again that the velocity just after rebounding from the floor is $v_a=\sqrt{2gh_2}$. Substituting the numerical values of heights into the above expressions, we get \begin{gather*} v_b=\sqrt{2(9.8)(1.25)}=-5\,\rm m/s \quad \downarrow \\\\ v_a=\sqrt{2(9.8)(0.96)}=+4.3\,\rm m/s \quad \uparrow \end{gather*} But keep in mind that velocity is a vector quantity with a direction. If we assume up as the positive direction, then we must choose the minus sign for $v_b$. 

(a) Now that we have the velocities before and after striking, we can construct the change in momentum of the ball, which is equal to the impulse ($\vec{J}$). \begin{align*} \vec{J}&=\Delta \vec{p} \\ &=m(\vec{v}_a-\vec{v}_b) \\&= (0.15)(4.3-(-5)) \\ &=1.25 \quad\rm N\cdot s\end{align*} (b) Using Newton's second law the average force exerted on an object during a short time interval of a collision is related to the change in momentum of that object as below \begin{align*} F_{av}&=\frac{\Delta p}{\Delta t} \\\\ &=\frac{1.25}{0.2} \\\\&=6.25 \quad \rm N \end{align*}

Impulse and force-time graphs problems

One of the most important problems in the momentum and impulse sections is related to the force-time graphs. 

Problem (13): In the graph below, a time-varying force on an object of mass $3\,\rm kg$ is plotted against time.  (a) What impulse is delivered to the object from $t=0$ to $t=15\,\rm s$?  (b) What is the final speed of the object, assuming its initial speed was zero? (c) Now suppose we want the final speed of the object to be zero. In this case, what should be the initial speed?

Solution : (a) The area under the force-time graph shows the impulse delivered to an object. In this problem, the area is a trapezoid whose value is one-second times the sum of the lengths of the parallel sides times the perpendicular distance between parallel sides, or \[\text{area}=\frac{(\text{lower side+upper side})\times height}{2}\] Therefore, the impulse equals to \[J=\frac{(9+15)\times 5}{2}=60\,\rm N\cdot s \]

Problem (14): The net force on a $4-\rm kg$ object varies in time as shown in the figure below. Find: (a) The initial acceleration of the object.  (b) The total impulse of the force delivered to the object over the entire time interval. (c) The final velocity of the object, assuming its initial velocity is $-3\,\hat{i}$. 

Solution : Here, the force vs. time graph exerted on an object is given.  (a) The initial acceleration is the force at time zero $t=0$ divided by the object's mass \[a=\frac{F}{m}=\frac{2}{4}=0.5\,\rm m/s^2\] (b) By definition, impulse ($J$) is the area under the force-time graph. In this case, the area is a trapezoid whose area is obtained as below \begin{align*} J&=\frac{(\text{lower side+upper side})\times height}{2} \\\\ &=\frac{(5+2)\times 4}{2} \\\\ &=14\,\rm N\cdot s \end{align*}  (c) In this part, it is assumed that the object initially traveled to the negative $x$ axis so a negative sign was added. Keep in mind that the area under the force vs. time graph represents the impulse or the change in momentum of the object. \[\text{area=impulse}=\Delta \vec{p}\] In the previous part, the impulse over the entire time period was computed. Now, equate that with the change in the object's momentum and solve for the final velocity \begin{gather*} \vec{J}=m(\vec{v}_f-\vec{v}_i) \\\\ 14\,\hat{i}=4(v_f-(-3) \,\hat{i}) \\\\ 14\,\hat{i}=4\vec{v}_f +12\,\hat{i} \\\\ \rightarrow \quad 4\vec{v}_f=14\,\hat{i}-12\,\hat{i} \\\\ \Rightarrow \quad \boxed{\vec{v}_f=0.5\,\hat{i}\quad \rm m/s} \end{gather*} Hence, after applying such a force varying in time, the object finally has a speed of $2\,\rm m/s$ in the positive $x$ direction. 

Problem (15): A varying force in time is applied to a body of mass $2\,\rm kg$ initially at rest. Below, the force-versus-time graph is shown. Find the speed of the object at $7\,\rm s$ and $12\,\rm s$. 

Solution : The object is initially at rest, so its initial momentum is zero, i.e., $p_0=0$. The momentum of the object at a later time $t=7\,\rm s$ is $p_7=mv_7$. Hence, the momentum change of the object in this time interval is \[\Delta p=m(v_7-v_0)=2v_7\] Recall that the area under the force-time graph gives us the impulse ($J$) of the force, which, on the other hand, is equal to the momentum change. The area between $0\,\rm s$ and $7\,\rm s$ is a triangle whose magnitude is half the product of base and height. \[J=\frac 12 \times 7\times 12 =42\,\rm N\cdot s\] Equating these two above expressions and solving for the speed at time $7\,\rm s$, we will get \begin{gather*} J=\Delta p \\\\ 42=2v_7 \\\\ \Rightarrow \quad v_7=21\,\rm m/s \end{gather*} Again, the impulse from $0\,\rm s$ to $12\,\rm s$ is the area under the graph, but notice that here the total area is composed of two areas: one is below the axis and the other is above the axis. That part below the axis always has a negative area!

Thus, the impulse or area in this time interval is found as \begin{align*}Area &= area \uparrow + area \downarrow \\\\ &=\frac 12 (7)(12)+\frac 12 (5)(-8) \\\\ &=22\quad \rm N\cdot s \end{align*} Equating this with the momentum change in the whole time interval and solving for the speed at the instant of $t=12\,\rm s$, we will get \begin{gather*} J=\Delta p=m(v_{12}-v_0) \\\\ 22=2(v_{12}-0) \\\\ \Rightarrow \quad v_{12}=11\,\rm m/s \end{gather*}  

Elastic Collisions in one Dimension: 

Problem (16): Two $4-\rm kg$ and $12-\rm kg$ balls traveling at speeds of $24\,\rm m/s$ and $-2\,\rm m/s$, respectively, collide head-on with each other. After the collision, the lighter ball moves at $-3\,\rm m/s$ in the opposite direction. What is the velocity (magnitude and direction) of the heavier object after the collision? 

Solution : Assume the positive direction to be rightward and label the after-collision velocities with a prime. The given information are $m_1=4\,\rm kg$, $m_2=12\,\rm kg$, $v_1=24\,\rm m/s$, $v_2=-2\,\rm m/s$, and $v'_1=-3\,\rm m/s$. Keep in mind that in all collisions, first of all, apply the conservation law of momentum as below  \begin{gather*} \vec{p}_{before}=\vec{p}_{after} \\\\ m_1v_1+m_2 v_2 =m_1 v'_1+m_2 v'_2 \\\\ (4)(24)+(12)(-2)=(4)(-3)+(12)v'_2 \\\\ \Rightarrow \boxed{v'_2=+7\quad \rm m/s} \end{gather*} Therefore, the heavier ball moves at a speed of $7\,\rm m/s$ to the right in the positive $x$ direction.

Problem (17): A $0.44-\rm kg$ golf ball moving to the right with a speed of $3.8\,\rm m/s$ collides head-on with another golf ball of mass $0.22\,\rm kg$ initially at rest. Assume the collision to be perfectly elastic. What are the magnitude and direction of each ball's velocity after the collision? 

Solution : Since the collision is perfectly elastic, we must apply the conservation of kinetic energy and momentum. Let $v'_1$ and $v'_2$ be the velocities after the collision. We are given $m_1=0.44\,\rm kg$, $m_2=0.22\,\rm kg$, $v_1=3.8\,\rm m/s$, and $v_2=0$. The law of conservation of momentum tells us that the sum of the momenta before the collision must be equal to the sum of the momenta after the collision. \begin{gather*} \vec{p}_{before}=\vec{p}_{after} \\\\ m_1v_1+m_2 v_2 =m_1 v'_1+m_2 v'_2 \\\\ (0.44)(3.8)+0=(0.44)v'_1+(0.22)v'_2 \\\\ \Rightarrow \boxed{v'_2+2v'_1=7.6} \end{gather*} where in the last step we divided both sides by the common factor of $0.22$. 

Since the collision is assumed to be perfectly elastic, kinetic energy is also conserved \begin{gather*} K.E._{before}=K.E._{after} \\\\ \frac 12 m_1 v_1^2+\frac 12 m_2 v_2^2 = \frac 12 m_1 {v'_1}^2 +\frac 12 m_2 v'_2 \\\\ \frac 12 (0.44)(3.8)^2+0=\frac 12 (0.44){v'_1}^2+\frac 12 (0.22){v'_2}^2 \\\\ \Rightarrow \boxed{{v'_2}^2 +2{v'_1}^2=2(3.8)^2} \end{gather*} Consider the right as a positive direction. Now, we have two equations and two unknowns. Solve for $v'_2$ from the first equation, put it into the second one, and solve for $v'_1$ using a quadratic equation online solver. For this equation we get two solutions, $\boxed{v'_1=+1.27\,\rm m/s}$ or $v'_1=+3.8\,\rm m/s$. It is obvious that the second cannot be true since it tells us that the velocity of the incoming object after the collision must be equal to its initial velocity in the same direction, which does not make sense. Now substitute this value into the first equation again and solve for $v'_2$ which gets $\boxed{v'_2=+5.06\,\rm m/s}$. Thus, the second ball moves in the same direction as the incoming ball, which is in the positive $x$ direction. 

In order to help students prepare for the AP Physics 1 exam, this page includes several practice problems on impulse and momentum that are solved. 

As you learned, an impulse is a change in the momentum of an object or the product of the average force exerted on the object times the short interval of time that force exerted.

You can also check this out for solving simple problems on Momentum .

Author : Dr. Ali Nemati Published : 2/1/2022 

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