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?
Solution:
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 \[J=\frac{(9+15)\times 5}{2}=60\,\rm N\cdot s \]
(b) By definition, impulse ($J$) is equal to the change in momentum of a body, i.e., \begin{gather*} \vec{J}=\Delta \vec{p} \\\\ \Delta \vec{p}=m(\vec{v}_f-\vec{v}_i) \end{gather*} where the subscripts $i$ and $f$ denote initial and final momenta. The object is at rest initially, so its initial momentum is zero, $p_i=0$, thus, we have \begin{gather*} J=\Delta p=m(v_f-v_i) \\\\ 60=3(v_f-0) \\\\ \Rightarrow \quad v_f=20\,\rm m/s \end{gather*}
(c) The impulse imparted to the object does not change since the area under the $F-t$ graph hasn't changed. Again, equate impulse to the change in momentum of the object, but this time set the final velocity to zero, \begin{gather*} J=m(v_f-v_i) \\\\ 60=3(0-v_i) \\\\ \Rightarrow \quad v_i=-20\,\rm m/s \end{gather*} The minus sign indicates the direction of the object, which is, say, in the negative $x$ direction.
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.
Summary:
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
© 2015 All rights reserved. by Physexams.com
