# 6.2: Conservation of Momentum in One Dimension

- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For this whale to leap out of the water, something underwater must be moving in the opposite direction, and intuition tells us it must be moving with relatively high velocity. The water that moves downward is pushed downward by the whale's tail, and that allows the whale to rise up.

## Conservation of Momentum in One Dimension

When impulse and momentum were introduced, we used an example of a batted ball to discuss the impulse and momentum change that occurred with the ball. At the time, we did not consider what had happened to the bat. According to Newton’s third law, however, when the bat exerted a force on the ball, the ball also exerted an equal and opposite force on the bat. Since the time of the collision between bat and ball is the same for the bat and for the ball, then we have equal forces (in opposite directions) exerted for equal times on the ball AND the bat. That means that the impulse exerted on the bat is equal and opposite (-Ft) to the impulse on the ball (Ft) and that also means that there was a change in momentum of the bat [−Δ(mv)BAT] that was equal and opposite to the change in momentum of the ball [Δ(mv)BALL].

The change in momentum of the ball is quite obvious because it changes direction and flies off at greater speed. However, the change in momentum of the bat is not obvious at all. This occurs primarily because the bat is more massive than the ball. Additionally, the bat is held firmly by the batter, so the batter's mass can be combined with the mass of the bat. Since the bat's mass is so much greater than that of the ball, but they have equal and opposite forces, the bat's final velocity is significantly smaller than that of the ball.

Consider another system: that of two ice skaters. If we have one of the ice skaters exert a force on the other skater, the force is called an **internal force** because both the object exerting the force and the object receiving the force are inside the system. In a closed system such as this, momentum is always conserved. The total final momentum always equals the total initial momentum in a closed system. Conversely, if we defined a system to contain just one ice skater, putting the other skater outside the system, this is not a closed system. If one skater pushes the other, the force is an external force because the receiver of the force is outside the system. Momentum is not guaranteed to be conserved unless the system is closed.

In a closed system, momentum is always conserved. Take another example: if we consider two billiard balls colliding on a billiard table and ignore friction, we are dealing with a closed system. The momentum of ball A before the collision plus the momentum of ball B before collision will equal the momentum of ball A after collision plus the momentum of ball B after collision. This is called the law of **conservation of momentum** and is given by the equation

p_{Abefore}+p_{Bbefore}=p_{Aafter}+p_{Bafter}

Example 6.2.1

Ball A has a mass of 2.0 kg and is moving due west with a velocity of 2.0 m/s while ball B has a mass of 4.0 kg and is moving west with a velocity of 1.0 m/s. Ball A overtakes ball B and collides with it from behind. After the collision, ball A is moving westward with a velocity of 1.0 m/s. What is the velocity of ball B after the collision?

**Solution**

Because of the law of conservation of momentum, we know that

p_{Abefore}+p_{Bbefore}=p_{Aafter}+p_{Bafter}

m_{A}v_{A}+m_{B}v_{B}=m_{A}v′_{A}+m_{B}v′_{B}

(2.0 kg)(2.0 m/s)+(4.0 kg)(1.0 m/s)=(2.0 kg)(1.0 m/s)+(4.0 kg)(v′_{B} m/s)

4.0 kg⋅m/s+4.0 kg⋅m/s=2.0 kg⋅m/s+4v′_{B}kg⋅m/s

4v′_{B}=8.0−2.0=6.0

v′_{B}=1.5 m/s

After the collision, ball B is moving westward at 1.5 m/s.

Example 6.2.2

A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars?

**Solution**

A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars?

Note that since the two trains stick together, the final mass is m_{A}+m_{B}, and the final velocity for each object is the same. Thus the conservation of momentum equation, m_{A}v_{A}+m_{B}v_{B}=m_{A}v′_{A}+m_{B}v′_{B}, can be rewritten m_{A}v_{A}+m_{B}v_{B}=(m_{A}+m_{B})v′

(30,000. kg)(2.2 m/s)+(30,000. kg)(0 m/s)=(60,000. kg)(v′ m/s)

66000+0=60000v′

v′=6600060000=1.1 m/s

After the collision, the two cars move off together toward the east with a velocity of 1.1 m/s.

Use the simulation below to observe how energy and momentum are exchanged in a bumper car collision. Begin by adjusting the sliders to make an elastic collision with cars of different masses. Then, see what happens to the kinetic energy in an inelastic collision with the same cars. Comparing these two scenarios should help you to gain a deeper understanding of why we always rely on momentum to analyze collisions.

Interactive Element

## Summary

- A closed system is one in which both the object exerting a force and the object receiving the force are inside the system.
- In a closed system, momentum is always conserved.

## Review

1. A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. With what speed does the goalie slide on the (frictionless) ice?

2. A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off together at 9.0 m/s. What was the original velocity of the bullet?

3. A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s. After the collision, the 0.50 kg ball moves away at -14 m/s. Find the velocity of the second ball after the collision.

4. Two carts are stationary with a compressed spring between them and held together by a thread. When the thread is cut, the two carts move apart. After the spring is released, one cart m=3.00 kg has a velocity of 0.82 m/s east. What is the magnitude of the velocity of the second cart (m=1.70 kg)after the spring is released?

5. Compared to falling on a tile floor, a glass may not break if it falls onto a carpeted floor. This is because

- less impulse in stopping.
- longer time to stop.
- both of these
- neither of these.

6. A butterfly is hit by a garbage truck on the highway. The force of the impact is greater on the

- garbage truck.
- butterfly.
- it is the same for both.

7. A rifle recoils from firing a bullet. The speed of the rifle’s recoil is small compared to the speed of the bullet because

- the force on the rifle is small.
- the rifle has a great deal more mass than the bullet.
- the momentum of the rifle is unchanged.
- the impulse on the rifle is less than the impulse on the bullet.
- none of these.

## Explore More

Use this resource to answer the questions that follow.

- What is Newton's Cradle?
- How does Newton's Cradle work?
- How does a Newton's Cradle show conservation of momentum?

## Additional Resources

PLIX: Play, Learn, Interact, eXplore: Bumper Car Collisions

Real World Application: Body Armor

Video:

Study Guide: Momentum Study Guide