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Lecture Notes: One-Dimensional Collisions
Newport AP Physics—C. Appel
Sections 9.3 – 9.4
Need Air Track, Swinging Ball Thing and LabPro attached to laptop.
Conservation of Momentum



Wordy: When the net external force acting on a system is zero, the total vector momentum of the system
remains constant.
𝑑𝑃⃑
Mathematical: Let 𝑃⃑ = ∑ 𝑝 then = 0 or 𝑃⃑ = constant
𝑑𝑡
If there are two objects in a system and the momentum of one changes, then the momentum of the other
must also change, so the sum remains constant.
Questions


Consider a person standing still on a frictionless, frozen lake. How does he/she get to the edge?
A man of mass 75 kg and a boy of mass 35 kg are standing together on a smooth ice surface for which
friction is negligible. If they push each other apart and the man moves away with a speed of 0.3 m/s relative
to the ice, how far apart are they after 5 s?
 The total momentum before the push is zero, so the total after the push must also be zero.
Let the direction the man moves be positive. His momentum is: p = mv = (75)(0.3) = 21 kgm/s
Therefore, the boys momentum must be –21 kgm/s; p = mv = (35)vboy = -21 kgm/s, so vboy = -0.6 m/s.
Hence, there speed of separation is 0.9 m/s. After 5 s, they are 4.5 m apart.
One-dimensional collisions

During collisions momentum is conserved, in other words, momentum is constant before, during and after a
collision (provided there are no external forces acting on the system).
Newton’s Cradle


Let’s consider the swinging ball thing. If you lift one ball and drop it what happens? Why?
Conservation of momentum.
 Let the mass of one ball be m and the speed of the ball just before the collision be v. So the momentum
of the system just before the collision is mv. Therefore the momentum after the collision must also be
mv. But, why can’t 2 balls fly off at half of the speed? The momentum after the collision would be
(2m)(v/2) = mv. Momentum would be conserved.
Elastic collisions



During an elastic collision momentum is conserved and mechanical energy is conserved.
If the KE before is equal to the kinetic energy after a collision then the collision is said to be elastic.
Elastic collisions occur when the objects completely bounce off of one another. There is no (or very little)
energy lost to deformation of the objects or to heat.
Newton’s Cradle (again)

The collision is elastic. If 2 balls flew off at half of the speed would energy be conserved?


1
1
𝑣 2
1
No. The energy before would be 2 𝑚𝑣 2 . After the collision the energy would be 2 (2𝑚) (2) = 4 𝑚𝑣 2 .
Therefore both conservation of momentum and energy tell us what will happen.
Collisions
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So if there is an elastic collision between two objects with the same mass and one at rest and the other
moving, one will stop the other will go off at the initial speed.
What if both objects are moving, and/or both have different masses.
Collision between an object at rest and an object with a different mass moving.
 Momentum is conserved: 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓 ⇒ 𝑚1 𝑣1𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
1

1
1
1
2
2
2
2
2
2
2
 Kinetic energy is conserved: 2 𝑚1 𝑣1𝑖
+ 2 𝑚2 𝑣2𝑖
= 2 𝑚1 𝑣1𝑓
+ 2 𝑚2 𝑣2𝑓
⇒ 𝑚1 𝑣1𝑖
= 𝑚1 𝑣1𝑓
+ 𝑚2 𝑣2𝑓
 Given the initial velocity find both of the final velocities (perform on air track)
 Yuck.
What about a collision between two moving objects of different mass.
 Really ugly. But there is an easier way. To save some minds spinning, I’ll skip the algebra. (If you
want to see it, just read my online notes.
 If we start with conservation of momentum and energy we would have:
1
1
1
1
2
2
2
2
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓 and 2 𝑚1 𝑣1𝑖
+ 2 𝑚2 𝑣2𝑖
= 2 𝑚1 𝑣1𝑓
+ 2 𝑚2 𝑣2𝑓
These two equations can be combined to get:
(𝑣1𝑖 + 𝑣1𝑓 ) = (𝑣2𝑓 + 𝑣2𝑖 ) or (𝑣2𝑖 − 𝑣1𝑖 ) = −(𝑣2𝑓 − 𝑣1𝑓 )
In other words, the relative speed of approach equals the relative speed of separation. This is quite
useful.
So now, instead of using momentum and energy as two equations and two unknowns, we can use this
equation with momentum as two equations two unknowns. Much easier.

For those of you that are curious, here’s how it works out:
Start with conservation of momentum and energy, gather the m1 and m2 terms.
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓 ⇒ 𝑚1 (𝑣1𝑖 − 𝑣1𝑓 ) = 𝑚2 (𝑣2𝑓 − 𝑣2𝑖 )
1
2
1
1
1
2
2
2
2
2
2
2
2
𝑚1 𝑣1𝑖
+ 2 𝑚2 𝑣2𝑖
= 2 𝑚1 𝑣1𝑓
+ 2 𝑚2 𝑣2𝑓
⇒ 𝑚1 (𝑣1𝑖
− 𝑣1𝑓
) = 𝑚2 (𝑣2𝑓
− 𝑣2𝑖
)
Looking at the energy expression I see something familiar…… the difference between two squares.
𝑚1 (𝑣1𝑖 + 𝑣1𝑓 )(𝑣1𝑖 − 𝑣1𝑓 ) = 𝑚2 (𝑣2𝑓 + 𝑣2𝑖 )(𝑣2𝑓 − 𝑣2𝑖 )
Let’s divide this by the momentum equation. We are then left with:
(𝑣1𝑖 + 𝑣1𝑓 ) = (𝑣2𝑓 + 𝑣2𝑖 )
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
The catch: We can only use this for elastic collisions.
Try an air track problem again.
Inelastic collisions
Perfectly inelastic collisions:
 The objects stick together after the collision.
 Momentum is conserved, but mechanical energy is not conserved.
 Since momentum is conserved 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = (𝑚1 + 𝑚2 )𝑣𝑓
Inelastic collisions:
 The objects do not stick together, but kinetic energy is not conserved. Some energy is lost to heat or to
deforming the objects.
Question

A bullet of mass 0.01 kg moves horizontally with a speed of 400 m/s and embeds itself in a block of mass
0.39 kg that is initially at rest on a frictionless table. Find (a) the final velocity of the bullet and block and
(b) the initial and final mechanical energy of the bullet-block system.
(a)
Since there are no external horizontal forces on the bullet-block system, the horizontal component of
momentum is conserved.
Since the block is at rest its momentum is zero, so the momentum of the system is the momentum of
the bullet.
p = mv = (0.01 kg)(400 m/s) = 4 kgm/s
After the bullet hits the block, the block moves. The bullet embeds itself in the block and the two
become one.
Since momentum is conserved the final momentum must also be 4 kgm/s
p = (mbullet + mblock)v = (0.40 kg)v = 4 kgm/s
So, v = 10 m/s
(b)
Since everything is at the same height call gravitational potential energy zero.
1
1
𝐸𝑖 = 𝐾𝑖 = 𝑚𝑣 2 = (0.01)(400)2 = 800 J
2
2
1
1
𝐸𝑓 = 𝐾𝑓 = (𝑚total )𝑣 2 = (0.40)(10)2 = 20 J
2
2
So most of the energy is lost to non-conservative forces.