COLLISIONS

1. Collisions are extremely important to understand because they happen all the time -
electrons collide with atoms, a bat with a ball, cars with trucks, star galaxies with other
galaxies,...In every case, the sum of the initial momenta equals the sum of the final
momenta. This follows directly from Newton's Second Law, as we have already seen.
2. Take the simplest collision: two bodies 1 2 1 2
1 2
1 1 2 2 1 1 2 2
of mass and moving with velocities and .
After the collision they are moving with velocities v and v . From momentum conservation,
v v
m m u u
mu +m u = m +m ⇒ 1 1 1 2 2 2
1 2 1 2 1 2 1 2 1
2 1 1 2 2 2 2 1 1 2 2 2
( v) (v )
This is as far as we can go. There are two unknowns but only one equation. However, if the
collision is elastic then,
v v
m u m u
mu m u m m
− = −
+ = + ⇒ 2 2 1 2 2
21 1 1 2 2 2 2
1 1 2 2 1 2 2 1
( v ) (v ).
Combine the two equations above,
v v v v .
In words, this says that in an elastic collision the relative speed of
m u m u
u u u u
− = −
+ = + ⇒ − = −
1 2
1 2
1
the incoming particles
equals the relative speed of the outgoing particles.
3. One can solve for v and v (please do it!) easily and find that:
v (m m
m
−
= 2
1 2
1 2 1 2
1 2 1
2 1 2
1 2 1 2
1 2 1 2 2 1
) ( 2 )
v ( 2 ) ( )
Notice that if = , then v and v . So this says that after the collision, the
bodies will
u m u
m m m
m u m mu
m m m m
m m u u
+
+ +
−
= +
+ +
= =
2 1 2
just reverse their velocities and move on as before.
4. What if one of the bodies is much heavier than the other body, and the heavier body is at
rest? In this case, m>>m andu = 0. We can immedi 1 1 2 ately see that v and v 0.
This makes a lot of sense: the heavy body continues to stay at rest and the light body just
bounces back with the same speed. In the lecture, you saw a demonstr
= −u =
2 1 2 1 1 2 1
ation of this!
5. And what if the lighter body (rickshaw) is at rest and is hit by the heavier body (truck)? In
this case, and 0. From the above equation we see that v and v 2 .

So the truck's speed is unaffected, but the poor rickshaw is thrust in the direction of the
truck at twice the truck's speed!
6. Sometimes we wish to slow down particles by making them collide with other particles.
In a nuclear reactor, neutrons can be slowed down in this way. let's calculate the
fraction by which the kinetic energy of a neutron of mass decreases in a head-on
collision with an atomic
m
2
2
f f
2
i i
1 2 1 2
f i 2
1 2 1 2
nucleus of mass that is initially at rest:
m M M + m
Solution: 1 1 v
v
For a target at rest: v ( )v 4 .
( )
7. A bullet with mass m, is fired into a b
i f
i
i f
i
m
K K K
K K
m m K K mm
m m K m m
−
= − = −
− −
= ∴ =
+ +
lock of wood with mass M, suspended like a
pendulum and makes a completely inelastic collision with it. After the impact, the
block swings up to a maximum height y. What is the initial speed of the bullet?
Solution:
By conservation of momentum in the direction of the bullet,
v ( ) , v ( )
The block goes up by distance y, and so gains potential energy. Now we can use the
m m MV m MV
m
+
= + ⇒ =
conservation of energy to give, 1 ( ) 2 ( ) , where V is the velocity
2
acquired by the block+bullet in the upward direction just after the bullet strikes. Now
use 2 . So finally, th
m MV m M gy
V gy
+ = +
= e speed of the bullet is: v ( ) 2 .
8. In 2 or 3 dimensions, you must apply conservation of momentum in each direction
separately. The equation looks as if it is one equation, but it i i f
m M gy
m
P P
+
=
=
􀁇 􀁇
s actually 3
separate equations: , , . On the other hand, suppose you had
an elastic collision. In that case you would have only one extra equation coming from
energy co
ix fx iy fy iz fz p =p p =p p =p
nservation, not three.
9. What happens to energy in an inelastic collision? Let's say that one body smashes into
another body and breaks it into 20 pieces. To create 20 pieces requires doing work
against the intermolecular forces, and the initial kinetic energy is used up for this.
φ
r
O
y
x
(x, y)
s