# collisions_ by wanghonghx

VIEWS: 2 PAGES: 3

• pg 1
```									The Law of conservation of momentum
The law of conservation of momentum states that :-

Momentum is conserved in ALL collisions or explosion in an isolated system where no
external forces act. In other words the momentum before the collision or explosion is the same
as that after it even if the kinetic energy is not conserved.

Collisions
Elastic and inelastic collisions
In a collision the kinetic energy is sometimes the
same before the collision as it is afterwards - such a
collision is called an elastic collision. Perfectly
elastic collisions are rare - usually some of the kinetic
energy is converted to sound or heat. Even a
trampolinist colliding with a trampoline bed loses
energy as the bed stretches and heats up. A perfectly
inelastic collision is one where all the kinetic energy
is converted into other forms - a piece of soft putty
falling onto the floor is an almost perfectly inelastic
collision, the kinetic energy of the moving putty is
converted into heat and a little sound as the putty is
deformed on impact. Most real collisions are usually
somewhere between the two extremes.

In all collisions the law of conservation of momentum applies. If a mass m1 moving at a velocity u1
collides with a mass m2 moving at a velocity u2 such that after the collision m1 moves at v1 and m2
moves at v2 then:-

momentum before collision = momentum after collision

m1u1 + m2u2 = m1v1 + m2v2

The law of conservation of momentum applies whether the collisions are elastic or not. (See
Figure 1)
After
Before
Partly
m1                                 m2        elastic             m1               m2
u1               v1                           u2                            v2
Figure 1

The special case of two equal masses making a completely elastic collision is shown in Figure 2.

Before                                                   After
Perfectly
m                                 m        elastic                m              m
u                v                            v                             u
A                                 B                                 A            B
Figure 2
Notice that the velocities of the two bodies A and B are swapped over.

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Inelastic collisions

As we have said in an inelastic collision some, or all, the kinetic energy of the colliding bodies is
lost - this may be converted into heat or used to deform the bodies. Imagine two balls colliding
head on, sticking together and then moving off after the collision (See Figure 3).

Perfectly
m1                                   m2                               m1 + m2
v1                   inelastic                                 v2
u1
Figure 3

The total momentum before collision must be equal to the total momentum after collision and if we
know the masses and velocities of the colliding objects before collision we can use this principle to
find out how fast they are moving - and in which direction - after the collision.

Example problem
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If a mass of 3.5 kg is moving left to right at 5 ms collides with a mass of 4.0 kg moving right to left at 3.0
-1
ms and they stick together find the final velocity of the combined masses.

Momentum before impact = 3.5x5.0 + [-4.0x3.0] = 5.5 Ns but this must equal the momentum after the
collision i.e. total mass x final velocity
Notice that one of the velocities is negative showing that the ball was moving right to left.

Mass afterwards = 7.5 kg
-1
Therefore velocity afterwards = 5.5/7.5 = 0.73 ms    this is positive showing that after collision the
combined two balls move from left to right.

The amount of kinetic energy lost in a collision depends on the nature of both objects involved in
the collision. For example a different amount of energy will be converted into heat and sound if
putty is dropped onto putty than if putty is dropped onto steel. The energy conversions will also be
different for steel on steel, velcro on velcro etc.

Remember that velocity and momentum are vector quantities and must be treated as such
in all collision problems.

2
Example problems
-1
1. A rifle of mass 3 kg fires a bullet of mass 50 gm (0.05 kg) at 100 ms .
(a) Calculate the recoil velocity of the rifle
(b) explain why it is better to hold the rifle into your shoulder when firing it.
-1
(a) recoil velocity (V) = -vm/M = 100x0.05/3 = 1.67 ms
(b) if the rifle is in contact with your shoulder when it is fired the mass that recoils is the mass of the
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Recoil velocity = - 0.05x100/53 = 0.09 ms much less then before!

-1
2. A bullet of mass 0.05 kg moving at 100 ms strikes a block of wood and is embedded in it to a depth
of 5 cm. If the wood has a mass of 2.5 kg calculate:-

(a) the velocity of the block and bullet after the collision
(b) the mean retarding force of the bullet
-1
(a) Using: velocity (V) = vm/M = 100x0.05/2.5 = 2 ms

(b) Using: - Kinetic energy lost = Retarding force (F) x distance
Therefore 250 -10.2 = 239.8 = Fx0.05

Retarding force = 4796 N
-1
3. A car of mass 1000 kg travelling from west to east along a road at 15 ms collides with a van of
mass 3000 kg travelling in the opposite direction. If they both come to rest after the collision what was
the initial velocity of the van?

Momentum after the collision = 0      so momentum before collision must also equal 0. Therefore :

Momentum before collision = momentum of car + momentum of van = 1000x15 + 3000v where v is the
-1
velocity of the van. This gives 1000x15 + 3000v = 0 and so v = - 1000x15/3000 = -5 ms        the
negative sign meaning that it was travelling in the opposite direction i.e. east to west.

15 ms-1

3

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