Conservation of Momentum: | 1
| 2 | 3 | 4 | 5
| 6 | 7 | Go
up

** - by Bryan Holman, 2002**

**1. A 1200 Kg car going 13 m/s collides with a 4200 Kg
truck at rest. Their bumpers
lock. What is their speed
afterwards?**

So, the 1200 Kg car is
going at 13 m/s before it collides with the 4200 Kg truck.
We know that p=mv, and that since there are two different cars, the
formula would be mv_{car} + mv_{truck} = mv_{both}.
So:

mv_{car} + mv_{truck}
= mv_{both}

(1200 Kg)(13 m/s) + (4200 Kg)(0 m/s) = (1200 Kg + 4200 Kg)v

15600 = (5400 Kg)v

**2.89 m/s = v**

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**2. A 60 Kg person running 3 m/s collides head on with a
100 Kg person running -2 m/s (The other way)
What is their final velocity if they stick together?**

First, I will call right
positive, so here two people run into each other and end up going to the left.
We know that p=mv. If you
look at the first frame, the two people are running at each other (don’t ask
me why), then collide in the second. To
find the momentum of the entire system, the formula would be: mv_{person a}
+ mv_{person b} = mv_{both}.
So: (The question states that they stick together but I didn’t want
to make a picture of that because that would just look really weird)

mv_{person a} + mv_{person
b} = mv_{both}

(60 Kg)(3 m/s) + (100 Kg)(-2 m/s) = (60 Kg + 100 Kg)v

-20 = (160 Kg)v

**-.125 m/s = v** (the
answer is negative which means their final velocity is going to the left)

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**3. A 50 Kg ice skater at rest throws a 5 Kg shot put at
a velocity of 3.5 m/s. What is
the recoil velocity of the skater?**

This problem is very
similar to the first two, except it starts out as a single object at rest,
then goes to two separate objects moving at different speeds.
Still, p=mv as before, so the formula would be: mv_{both} = mv_{ice
skater} _{+} mv_{shot put}.
So: (If you couldn’t tell, those are ice skates on the bottom of the
ice skater’s feet. Don’t
laugh at my drawings!)

mv_{both} = mv_{ice
skater} _{+} mv_{shot put}

(55 Kg)(0 m/s) = (50 Kg)v + (5 Kg)(3.5 m/s)

0 = (50 Kg)v + 17.5

-7.5 = (50 Kg)v

**-.35 m/s = v** (The
negative implies that the ice skater is going to the left because I like right
to be positive)

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**4. A 1200 Kg car going 15 m/s rear-ends with a 800 Kg
car going 5 m/s in the same direction. Their
bumpers lock. What is their speed
afterwards?**

This problem is exactly
like question #1. Therefore the
same formula will be needed. We
know that p=mv, so mv_{car} + mv_{truck} = mv_{both}.

mv_{car} + mv_{truck}
= mv_{both}

(1200 Kg)(15 m/s) + (800 Kg)(5 m/s) = (1200 +800)v

22000 = 2000v

**v = 11 m/s**

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**5. A 2000 Kg airplane going 45 m/s fires a 2 Kg shell
forward at a speed of 1200 m/s. What
is the final velocity of the plane? (Planes
crashed because of this!)**

The airplane and the shell
are together at first, so because of the law of conservation of momentum the
momentum of both the airplane and the shell when they are together will equal
the momentum of both of them when they are separate.
P=mv, so the equation would be mv_{both} = mv_{plane} _{+}
mv_{shell}.

mv_{both} = mv_{plane}
_{+} mv_{shell}

(2000 Kg + 2 Kg)(45 m/s) = (2000 Kg)v + (2 Kg)(1200 m/s)

90090 = 2000v + 2400

87690 = 2000v

**v = 43.8 m/s**

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**6. A 14.5 g bullet traveling 783 m/s horizontally
strikes an 9.24 Kg block of wood at rest on a level frictionless table.
The bullet goes through the block, but is traveling only 382 m/s in the
same direction after the collision. What
is the velocity of the block after the collision? (Assume the block loses no
mass)**

In this problem, the bullet
travels through the block, and the block moves afterwards.
This is a before and after question.
We know that p=mv, and because momentum is conserved, we can figure out
this problem. The momentum of the
bullet plus the block before will equal the momentum of both after: so the
equation is mv_{bullet} _{+} mv_{block} = mv_{bullet}
_{+} mv_{block}. Remember
that the block does not lose any mass. Also
remember to convert grams to kilograms.

mv_{bullet} _{+}
mv_{block} = mv_{bullet} _{+} mv_{block}

(.0145 Kg)(783 m/s) + (9.24 Kg)(0 m/s) = (.0145 Kg)(382 m/s) + (9.24 Kg)v

11.3535 = 5.539 + 9.24v

5.8145 = 9.24v

**v = .629 m/s**

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**7. Bumper car A (326 Kg) with velocity 3.7 m/s collides
with the rear of car B (536 Kg) which has a velocity of 2.4 m/s in the same
direction. After the collision,
car A has a velocity of -1.2 m/s. What
is the velocity of car B after the collision.**

This question is similar to
the last one, a before and after problem.
The total momentum before the collision is the momentum of both cars,
same with after. So the total
momentum before equals the total momentum after.
P=mv, so mv_{car A} _{+} mv_{car B} = mv_{car
A} _{+} mv_{car B}.

mv_{car A} _{+}
mv_{car B} = mv_{car A} _{+} mv_{car B}

(326 Kg)(3.7 m/s) + (536 Kg)(2.4 m/s) = (326 Kg)(-1.2 m/s) + (536 Kg)v

2492.6 = -391.2 + 536v

2883.8 = 536v

**v = 5.4 m/s**

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