BEFORE Conservation of Momentum Problems 1 =9000 m …

v1=3

m1=9000

v2=0

m2=64000

BEFORE

v’

m1+m2=73000

AFTER

Conservation of Momentum Problems

Do your work on a separate sheet of paper or notebook. For each problem,

- draw clearly labeled diagrams showing the masses and velocities for each object before and after the

collision. The diagrams are essential (they can be very simple – just draw all objects as boxes )

- Define the system you are analyzing and justify why conservation of momentum of the system can be

used.

Don’t forget about directions – momentum, velocity and impulse are VECTORS.

1. In a railroad yard, a train is being assembled. An empty boxcar, coasting at 3 m/s along a frictionless track,

strikes a loaded car that is stationary, and the cars couple together. Each of the boxcars has a mass of 9000 kg

when empty, and the loaded car contains 55,000 kg of lumber.

The 2 cars together are an isolated system because there are no net external forces acting on the system (no

friction, gravity and normal add to zero, and the internal collision forces sum to zero). Therefore, the

momentum of the system of both cars is conserved (psys=0 because Fsys = 0).

a. Write the momentum conservation equation for this system in terms of variables (m,v).

b. Find the speed of the coupled boxcars.

c. Calculate and compare the kinetic energy of the system before and after the collision. Use this comparison

to determine the type of collision (elastic, inelastic, or explosion).

Kinetic energy is lost and trains stick together after collision so collision is perfectly inelastic.

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2. An 80 kg astronaut is on a spacewalk away from the shuttle when his tether line breaks. Starting at rest

(relative to the shuttle) he throws the 9.5kg oxygen tank away from him with a velocity of -2.0 m/s so that he

recoils in the opposite direction, towards the shuttle.

vT=vA=0

mT=10mA=80

vA’=?

mT=9.5

mA=80vT’=2

BEFORE AFTER

a. Write the momentum conservation equation for this system in terms of variables (m,v).

b. Find the speed with which the astronaut moves towards the shuttle.

c. Determine the impulse exerted on the oxygen tank by the astronaut. The impulse ON the TANK

is the momentum change OF THE TANK

d. Determine the impulse exerted on the astronaut by the oxygen tank.

Since the force the astronaut exerts on the tank is equal and opposite to the force the tank exerts on the

astronaut (Newton’s 3rd

), and since the forces act over the same duration of time, the impulses are also equal

and opposite.

Therefore, the impulse on the astronaut is FAt = +19 N s

Because there are no other forces acting on the astronaut and tank (astronaut and tank are an isolated system),

these equal and opposite forces and impulses cause equal and opposite momentum changes which is why the

momentum of the whole system does not change (no Fnet, no p)

e. Calculate and compare the total kinetic energy before and after the event. Use this comparison to determine

the type of collision (elastic, inelastic, or explosion).

K gained so explosion f. Determine the maximum distance the astronaut can be from the shuttle when the line breaks and still return

within 60.0 s (the amount of time he can hold his breath).

Once the astronaut throws the tank and recoils due to the thrust of the tank, the astronaut does not accelerate

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This is the distance traveled in 60 s at constant velocity of 0.24m/s

3. A 50 kg cart is moving across a frictionless floor at

2.0 m/s. A 70 kg boy, riding in the cart, jumps off

so that he hits the floor with zero velocity.

The boy and the cart together are an isolated

system because there are no net external forces

acting on the system (no friction, gravity and

normal add to zero and the internal interaction

forces sum to zero). Therefore, the momentum of

the system of boy and cart is conserved (psys=0

because Fsys = 0).

a. Write the momentum conservation equation in

terms of variables (m,v).

b. What was the velocity of the cart after the boy

jumped?

c. How large an impulse did the boy give to the cart?

finding the impulse ON the CART

which is the momentum change OF THE CART (if the

impulse is the net impulse.

d. Calculate and compare the total kinetic energy before and after the event. Use this comparison to

determine the type of collision (elastic, inelastic, or explosion).

Kinetic energy gained so explosion

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m1=70

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4. Two children with masses of 50 kg and 70 kg are at rest on frictionless in-line skates. The larger child pushes

the smaller so that the smaller child rolls away at a speed of 10 m/s.

The boy and girl together are an isolated system because there are no net external forces acting on the

system (no friction, gravity and normal add to zero and the internal interaction forces sum to zero). Because

there is no net force on the system, the momentum of the system of boy and girl is conserved. (psys=0

because Fsys = 0).

a. Write the momentum conservation equation in terms of variables (m,v).

b. Calculate the velocity of the larger child.

c. Calculate the impulse that each child imparts to the other.

d. Calculate and compare the total kinetic energy of the system before and after the event. Use this

comparison to determine the type of collision (elastic, inelastic, or explosion).

KE gained so explosion

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v1=45m1=0.050

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y=1

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Projectile Motion of Apple

v2’

x=?

5. A 0.8 kg apple is balanced on a circus performer's head 1.9 m above the ground. An archer shoots a 50 g

arrow at the apple with a speed of 45 m/s. The arrow passes through the apple and emerges with a speed of

12 m/s.

The arrow and apple together can be considered an isolated system if we examine the collision immediately

before the collision takes place and immediately after so that the impulse of gravity on the arrow is

negligible. Then no net external forces act on the system (neglect the friction between apple and head, no air

resistance apple gravity and normal add to zero and the internal interaction forces sum to zero). Therefore,

the momentum of the system of arrow and apple is conserved (psys=0 because Fsys = 0).


b. Find the speed of the apple as it flies off the performer's head.

c. What is the impulse exerted on the apple by the arrow? Impulse ON the APPLE is the momentum change

OF THE APPLE

d. What is the impulse exerted on the arrow by the apple?

Equal and opposite impulses since the collision forces are equal and opposite by Newton’s 3rd

Law and the

time during which they act is also the same.

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v1=8m/sv2=0

Before

m2=12kgm1= 6kg

v1’=? v2’=5.33

Afterm2=12kgm1= 6kg

ptotal=+48 kg m/s

e. Calculate and compare the K before and after the event. Use this comparison to determine the type of

collision (elastic, inelastic, or explosion).

K lost so inelastic

f. When the apple hits the ground, how far behind the clown is it? Neglect friction when the apple is atop

the clown’s head. Once the collision is over, the arrow has transferred momentum and energy to the apple

and so the apple has horizontal velocity and it undergoes projectile motion (if we neglect resistance). It is

horizontally launched from a height of 1.9m. (As the apple undergoes projectile motion, its momenti=um

changes in magnitude and direction; momentum is NOT conserved during the projectile motion unless we

include the Earth in the system.)

x y

6. An object, which has a mass of 6.0 kg, is moving to the right with a velocity of 8.0 m/s when it collides with

a second mass of 12.0 kg which is initially at rest. After the collision, the 12.0 kg mass moves off to the right

with a new velocity of 5.33 m/s. Neglect friction. Momentum is conserved in the system of the 2 blocks

because there is no net external force on the system; the internal collision forces do not change momentum

of the system because they sum to zero (psys=0 because Fsys = 0)


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m2=6kgm1= 2kg

v1’=? v2’=?

Afterm2=6kgm1= 2kg

(ptotal=+24 kg m/s)

b. Determine the velocity of the 6.0 kg object after the collision

c. Is the collision elastic? How do you know? Support your answer with evidence.

Kbefore = Kafter. Therefore the collision is elastic

7. An object of mass m1=2.0 kg, moving with a velocity of v1=12.0 m/s collides head-on with a stationary

object whose mass is m2=6.0 kg. The collision is elastic. What are the velocities of the two objects after the

collision? Neglect friction. (Note: there are two unknowns here. Elastic collision means that you have 2

conservation equations to use: conservation of momentum and conservation of mechanical energy. You can

also use conservation of momentum along with another equation involving relative velocities; finish and look

at the geogebra simulation to find this third relationship). Momentum is conserved in the system of the 2

blocks because there is no net external force on the system; the internal collision forces do not change

momentum of the system because they sum to zero (psys=0 because Fsys = 0)

There are two unknown variables here, v1’ and v2’. Because the collision is elastic, we have two equations to

use to solve for the two unknowns

1. Conservation of momentum: the 2 block system has no net force on it so there is no change in

momentum

2. Conservation of energy: because the collision is elastic, kinetic energy of the system is also conserved

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There are many ways to use these 2 equations to solve for the unknowns. One way is to put expression for v1’

from 1st equation into 2nd

There are 2 solutions to the equation

- v2’ = 0: this makes no sense because it would mean that after being struck, object 2 would not move

- v2’ = 6.0 m/s. This is the one that makes physical sense

Check your answer by using these values to make sure that with them pbefore = pafter and Kbefore = Kafter

2nd

METHOD (easier) There is another equation that can be used in elastic equations and therefore another

set of 2 equations that can be used. For elastic collisions only, the magnitude of the relative velocity does not

change before and after the collision. In fact the relative velocity before an elastic collision is equal in mag

and opposite in direction to the rel velocity after collision:

v1 - v2 = -(v1’ – v2’)

This equation can be used along with the conservation of momentum equation to solve for the two

unknowns.

There are many ways to use these 2 equations to solve for the unknowns. One way is to add the two

expressions together to remove v1’ and solve for v2’. Then use either of the two equations to get v1’.

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8. A particle of mass 4.0 kg, initially moving with a velocity of 2.0 m/s collides elastically with a particle of

mass 6.0 kg. initially moving with a velocity of -4.0 m/s. What are the velocities of the two particles after the

collision? There is no friction or air resistance. Momentum is conserved in the system of the 2 particles

because there is no net external force on the system; the internal collision forces do not change momentum

of the system because they sum to zero.

There are two unknown variables here, v1’ and v2. Therefore, we need to set up a system of 2 equations.

METHOD 1: In elastic collisions of an isolated system, both momentum and kinetic energy are conserved.

There are many ways to use these 2 equations to solve for the unknowns. One way is to put expression for v1’

from 1st equation into 2nd

Find the roots of the quadratic equation to solve this and you get 2 possible solutions:

- v2’ = 0.80 m/s: this is the only solution that makes physical sense

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Collision

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After

collision

Slides with

constant

acceleration

m2=2kg

m2+m2

m1=0.025kg

- v2’ = -4.0 m/s: this makes no sense because it would mean that after being struck, the velocity of object

2 would not change


2nd

METHOD (easier) In elastic collisions of an isolated system, momentum is conserved and the relative

velocity of the 2 objects before the collision is equal and opposite to the relative velocity after the collision


9. A 25 g bullet traveling horizontally at

140 m/s is fired into 2.0 kg block,

initially at rest on a level surface

with a coefficient of kinetic friction

of 0.28. The bullet gets imbedded in

the block. The momentum

transferred to the block causes it to

slide along the surface. Note: There

are 2 parts to this problem: collision

(before to immediately after)

followed by the sliding of the block.

It is important that you draw 3

diagrams here: before collision,

immediately after collision, and the

sliding with constant acceleration

along surface.

a) Determine the velocity of the block-bullet combination immediately after the bullet collides with the

block (before it starts to slide). The bullet and block together can be considered an isolated system (no

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net force or impulse) if we examine the collision immediately before and immediately after it happens so

that before the collision, the impulse of gravity on the bullet is negligible, and right after the collision,

the impulse of friction on the block is also negligible. Then no net external forces act on the system

(block weight and normal add to zero, and the internal interaction forces sum to zero). Therefore, the

momentum of the block+bullet system is conserved from immediately before to immediately after the

collision (psys=0 because Fsys = 0) .

b) Determine how far the bullet and the block slide along the surface before coming to rest. Once the

collision is over, the bullet has transferred its momentum and some energy to the block and so the block

and bullet together have horizontal velocity, v’, causing the block-bullet to slide along the surface. Since

there is friction, the block/bullet will slow down with constant acceleration.

We can find the distance it slides in 2 ways

METHOD 1: Newton’s Laws and kinematics. Once the collision is over, all the forces acting on the

sliding block are constant and therefore the acceleration is constant (we can use kinematics to find x

once we find acceleration (we know v0 and vf)). If you use Newton’s Laws to find acceleration, you must

follow the steps you have learned: FBD, apply Newton’s 2nd

. Diagram above has the FBD

Now we know three variables for kinematics

METHOD 2: Conservation of Energy. Once the collision is over, the bullet and block have kinetic

energy. However, the system of mechanical energy, which includes the bullet+block+Earth are is NOT

isolated because friction does negative work on the system removing kinetic energy from it and

converting the lost kinetic energy to nonmechanical energy such as heat and sound. Although

mechanical energy is not conserved, total energy is (as always) conserved

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BEFORE

AFTER

mS+mC

(ptotal=+6 kg m/s)

c) How much energy was lost as the bullet was lodged in the block? Because the collision was perfectly

inelastic (the 2 objects stuck together), kinetic energy was lost.

d) The collision of the bullet with the block took place over 0.0022s. Determine the average net force

exerted on the bullet. The impulse ON THE BULLET caused the momentum change OF THE BULLET

10. A 1.20-kg skateboard is coasting along the

pavement at a speed of 5.00 m/s when a 0.800-

kg cat drops from a tree vertically downward

onto the skateboard. What is the speed of the

skateboard-cat combination? (Remember that

momentum is a vector and that conservation of

momentum is a vector equation that is applied

separately in x- and y- directions)

The cat falls vertically so it carries no horizontal

momentum. Therefore the initial momentum of

6 kg m/s carried by the skateboard does not

change. What does change is the increased mass

carrying the 6 kg m/s momentum and so the

velocity goes down once the cat’s mass is added

to the skateboard.

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11

Horizontal momentum: The cat and skateboard together can be considered an isolated system (no net force or

impulse) because

- the impulse of gravity on the cat is vertical and does not affect horizontal momentum and

- the skateboard moves with constant velocity which means there is no friction acting on it (no net

horizontal force).

Then no net horizontal external forces act on the system (the internal interaction forces sum to zero).

Therefore, the horizontal momentum of the cat+skateboard system is conserved

(psys x=0 because Fsys x = 0)

Vertical momentum (which we are not concerned about in this problem but it is important to consider the

concepts): The impulse of gravity on the cat is a large net impulse that results in a transfer of momentum to

the Earth (py cat = -py Earth) so in order to conserve momentum vertically, we would have to add the earth to

the system of cat and skateboard. Since the force of gravity on the cat due to the earth is equal and opposite

to the force of gravity on the earth due to the cat, then there is no net external force (impulse) on the

cat+skateboard+Earth system

In the x-direction: Before the collision, the cat has no horizontal momentum. All the px is carried by the skateboard. After the

collision, the horizontal momentum is conserved and since it is now carried by the increased mass of

skateboard and cat, the speed of the skateboard goes down (and the cat gains horizontal speed). The both

travel together at v’ (this is a perfectly inelastic collision)

11. A 65.0-kg person throws a 0.045 0-kg snowball forward with a ground speed of 30.0 m/s. A second person,

with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving

forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two

people after the snowball is exchanged? Disregard the friction between the skates and the ice.

The two skaters + snowball together can be considered an isolated system (no net horizontal force or

impulse) because there is no net external force on the system: no air resistance, no friction between skates

and ice, normal and weight on skaters cancel, and the vertical impulse on the snowball does not affect the

horizontal momentum. Therefore, the horizontal momentum of the two skaters + snowball system is

conserved.

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v2=0

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v2=0mS

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m1

v2'v1'

m2+mS

Before

After

(ptotal=+163 kg m/s)

After

Before

12. A 2.00 kg block is sliding at 15.0 m/s to the right. It collides with and sticks to a 1.00 kg block attached to a

spring connected to a wall as shown. The horizontal surface is frictionless. The spring has a spring constant

of 120 N/m. Note: There are 2 parts to this problem: collision (before to immediately after) followed by

blocks compressing spring.

a) Determine the velocity of the double block combination immediately after the collision.

To analyze the collision part of the problem, we must look at momentum (mechanical energy is not

conserved in this collision and the amount of energy lost during the collision is not known). The two

blocks together can be considered an isolated system (no net force or impulse) if we examine the

collision before and immediately after it happens before the compression of the spring begins and the

spring force acts on the blocks as an external force. Then no net external forces act on the system (block

weights and normal add to zero, there is no friction, and the internal interaction forces sum to zero).

Therefore, the momentum of the two block system is conserved from before to immediately after the

collision.

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v1=15m/s

Before

collision

(ptotal=+30 kg m/s) m1=2kg

m1 m2

m2=1kg

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collision

k=120N/m

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compression

b) What will be the maximum distance the spring is compressed? Once the collision is over, the first

block has transferred some of its momentum to the second block so that the two blocks move together

horizontally causing the spring to compress.

(Note: cannot use conservation of momentum for the second part since the spring is attached to the

wall/earth which exerts a large external net force on the blocks/spring system. In order to have

conservation of momentum, the Earth would have to be included in the system making the problem

intractable.)

To analyze the second part of the problem in which the blocks compress the spring, it is easiest to use

conservation of mechanical energy; the system of the two blocks and the spring is energetically isolated

since there are no external forces doing work on the system (no friction, normal force and gravity do no

work). Therefore, the mechanical energy of the blocks/spring system is conserved. Initial point is when

spring is not compressed and blocks have kinetic energy. Final point is when spring is maximally

compressed (all the K was transformed into US)

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13. A rifle, which has a mass of 5.50 kg, fires a bullet, which has a mass of m=65.0grams, at a ballistic

pendulum. The ballistic pendulum consists of a block of wood, which has a mass M = 5.00 kg, attached to

two strings of length L=1.25m. When the block is struck by the bullet, the block swings backward until the

angle between the ballistics

pendulum and the vertical reaches a

maximum angle of 38o. Note: There

are 2 parts to this problem: collision

(before to immediately after)

followed by the swinging of

pendulum.

a) Determine the maximum gravitational potential energy of the ballistic pendulum-Earth system when it

reaches its maximum angle. If we define the top of the block as the height of zero gravitational potential

energy, then the max height the block swings to is h = L – Lcos38 = 1.25(1-cos38) = 0.265 m. Potential

energy at that max height is

b) Determine the velocity of the block of wood+bullet immediately after being struck

After being struck, the collision is over. Once the collision is over, the bullet has transferred its

momentum (and kinetic energy) to block+bullet so that it swings up.

(Note: we cannot use conservation of momentum for this second part because the block-bullet is

attached to a string which is attached to the wall/earth; the Earth (via the string) therefore exerts an

external net force on the blocks/bullet system which changes its momentum. In order to have

conservation of momentum, the Earth would have to be included in the system making the problem

intractable.)

To analyze the second part of the problem in which the block/bullet swings as a pendulum, it is easiest

to use conservation of mechanical energy; the system of the block+bullet+Earth is energetically isolated

since there are no external forces doing work on the system (no friction, tension force is centripetal and

does no work). Therefore, the mechanical energy of the block+bullet+Earth system is conserved. Initial

point is bottom of pendulum (UG=0). Final point is top at max height (max UG) (all the K was

transformed into US)

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c) Determine the velocity of the bullet, vB, immediately before colliding with the block of wood.

To analyze the collision part of the problem, we must look at momentum (mechanical energy is not

conserved in this collision and the amount of energy lost during the collision is not known). The

bullet+block together can be considered an isolated system (no net force or impulse) if we examine the

collision before and immediately after it happens before the block swings and the tension force acts on

the block+bullet as an external force. Then no net external forces act on the system (block weights and

tension add to zero, the impulse of gravity on bullet does not affect horizontal momentum, and the

internal interaction forces sum to zero). Therefore, the momentum of the bullet+block system is

conserved from before to immediately after the collision.

d) How much work was done on the bullet as it got stopped by the block of wood (in other words, how

much energy was lost in the collision?)

The collision was perfectly elastic, so kinetic energy was lost

The amount of mechanical energy lost in the collision is due to the work done by the block on the bullet

in slowing it down

e) Determine the recoil velocity of the rifle. The interaction between rifle+bullet is a “collision” or

explosion. The system of rifle + bullet is isolated because there are no net forces acting on the system.

Therefore, momentum of the system is conserved.

f) How much energy was released when the bullet was fired (in other words, how much energy was gained

in explosion?)

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Documents

BEFORE Conservation of Momentum Problems 1 =9000 m …