Take Home Test

Impulse Momentum, and Collisions

Take Home TestAn 8.00 g bullet is fired into a 250 g

block that is initially at rest at the edge of a table of height 1.00 m. The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

What type of collision is this?

Perfectly Inelastic Collision

fii vmmvmvm )( 212211

Knowsm1= .008 kgm2 = 0.250kgV2i = 0V1f = V2f

h=Δy =1.00mΔx = 2.00m

UnknownsV1i = ?V1f = V2f = ?

2

2

1tatvy giy

We must first figure out the combined Vf, how can we do that?

What else about this situation do we know? 0iyv2/81.9 smag

What else about this situation don’t we know? ?t

yavv gif 222 2/43.42 smyav gf

Can we figure out time, then Vf?

sa

yt

g

452.02

Now what?

2

2

1tatvx xix

smt

xvvtvx fixix /43.4

Vix = Vf, because velocity is Constant in the x-axis, no accel. In the X. Projectile

1

22211

)(

m

vmvmmv ifi

But wait! There’s more

Take Home TestAn 8.00 g bullet is fired into a 250 g

block that is initially at rest at the edge of a table of height 1.00 m. The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

What type of collision is this?

Perfectly Inelastic Collision

fii vmmvmvm )( 212211

Knowsm1= .008 kgm2 = 0.250kgV2i = 0V1f = V2f

h=Δy =1.00mΔx = 2.00m

UnknownsV1i = ?V1f = V2f = ?

2/43.42 smyav gf 1

22211

)(

m

vmvmmv ifi

smx

xxv i /143

1000.8

043.4)102501000.8(3

33

1

Take Home Test A 1200 kg car traveling initially with a

speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9000 kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east.

What is the velocity of the truck right after the collision?

How much mechanical energy is lost in the

collision? Account for this loss in energy.

What type of collision is this?

Elastic Collision

ffii vmvmvmvm 22112211

smvm

vmvmvma f

fii /9.20) 22

112211

222

211

222

211 2

1

2

1

2

1

2

1

00)

iiff vmvmvmvmE

KEPEKEEb

Take Home TestA 1200 kg car traveling initially with a

speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9000 kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east.

What is the velocity of the truck right after the collision?

How much mechanical energy is lost in the

collision? Account for this loss in energy.

What type of collision is this?

Elastic Collision

ffii vmvmvmvm 22112211

smvm

vmvmvma f

fii /9.20) 22

112211

222

211

222

211 2

1

2

1

2

1

2

1

0)

ffii

fiif

vmvmvmvmE

KEKEEKEKEE

KEEPEKEEb

lostJxE ,105.1 4

2222 9.20*90002

10.18*1200*

2

10.20*9000

2

10.25*1200

2

1E

Take Home TestA block of mass m1 = 1.60 kg, initially moving

to the right with a velocity of 4.00 m/s on a frictionless horizontal track, collides with a massless spring attached to a second block of mass m2 = 2.10 kg moving to the left with a velocity of -2.50 m/s. The spring has a spring constant of 6.00 X 102 N/m.

Determine the velocity of block 2 at the instant when block 1 is moving to the right with a velocity of +3.00 m/s.

Find the compression of the spring.

ffii vmvmvmvm 22112211

smvm

vmvmvmf

fii /74.122

112211

Now we have solved for a). What is the next step to solving for the compression of the spring xf?

sg PEPEKEE

sfgffsigii PEPEKEPEPEKE

Since there is no friction, hence no loss of energy how could we simplify this?

Elastic Collision

Conservation of Momentum

Conservation of Energy ->

Take Home TestA block of mass m1 = 1.60 kg, initially

moving to the right with a velocity of 4.00 m/s on a frictionless horizontal track, collides with a massless spring attached to a second block of mass m2

= 2.10 kg moving to the left with a velocity of -2.50 m/s. The spring has a spring constant of 6.00 X 102 N/m.

Find the compression of the spring.

2222

211

2222

211 2

1

2

1

2

1

2

1

2

1

2

1fffiii

sfgffsigii

kxvmvmkxvmvm

PEPEKEPEPEKE

2222

211

222

211 2

1

2

1

2

10

2

1

2

1fffii kxvmvmvmvm

2)2

1()

2

1

2

1

2

1

2

1(2 22

22211

222

211 fffii kxvmvmvmvm

mxk

vmvmvmvmf

ffii 173.0222

211

222

211

Conservation of Xi = 0,

now solve for Xf

Distribute the 2 canceling all ½

Divide out k and √ both sides

Take Home Test A car with mass 1.50 x 103 kg traveling east at a speed

of 25.0 m/s collides at an intersection with a 2.50 x 103

kg van traveling north at a speed of 20.0 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that frictions between the vehicles and the road can be neglected.

fi pp

yfvcxfvcviyvcixc vmmvmmvmvm )()(

The pf is moving in two dimension of space, oh no!

Δp, but we don’t know the θ

Conservation of momentum, Expand this equation out for perfectly inelastic collision

What type of collision?

xyf pp

yfxff ppp The combined velocities are 2D

cosfxf vv sinfyf vv

sin)(cos)( fvcfvcviyvcixc vmmvmmvmvm

yfxfi ppp

cos)( fvccicx vmmvmp

sin)( fvcvivy vmmvmp

What do we know?We know what cosθ is

We know what sinθ is

pxf pyf

The definition of tangent is the ratio of sinθ to cosθ

Take Home Test A car with mass 1.50 x 103 kg traveling east at a speed

of 25.0 m/s collides at an intersection with a 2.50 x 103

kg van traveling north at a speed of 20.0 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that frictions between the vehicles and the road can be neglected.

fi pp

yfvcxfvcviyvcixc vmmvmmvmvm )()(

The pf is moving in two dimension of space, oh no!

Δp, but we don’t know the θ

Conservation of momentum, Expand this equation out for perfectly inelastic collision

What type of collision?

xyf pp

yfxff ppp The combined velocities are 2D

cosfxf vv sinfyf vv

sin)(cos)( fvcfvcviyvcixc vmmvmmvmvm

yfxfi ppp

cos)( fvccicx vmmvmp

sin)( fvcvivy vmmvmp

What do we know?We know what cosθ is

We know what sinθ is

pxf pyf

The definition of tangent is the ratio of sinθ to cosθ

Take Home TestA car with mass 1.50 x 103 kg traveling east at a speed

of 25.0 m/s collides at an intersection with a 2.50 x 103 kg van traveling north at a speed of 20.0 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that frictions between the vehicles and the road can be neglected.

What type of collision?

cos)( fvccicx vmmvmp

sin)( fvcvivy vmmvmp

The definition of tangent is the ratio of sinθ to cosθWhat can I do with this information?

fvc

cic

vmm

vm

)(cos

fvc

viv

vmm

vm

)(sin

Do I know what cosθ is equal to?

Do I know what sinθ is equal to?What can I do with this information?

cos

sintan

cos

1*

1

sintan

cic

fvc

fvc

viv

vm

vmm

vmm

vm )(*

)(tan

cic

viv

vm

vmtan 1.53tan 1

cic

viv

vm

vm

smvmm

vmf

vc

cic /6.15cos)(

smvmm

vmf

vc

viv /6.15sin)(

Take Home TestConsider a frictionless track. A block of mass m1 = 5.00

kg is released from A. It makes a head-on elastic collision at B with block of mass m2 = 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision m2 moves to the right with a velocity of 3.3 m/s.

02

10 2

11

fi

gffgii

vmmgh

PEKEPEKEsmvgh fi /9.92 1

smvm

vmvmf

fi /3.30

11

2211

ffii vmvmvmvm 22112211

2112

1ff vmmgh m

g

vh ff 556.02

21

The End