momentum

Question 1

Equation for momentum

Answer 1

p = mv
(units in kg*m/s)

DELTA p = m * DELTA v

Question 2

Define momentum (roughly)

Answer 2

“inertia in motion,” the way to quantify how hard it is to stop something

Question 3

how hard it is to stop something

Answer 3

how hard it is to stop something how hard it is to stop something how hard it is to stop somethinghow hard it is to stop something how hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop something how hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop somethinghow hard it is to stop something

Question 4

units of p

Answer 4

kg * m/s

Question 5

define impulse

Answer 5

capital I = delta p

the change in momentum

Question 6

break down F = ma into delta p using time

Answer 6

F= m * (delta v/ delta t)
multiply by delta t on both sides
F* delta t = m* delta v
F* delta t = delta p

Question 7

What does the area of graph of force vs. time equal? (force is changing over time)

Answer 7

change in momentum

Question 8

units of change in momentum, or impulse I

Answer 8

F * delta t = delta p

N * s

the same as units of momentum (kg * m/s)

Question 9

units of newton

Answer 9

kg * m/s^2

Question 10

define the impulse approximation

Answer 10

DURING the time of impact, the force of impact is so much larger than the other forces that they can be considered negligible, and thus force of impact = net force

this only works when time of collision is very very small

Question 11

use Newton’s second law and Impulse to explain how an airbag stops you from snapping your neck and cracking your skull and dying on the spot

Answer 11

Net F = ma (kgm/s^2)

this equation can be converted to the “momentum version” as so:
Net F = mchange in speed/change in time
Net F = m(Vfinal-Vinitial)/delta t

Net F = force of impact via impulse approximation

F(impact) * delta t = mVf - mvi

neither mass nor speed (with Vf = 0 and Vi and mass being the same regardless of airbag)

also, set m*delta v = delta p, or change in momentum, or IMPULSE

F(impact)* delta t = impulse

so impulse does not change, thus an increase time of impact (the helmet gives your head more time to transfer energy with the ground) results in a decreased force of impact

Question 12

law of conservation of momentum

Answer 12

w/o an external force (often, but NOT ALWAYS gravity/friction), momentum is conserved from before and after ALL collisions

Question 13

mathematically prove that momentum is conserved before and after a collision (assuming no external force)

Answer 13

NEWTONS 3RD LAW: forces are equal and opposite, thus F1 =
-F2, (note the negative as force is a vector) and as they share the same collision, t1 = t2

if Impulse = Ft, then: F1t1 = -F2t2

and if the impulese are equal and opposite, then:

m1(delta v1) = -m2(delta v2)

m1(v1f) - m1(v1i) = -m2(v2f) + m2(v2i)

and, adding the initial momentum of object 1 and final momentum of object 2 to both sides:

m1(v1f) + m2(v2f) = m1(v1i) + m2(v2i)

in words, this means that object 1’s initial momentum + object 2’s initial momentum = 1’s final + 2’s final momentums

or, in simplest terms,

the TOTAL initial momentum = TOTAL final momentum

Question 14

WHEN does the law of conservation of momentum take effect?

Answer 14

during ALL collisions (elastic AND inelastic) and explosions

Question 15

is kinetic energy or potential energy a vector?

Answer 15

no, this is why momentum can be conserved (net 0 if different directions) and ke might not be

Question 16

Newton’s 3rd law of equal and opposite forces, alongside the conservation of momentum implies that impulse is……

Answer 16

also equal and opposite, with equal opposite forces and (as the forces are stated by the 3rd law to be equal and opposite over TIME) time is equal, -Ft = Ft and thus -I = I and impulse is equal if momentums are equal

Question 17

what does the equation m1v01 + m2v02 = mtotal*vtotal represent physically?

Answer 17

an explosion into multiple pieces, like a rocket firing off hot gas downwards and creating greater internal momentum forces

Question 18

elastic and inelastic in problems

Answer 18

elastic = objects split apart

inelastic = objects stick together

Question 19

when an object falls towards the earth, the object gains negative momentum. but momentum must be equal in the earth-object force. what happens to keep the net momentum 0?

Answer 19

the earth gains momentum (gravity attracts BOTH objects towards each other, thus the object’s velocity increases causes the earth to increase velocity as well, though much less as the mass is substantially higher)

Question 20

explain how momentum is conserved and kinetic energy isn’t always on a CONCEPTUAL level (in regards to collisions)

Answer 20

1. momentum is directly derived from Newton’s 3rd Law. as every force within parts of the system result in an equal and opposite change of another part of the system (think a rocket pushing hot gas downwards, and thus the hot gas pushing it upwards), momentum is conserved (with no external forces).
2. kinetic energy is not always conserved (both on objects and in net KE) in anything other than perfect elastic collisions because energy is lost to heat, sound, deformation, etc.

Question 21

a berd perched on an 8.00 cm tall swing has a mass of 52.0 g, and the base of the swing has a mass of 153 g. assume that the swing and bird are originally at rest, and that the bird takes off horizontally at 2.00 m/s. if the base can swing freely (w/o friction) around the pivot, how high will the base of the swing rise above its original level?

Answer 21

Its an EXPLOSION, so momentum is conserved (though ke is not here)
phase 1:
(m1 + m2)v0 = m1v1f + m2v2f
0 = 52(2.0) + 153(v2f)
0 = 104 + 53v2f
v2f = -0.679 m/s

phase 2: conservation of mechanical energy (kinetic energy gets converted to potential as the swing moves up)

0.5mv^2 = mgh
.5v^2 = gh
0.5(-0.679)^2 = 9.8*h

0.2305205/9.8 = h
h = 0.0235 m
h = 2.35 cm

Question 22

why do elastic/bouncing collisions transfer more momentum?

Answer 22

change in momentum will be far greater, ex.

20 ——> -20 means delta p is -40
20 ——-> 0 means delta p is only 20
(collision of balls at 20 m/s but one bounces and one doesnt respectively)

Question 23

why do elastic/bouncy collisions exert more force?

Answer 23

bcz the force not only needs to stop the force as in inelastic, but also turn it around, causing more total force transfer