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Impulse and Momentum Review
… This is what’s on the test
Basic Relations , momentum
• Momentum: a vector that is m•v
• Find the momentum of a 40kg bike that is going
10m/s South
• If the momentum of a go cart is 2500kg•m/s east
when it is going 10m/s east, what is its mass?
• If the momentum of a 100kg go cart is
2500kg•m/s east, find its velocity
Basic Relations , momentum
• Find the momentum of a 40kg bike that is going
10m/s South = 400kg•m/s South
• If the momentum of a go cart is 2500kg•m/s east
when it is going 10m/s east, what is its mass?
250kg
• If the momentum of a 100kg go cart is
2500kg•m/s east, find its velocity
25m/s East
Basic Relations , momentum
• If the momentum of a go cart is 2500kg•m/s east,
what would its momentum be if it doubled its
speed?
• If the momentum of a go cart is 2500kg•m/s east,
what would the momentum of another go cart be
if its mass is 25% greater but it has the same
velocity?
Basic Relations, momentum
• If the momentum of a go cart is 2500kg•m/s east,
what would its momentum be if it doubled its
speed? Double, 5000kg•m/s east
• If the momentum of a go cart is 2500kg•m/s east,
what would the momentum of another go cart be
if its mass is 25% greater but it has the same
velocity? 25% greater, 3250kg•m/s
Basic Relations, impulse
• Impulse = F•t
• Find the impulse of a 4000N average force to the
west that acts for 50s.
• Find the impulse of a 4000N average force to the
west that acts for half the time as above.
• Will the change in momentum be the same,
greater, or less for the 50s impulse?
Basic Relations, impulse
• Impulse = F•t
• Find the impulse of a 4000N average force to the
west that acts for 50s. = 200,000N•s west
• Find the impulse of a 4000N average force to the
west that acts for half the time as above. =
100,000N•s west
• Will the change in momentum be the same,
greater, or less for the 50s impulse? Greater (2x
greater)
Basic Relations, impulse is the
change in momentum
• Impulse = F•t = change in momentum
• If (and only if) the momentum change is the same, force
will decrease if time increases.
• An object has 400kg•m/s of momentum to the north
when a 10N force to the north is applied for 2s. Find the
change in momentum and the final momentum.
• An object has 400kg•m/s of momentum to the north
when a 10N force to the south is applied for 2s. Find the
change in momentum and the final momentum.
Basic Relations, impulse is the
change in momentum
• An object has 400kg•m/s of momentum to the
north when a 10N force to the north is applied for
2s. Find the change in momentum and the final
momentum. = 20N*s north; 420kg*m/s north
• An object has 400kg•m/s of momentum to the
north when a 10N force to the south is applied for
2s. Find the change in momentum and the final
momentum. = 20N*s south; 380kg*m/s north
Basic Relations, impulse is the
change in momentum
• An object has 400kg•m/s of momentum to the
north when a 10N force to the east is applied for
20s. Find the change in momentum and the final
momentum.
• An object has 400kg•m/s of momentum to the
north . Find the impulse required to stop it.
Basic Relations, impulse is the
change in momentum
• An object has 400kg•m/s of momentum to the
north when a 10N force to the east is applied for
20s. Find the change in momentum and the final
momentum.
Pythag = 44.7m/s
tan-1 (40/20) = 63 deg North of E
Basic Relations, impulse is the
change in momentum
• An object has 400kg•m/s of momentum to the
north. Find the impulse required to stop it.
Dmv = - 400kg•m/s = -400N*s = 400N*s south
Impulse and Momentum at an
angle
• An object has 400kg•m/s of momentum 30deg
east of south. Find the components in the north
and east.
• An object has 400kg•m/s of momentum 30deg
east of south when a 20N force is applied to it for
20s at 30deg east of south. Find its final
momentum and velocity.
Impulse and Momentum at an
angle
• An object has 400kg•m/s of momentum 30deg
east of south. Find the components in the north
and east.
• 400 sin(30deg) = 200kg•m/s
east
• 400 cos(30deg) = 346.4kg•m/s
east
• Vel = (400kg•m/s)/mass
30deg east of south.
Impulse and Momentum at an
angle
• An object has 400kg•m/s of momentum 30deg
east of south. Find the components in the north
and east.
• 400 sin(30deg) = 200kg•m/s
east
• 400 cos(30deg) = 346.4kg•m/s
east
Impulse and Momentum at an
angle
• An object has 400kg•m/s of momentum 30deg
east of south. If it strikes another object that is at
rest, what must the total momentum be after the
collision in the north and east directions? What
condition are we assuming for this to be true?
Impulse and Momentum at an
angle
• Since momentum must be conserved (if there are
no outside forces), it’s the same as before
• 400 sin(30deg) = 200kg•m/s
east
• 400 cos(30deg) = 346.4kg•m/s
east
Conservation of Momentum
The total linear momentum of an isolated system
remains constant. An isolated system is one for which
the vector sum of the external forces acting on the
system is zero. All of the momentum before an event
must be the same after the event.
m1v1f + m2v2f = m1v1i + m2v2i
Conservation of Momentum
•If a 80kg guy is rolling on his skates very fast
towards his girlfriend at 8m/s while she is rolling
towards him at 2m/s. Find their final velocity if
they hold on to one another after the collision
assuming that she is only half his mass.
•m1v1f + m2v2f = m1v1i + m2v2i
•8okg*v1f + 40kg*v2f = 80kg*8m/s+ 40kg*(-2m/s)
•Since they move together, v1f = v2f = vf
Conservation of Momentum
8okg*v1f + 40kg*v2f = 80kg*8m/s+ 40kg*(-2m/s)
•Since they move together, v1f = v2f = vf
•8okg*vf + 40kg*vf = 80kg*8m/s+ 40kg*(-2m/s)
•8okg*v1f + 40kg*v2f = 560kg*m/s
•120kg(v1f) = 560kg*m/s
•4.67m/s in the boyfriend’s initial direction (which
was made the pos direction above)
Conservation of Momentum
If he pushed her fast enough so that he stops, how
fast would the girlfriend have to be going?
• m1v1f + m2v2f = m1v1i + m2v2i
• 8okg*0m/s+ 40kg*v2f = 80kg*8m/s+ 40kg*(-2m/s)
• 0+ 40kg*v2f = 560kg*m/s
• v2f = (560kg*m/s)/40kg*v2f = 14 m/s in the orig direction
of the boyfriend
Conservation of Momentum
If a superball is 0.4kg and going 25m/s east before striking
a wall and returning at 25m/s, find the force that the wall
exerted if the time of contact was 0.20 sec.
Dmv = F*t =
0.4kg(-25m/s – 25m/s) = F*t
= 0.4kg(-50m/s) = -20kg*m/s = F*t
-20kg*m/s = F*(0.20s)
• F = -100N
Conservation of Momentum
• The drawing shows a collision between
two pucks on an air hockey table. Puck A
has a mass of 0.25 kg and is moving along
the x-axis with a velocity of 5.5 m/s. It
makes a collision with puck B, which has a
mass of 0.5 kg and is initially at rest. After
the collision, the two pucks fly apart with
the angles shown in the drawing. Find the
final speed of (a) puck A and (b) puck B.
• p initial, total = 5.5(.25) = 1.375kg*m/s to
the right.
• After collision in the y direction
mvAsin(65) - mvBsin(37) = 0
• After collision in the x direction
mvAcos(65)+ mvBcos(37) = 1.375kg*m/s
right.
• After collision in the y direction
.25vA(.9063)+ .5vB(-.60) = 0;
• vB = 0.753vA.
• After collision in the x direction
.25vA(.4226)+ .5vB(.7986) = 1.375kg*m/s r
• .25vA(.4226)+ .5(0.753vA)(.7986) = 1.375kg*m/s r
• 0.4063vA = 1.375kg*m/s rt
• vA = 3.4m/s; Now plug back into vB = 0.753vA. = 0.753(3.4m/s)
• vB = 0.753vA. = 0.753(3.4m/s) = 2.55m/s
Impulse again
• If two objects collide, Newton’s 3rd law tells us
that the forces must be equal but opposite
directions. Common sense tells us that if A is
in contact with B for some time, t, then B is
also in contact with A for time, t.
• Therefore, the impulse between two
interacting bodies will also be equal but
opposite (if there are no outside forces)
Test Taking Skills
• Tip 1: Take everything one step at a time and
include units for each step.
• Tip 2: Watch the signs/directions
• Tip 3: When adding vectors, don’t forget to
add head-to-tail
• Tip 4: If you put the correct answer, don’t
change it!! 
• Tip 5: Ask yourself if the answer makes sense.