Transcript m 1 v 1i + m 2 v 2i

Chapter 6
Momentum
Let’s start with the following situation:
A 70kg box travelling at 5m/s is given an extra push of
140N for 10s. What will be it’s new velocity?
F = ma
a = F = 140N
m
70kg
m = 70kg
a = 2 m/s2
o = 5m/s
F = 140N
v+ at
t = 10s
= 5m/s + 2m/s2(10s)
vf ?
= 25m/s
Force stops after 10s, so the final velocity is 25m/s.
Now v is constant because there is no more Force (F=0)
Now let’s combine Force & Kinematics:
If:
F = ma
&
a = ∆v = vf – vi
∆t
∆t
then
F = m(vf – vi)
∆t
so
F.∆t = m(vf – vi)
F.∆t = mvf – mvi
Impulse(J)
(think Impact!)
momentum(p) not to be confused with
KE (= ½ mv2 )
Now we can say
Impulse(J) = change in momentum(∆p)
J = ∆p
J = pf – pi
F.∆t = mvf – mvi
Impulse:
J = F.∆t
momentum: p = mv
Note:
∆t & m are scalar,
F, vf & vi are vectors.
vf & vi can differ in magnitude,
direction, or both.
Example:
A car at a stop sign starts then stops suddenly. The car behind it
hits the car at 5mph and dents the bumper in by 10cm before
coming to a halt. The second car has a mass of 2000kg.
Calculate the Force of impact.
A ball bouncing off the ground looks like this:
The Force of Impact can be tracked over time and plotted
on a Force-time (F-t) graph.
A variable Force is difficult to
do calculations with – unless
you use Calculus!
It would be better to know and
work with the Average Force.
Previously we have established that the derivative of a
graph is the slope of the line and it represents the
division of the y-axis and x-axis.
m = dy/dx
E.g. for a v-t graph, m = a
Well, the anti-derivative (integral) is the AREA under the
graph and it represents the the product of the y-axis and
x-axis.
Area = ∫y.dx
E.g. for a v-t graph, Area = x
But for an F-t graph, Area = J
Calculating the area is easier if
the graph consists of straight
lines and regular shapes.
When the area under the graph [Impulse(J)] is determined, it
can be divided by the time interval (∆t) to give us the
Average Force (F).
Favg = J = Area
∆t
∆t
The area under the graph, and the area under the rectangle
are the same, therefore the Impulse is the same.
Example:
A 100g Tee ball is hit with a bat and the Force of Impact is
tracked over time in the graph below. Calculate:
a.) The Impulse imparted onto the ball,
b.) The average Force by the bat, and
c.) The velocity at which the ball takes off.
Playing Pool:
If you hit the white ball straight onto the black ball – what happens?
- The black ball
- The white ball bounces back, but with a reduced velocity.
- So Energy is transferred from the white ball to the black ball.
vWi
vBi = 0
FW
FB
Newton’s 3rd Law applies: “..every action..equal but opposite rn”
So
FW = - F B
FW.∆t = -FB.∆t
∆pW = -∆pB
mW(vWf – vWi) = -mB(vBf – vBi)
} in same ∆t
} Impulse
} change in momentum
mWvWf – mWvWi = -mBvBf + mBvBi
mWvWi + mBvBi = mWvWf + mBvBf
Total momentum before = total momentum after
aka
“Conservation of Momentum”
Perfectly Elastic:
If
AND
i.e.
e.g.
No loss of Energy through sound, heat, light…
pool balls (near enough!)
so
AND
m1v1i + m2v2i = m1v1f + m2v2f applies
v1i – v2i = -(v1f – v2f) } a special case
v1i – v2i = v2f – v1f
} formula
(see p.173 Eqn. 6.12 for derivation)
Inelastic:
If
BUT
i.e.
total pbefore = total pafter
e.g.
Energy IS lost through sound, heat, light…
& Work Energy of Deformation (damage in a crash)
Almost all examples are Inelastic.
so
m1v1i + m2v2i = m1v1f + m2v2f applies
Perfectly Inelastic:
If
total pbefore = total pafter
BUT
KEbefore ≠ KEafter
AND the two objects stick together after the collision
i.e.
e.g.
rail (train) cars, car accidents…
so
m1v1i + m2v2i = m1v1f + m2v2f applies
but it now becomes
m1v1i + m2v2i = (m1 + m2)vf
combined mass
Examples:
Elastic Collision.
p.179 Q.37 A 25.0g ball moving to the right at 20cm/s
overtakes and collides elastically with a 10.0g ball moving in
the same direction at 15.0cm/s. Find the velocity of each ball
after the collision.
m1v1i + m2v2i = m1v1f + m2(v1i – v2i + v1f )
m1v1i + m2v2i = m1v1f + m2(v1i – v2i )+ m2v1f
m1v1i + m2v2i = m1v1f + m2v1f + m2(v1i – v2i )
m1v1i + m2v2i - m2(v1i – v2i ) = v1f (m1+ m2)
v1f = m1v1i + m2v2i - m2(v1i – v2i )
(m1+ m2)
v1f = 25g(20cm/s) + 10g(15cm/s) – 10g(20cm/s – 15cm/s)
(25g + 10g)
v1f =
v2f = v1i – v2i + v1f
v2f = 20cm/s – 15cm/s + 17.1cm/s
v2f =
Example:
InElastic Collision
Jack, a 75kg student is standing on a smooth floor with a pair of
roller skates on. He throws the 800g Physics textbook to Jill
at a velocity of 12m/s. What will be the velocity of Jack?
Example:
Perfectly InElatic Collision.
Jill (65kg), also on roller skates standing still, now catches the
800g textbook travelling at 12m/s. How fast will she be
going?
Example:
Perfectly Inelastic Collision.
A 40 tonne (40,000kg) freight car coasting at a speed of 5m/s
along a track strikes a 30 tonne stationary freight car and
couples to it. What will be their combined speed after impact?
The Ballistic Pendulum
A 15.0g bullet is fired at 280m/s horizontally into a 3.00kg block
of wood suspended by 2 cords. The bullet sticks in the block.
Compute the height the block will rise above the horizontal.
A gun fires a 10.0g bullet horizontally at a 1.90kg block of soft
wood resting on a frictionless table. The bullet emerges at the
other end of the block at a speed of 25.0m/s. The table is
1.20m high and the block lands 0.85m from the edge of the
table. Calculate the muzzle velocity of the gun.
A 7500kg truck is travelling East at 45mph. A 1500kg car is
moving North at 55mph. It approaches the intersection and
ignores the red light. The car crashes into the truck and get
tangled with it.
a.) What speed and direction will the wreckage begin to move?
b.) If the coefficient of friction between the tyres and the road is
0.350, then how far will the wreckage travel?