Transcript Lecture9_Collisions

A bowling ball and ping-pong
ball are rolling towards you with
the same momentum. Which
ball is moving toward you with
the greater speed?
A) the bowling ball
B) the ping pong ball
C) same speed for both
Even the lightest bowling ball (a 6-pounder) is
1000 heavier than a ping pong ball (~2.7 oz).
The ping ball would have to move ~1000 times faster!
A bowling ball and ping-pong
ball are rolling towards you
with the same momentum. If
you exert the same force in
stopping each, which takes a
longer time to bring to rest?
A) the bowling ball
B) the ping pong ball
C) same time for both
for which is the
stopping distance greater?
If a cue ball moving with velocity v strikes a stationary
v
v=0
billiard ball head-on and comes to an abrupt halt,
v
v=0
its target ball moves off with the same velocity v.
During their brief contact
Force of red
billiard ball
on cue ball
stopping cue ball
Force of
= cue ball on
billiard ball
launching target ball
Notice: mv + 0 = 0 + mv
During their brief contact
FA pushes B = - FB pushes A
FAB t = - FBA t
over the
same time!
mB vB - mB vB  -(m AvA - m Av A )
Here I use:
vB to represent B’s initial velocity v0
and v'B to represent B’s final velocity.
m Av A  mB vB  m AvA  mB vB
The total momentum remains unchanged!
We say: Momentum is conserved.
v
A fast moving car traveling with a
speed v rear-ends an identical model
(and total mass) car idling in neutral
at the intersection. They lock bumpers
on impact and move forward at
A)
B)
C)
D)
0 (both stop).
v/4
v/2
v
A heavy truck and light car both
traveling at the speed limit v, have
a head-on collision. If they lock bumpers
on impact they skid together to the
A) right
B) left
Under what conditions would they stop dead?
A heavy truck and light car have a head-on
collision bringing them to a sudden stop.
Which vehicle experienced
the greater force of impact?
the greater impulse?
the greater change in momentum?
the greater acceleration?
A) the truck
B) the car
C) both the same
A 100 kg astronaut at rest catches a
50 kg meteor moving toward him at
9 m/sec. If the astronaut manages
to hold onto the meteor after catching
it, what speed does he pick up?
A)
B)
C)
D)
E)
F)
3 m/sec
4.5 m/sec
9 m/sec
15 m/sec
18 m/sec
some other speed
(100 kg0)+(50 kg9m/s) = (150kg)v'
v' =450 kg·m/s
150kg
For these two vehicles to be stopped
dead in their tracks by a collision at
this intersection
A)
B)
C)
D)
They must have equal mass
They must have equal speed
both A and B
is IMPOSSIBLE
Car A has a mass of 900 kg and is travelling
east at a speed of 10 m/sec. Car B has a
mass of 600 kg and is travelling north at
a speed of 20 m/sec. The two cars collide,
and lock bumpers. Neglecting friction which
arrow best represents the direction the
combined wreck travels?
A
B
C
900 kg
10 m/sec
600 kg
20 m/sec
Car A has a mass of 900 kg and is travelling east at a speed of
10 m/sec. Car B has a mass of 500 kg and is travelling north at
a speed of 25 m/sec. The two cars collide and stick together.
Neglecting friction Which of the arrows best represents the
direction the combined wreck travels?
A
B
C
900 kg
10 m/sec
600 kg
20 m/sec
The answer is (A). Momentum is conserved, so the total momentum before the
collision must equal the total momentum after. But momentum is a vector, so we have
to add the vector arrows (by sliding them and placing them head to tail). The
momentum of car A is mass  velocity = 9000 kg m/s in the direction of east. The
momentum of car B is mass  velocity = 12000 kg m/s north. Draw these vectors,
making sure to make the lengths proportional to the momenta.
A projectile with initial speed v0 scatters off
a target (as shown) with final speed vf.
mvf
mv0
The direction its target is sent recoiling
is best represented by
B
C
A
D
T
E
G
F
A projectile with initial speed v0 scatters off
a target (as shown) with final speed vf.
mvf
mv0
The sum of the final momentum (the
scattered projectile and the recoiling
target) must be the same as the initial
momentum of the projectile!
F
SOME ANSWERS
Question 1
B) the ping pong ball
Even the lightest bowling ball (a 6-pounder) is
1000 heavier than a ping pong ball (~2.7 oz).
The ping ball would have to move ~1000 times faster!
Question 2
C) same time for both
v
V
So the bowling ball is not moving very fast, while the ping pong ball must be moving at a pretty high speed.
But we’re told both have the same momentum! To stop either one means to
remove its momentum completely.All it has (mv) to 0. So both must undergo the
exact same loss in momentum. The stopping time can be figured out from the
t must be the
momentum change needed:
same for each!
Ft  (mv)
same for each
Question 3
B) the ping pong ball
The bowling ball reaches you will a small speed, v, which you slow to zero.
During those t seconds, it travels with an average speed v/2, moving a distance
(v/2)t before stopping. In the same amount of time, the ping pong ball travels
much farther: (V/2)t.
Or we can note that stopping distance can be directly computed using:
Fd  mv  mv  v
1
2
2
1
2
same for each
bigger v
needs
bigger d
SOME MORE ANSWERS
and since they
m1v1  m2v2  m1v'1 m2v'2 lock bumpers and
Question 4 C) v/2
v2=0
move together
m1v1  (m1  m2 )v'
v’1=v’2
and since m1 = m2 we don’t need to distinguish them by different labels.
mv  (m  m)v'  2mv'
Question 5
mv  2mv'
v'  v / 2
A) right
Question 6
• Since forces are equal and opposite, both experience the same force.
• Since both experience the same force in the same time, they both receive
the same impulse.
• Since they both have the same impulse, they both must undergo the same
change in momentum.
• Since they both experience the same force, the less massive car has a
greater acceleration, since a = F/m.
Question 7
A) 3 m/sec
(100 kg0)+(50 kg9m/s) = (150kg)v'
v' =450 kg·m/s
150kg
D) is IMPOSSIBLE Momentum is conserved, so the total
momentum before the collision must
Question 9
A
equal the total momentum after. But
momentum is a vector - we have to add the vector arrows (by sliding them so they
meet head to tail). The momentum of car A is mass velocity = 9000 kg m/s EAST.
The momentum of car B is mass velocity = 12000 kg m/s NORTH. Just try
drawing these vectors, making sure their lengths are proportional to the momenta.
Question 8
Question 10
The question becomes:
mvf
plus WHAT? =
mv0
F
F