even the lightest bowling ball (a 6-pounder) is 1000 heavier than a ping pong ball (~2.7 oz)
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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 - PowerPoint PPT PresentationTRANSCRIPT
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?
AA) the bowling ball) the bowling ball
BB) the ping pong ball) the ping pong ball
CC) same speed for both) 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?
AA) the bowling ball) the bowling ball
BB) the ping pong ball) the ping pong ball
CC) same time for both) same time for both
for which is the
stopping distance greater?
stopping each, which takes a longer time to bring to rest?
v v = 0
v = 0
If a cue ball moving with velocity v strikes a stationary
billiard ball head-on and comes to an abrupt halt,
v
its target ball moves off with the same velocity v.
During their brief contact
Force of red Force of billiard ball = cue ball on on cue ball billiard ball
stopping cue ball
Notice: mv + 0 = 0 + mv
launching target ball
During their brief contact
FA pushes B = FB pushes A
FAB t = FBA t over the same time!
)(AAAABBBB
vmvmvmvm
Here I use:vB to represent B’s initial velocity v0
and v'B to represent B’s final velocity.
BBAABBAAvmvmvmvm
The total momentum remains unchanged!
We say: Momentum is conserved.
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) 0 (both stop).B) v/4C) v/2D) v
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) rightB) 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 truckB) the carC) both the same
A 100 kg astronaut at rest catches a 50 kg meteor moving toward him at 9 m/sec. If the astronaut managesto hold onto the meteor after catching it, what speed does he pick up?
A) 3 m/secB) 4.5 m/secC) 9 m/secD) 15 m/secE) 18 m/secF) some other speed
(100 kg0)+(50 kg9m/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) They must have equal massB) They must have equal speedC) both A and BD) is IMPOSSIBLE
A
B
C
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?
900 kg10 m/sec
600 kg20 m/sec
A
B
C
Car A has a mass of 900 kg and is travelling east at a speed of10 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?
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.
900 kg10 m/sec
600 kg20 m/sec
mv0
mvf
A projectile with initial speed v0 scatters off a target (as shown) with final speed vf.
The direction its target is sent recoiling is best represented by
A
T
B CD
E
G F
mv0
mvf
A projectile with initial speed v0 scatters off a target (as shown) with final speed vf.
The sum of the final momentum (the scattered projectile and the recoilingtarget) must be the same as the initialmomentum of the projectile!
F
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 momentum change needed: )(mvFt
same for each
t must be the same for each!
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.
v V
Or we can note that stopping distance can be directly computed using:
vmvmvFd 212
21
same for each
bigger vneeds
bigger d
CC) same time for both) same time for both
Question 2
BB) the ping pong ball) the ping pong ballQuestion 3
SOME ANSWERS
BB) the ping pong ball) the ping pong ballQuestion 1
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!
C) v/2Question 4
SOME MORE ANSWERS
22112211'' vmvmvmvm
v2=0
and since they lock bumpers and
move togetherv’1=v’2
')(2111
vmmvm and since m1 = m2 we don’t need to distinguish them by different labels.
'2')( mvvmmmv '2mvmv 2/' vv
A) rightQuestion 5
• 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 6
A) 3 m/secQuestion 7 (100 kg0)+(50 kg9m/s) = (150kg)v'
v' =450 kg·m/s 150kg
D) is IMPOSSIBLEQuestion 8
AQuestion 9
Momentum is conserved, so the total momentum before the collision must 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.
FF
Question 10
mvf
The question becomes:
plus WHAT? =mv0