AP Physics C
1.
2.
3.
4.
Momentum and Collisions
As fuel burns the mass of a rocket decreases. If at a certain
point the mass of a rocket is half its original mass and
velocity is twice the original velocity, how are momentum
and energy affected?
(A) momentum and kinetic energy are unchanged.
(B) momentum is unchanged and kinetic energy is
doubled.
(C) momentum is halved and kinetic energy is doubled.
(D) momentum is halved and kinetic energy is
unchanged.
(E) momentum is halved and kinetic energy is
quadrupled.
A 5 kg mass undergoes a change in momentum of 10
kg·m/s during a 4 s period of time. What is its impulse?
(A) need to know force
(D) 40 kg·m/s
(B) 10 kg·m/s
(E) 50 kg·m/s
(C) 20 kg·m/s
A 5000 kg rocket is at rest in deep space. The rocket burns
fuel pushing 10 kg of exhaust gases rearward at 4000 m/s.
This process takes 10 s. During the time interval what
average force is applied to the rocket?
(A) 1000 N
(D) 4000 N
(B) 2000 N
(E) 5000 N
(C) 3000 N
A 2 kg ball going 10 m/s follows a path that is
perpendicular to the surface of a wall. It impacts the wall
and looses 20 % of its momentum in the collision. What is
the balls impulse?
(A) 4 kg·m/s
(D) 36 kg·m/s
(B) 8 kg·m/s
(E) 40 kg·m/s
(C) 16 kg·m/s
4
F
3
1
6.
7.
A 5000 kg rocket is at rest in deep space. The rocket burns
fuel pushing 10 kg of exhaust gases rearward at 4000 m/s.
This process takes 10 s. What is the total momentum of the
system?
(A) 0 kg·m/s
(D) 50,000 kg·m/s
(B) 10,000 kg·m/s
(E) 90,000 kg·m/s
(C) 40,000 kg·m/s
2
3
4
v
m
9.
Two objects both with mass m and velocity v intersect
and collide perfectly inelastically as shown above.
Determine the total momentum of this system.
(A) mv
(D) 2 2mv
(E)
4mv
(B)
2mv
(C) 2mv
10. A 5000 kg rocket is at rest in deep space. The rocket burns
fuel, pushing 10 kg of exhaust gases rearward at 4000 m/s.
This process takes 10 s. How fast is the rocket going as a
result of this process?
(A) need more info
(D) 12.5 m/s
(B) 4 m/s
(E) 20 m/s
(C) 8 m/s
11. A large mass explodes into two smaller masses. Mass m
moves to the right at speed v. Mass M moves to the left
at speed V. Calculate the speed V of mass M.
mv
Mv
(D)
(A)
( M + m)
( M + m)
(B)
M
mv
(C)
mv
M
t
What would be the speed of a 3 kg mass, originally
traveling at 2 m/s, that undergoes the motion graphed
above?
(A) Need more info.
(D) 3 m/s
(B) 1 m/s
(E) 4 m/s
(C) 2 m/s
When no external forces act, momentum is conserved in
(A) elastic collisions.
(D) explosions.
(B) inelastic collisions.
(E) All are correct.
(C) perfectly inelastic collisions.
A 2 kg ball moving at 3 m/s collides with a 4 kg that is
stationary. If the collision is perfectly inelastic, what is the
speed of the 2 kg ball after the collision?
(A) −1 m/s
(D) 1.5 m/s
(B) −0.5 m/s
(E) 2.0 m/s
(C) 1 m/s
v
m
2
1
5.
8.
MC Sample
(E)
mv
M
( − m)
12. A large mass M explodes into two smaller masses. Mass
m moves to the right at speed v. What is the speed of the
remaining piece of mass that must go to the left?
mv
Mv
(A)
(D)
( M + m)
( M + m)
(B)
M
mv
(C)
mv
M
(E)
mv
( M − m)
13. Two objects have the same momentum. How can one of
the objects have more energy?
(A) It has a higher impulse than the other object.
(B) It has more mass than the other object.
(C) It weighs more than the other object.
(D) It is moving faster than the other object.
(E) This is not possible.
Answers
1.
B
In the formula p = mv , if mass is halved and speed doubled the changes cancel out. Momentum is unchanged. In the formula
1 2
mv , if mass is halved, then K would be halved. If speed doubles, then K is quadrupled due to the square. Combining
2
half and quadrupled results in a net doubling of K.
K=
2.
B
Impulse equals change in momentum. The other numbers in the question are just distracters. The other answers are multiples of
the various bogus numbers given.
3.
D
To solve for force try the impulse formula.
∆p = F ⋅ ∆t
m ( v − v0 ) = F ⋅ ∆t
F=
m ( v − v0 )
∆t
=
( 5000 )(8 − 0 )
=
(10 )
4000 N
4.
D
J = ∆p = m ( v − v0 ) = ( 2 ) ⎡⎣( −0.8 (10 ) ) − (10 ) ⎤⎦ = 36kg ⋅ m s
5.
E
In graphs look for intercepts, slopes, areas, data points, and general trends. Slope and area are the biggest in this class. Slope is
rise over run, or force divided by time. This does seem like anything. Area is base time height, but formulas always are in the
height times base order. Try force times time and we get the impulse formula. Also known as a change in momentum.
The area = ∆p = 6 kg·m/s
( 6)
∆p
v=
+ v0 =
+ ( 2) = 4 m s
∆p = m ( v − v0 )
m
( 3)
6.
E
As long as no external forces act momentum is conserved in all collisions and explosions.
7.
C
Perfectly inelastic means that they stick together, so the 2 kg mass becomes part of the larger object and has the same velocity as
the combined mass
m v + m2 v2i ( 2 )( 3) + ( 4 )( 0 )
m1v1i + m2 v2i = ( m1 + m2 ) v f
v f = 1 1i
=
= 1m s
( m1 + m2 )
( 2 + 4)
8.
A
In a simple two mass explosion each particle moves away with equal and opposite momentum. Momentum is a vector, so equal
and opposite add together for a total of zero.
9.
B
px = mv and p y = mv . Anytime that you have the x and y components of a vector simply Pythagorean Theorem them to
find the overall vector. p =
10. C
11. C
px 2 + p y 2 =
( mv )
2
+ ( mv ) = 2 ( mv ) = 2mv
2
2
This is an explosion, with exhaust going one way and the rocket going the other.
m
v
(10 )( 4000 )
vrocket = exhaust exhaust =
= 8m s
mrocket vrocket = mexhaust vexhaust
mrocket
( 5000 )
( m + M ) vi = mv f
+ MV f
0 = mv f + MV f
mv f = − MV f
But you need the initial speed of combined mass. Assume the simplest case: It must be stationary at the start.
Technically one mass must have negative momentum, but it is asking for speed (scalar: no direction).
mv
mv = MV
V=
M
12. E
At first glance this seems to be problem 11 slightly reworded, but it is subtly different. The original mass M breaks into to two
masses, m and another mass which is never given. The missing mass is what remains of M after m breaks off, M − m. Again
assume that the original mass M is standing still at the start, and that speed does not involve plus or minus signs.
mv
Mvi = mv f + ( M − m ) V f
0 = mv + ( M − m ) V f
mv = ( M − m ) V f
Vf =
( M − m)
Problems 11 and 12 are two versions of a similar problem. Are there more? Note the wording looks similar at a glance, and the
answers are the same for both. It is easy to make two versions of a test that are very different. Which version will you have?
13. D
If there are two objects the one going faster has to have less mass in order to have the same momentum. But, when it comes to
energy velocity is squared. So the object with more speed has a higher kinetic energy without having more momentum.