AP Physics 1 Practice Multiple Choice Dynamics 2
1 A Sum the forces for the system to solve for acceleration.
Σ𝐹!"! = 𝐹
(𝑚# + 𝑚$ )𝑎 = 𝐹
5𝑎 = 20 𝑎 = 4 m/s$
The surfaces of the masses are pressing against each other, which means that the internal horizontal forces acting on
each mass are the normal forces. These normal forces are equal and opposite (3rd law), which means you can solve for
the forces on either mass. Choose a mass and sum the horizontal forces acting on that mass only. The 2.0 kg mass only
has one horizontal force acting on it making it easier to solve. The normal force acting on mass 2 is positive (rightward).
Σ𝐹$ &' ()!! *+," = 𝑁
𝑚$ &' 𝑎 = 𝑁 substitute the acceleration from above
(2)(4) = 𝑁 𝑁 = 8N
2 C When solving for acceleration of compound bodies, sum forces for the entire system. Since the masses are all connected
and have the same acceleration treat all the masses like one big mass.
The 24 N force is moving all three blocks together. Add all three blocks.
ΣFsys = F ( m1 + m2 + m3 ) a = F (3 + 2 + 1) a = 24 a = 4 m s2
3 D When solving for the force between connected masses, normal force or tension, sum forces for all the masses that are
pushed by that normal force or pulled by the string.
The string between mass 2 and mass 3 is pulling on mass 1 and mass 2. Sum force for mass 1 and 2.
( )( )
ΣF1 = m1a = 3 4 = 12 N ( )( )
ΣF2 = m2 a = 2 4 = 8 N String must pull both masses T = 12 + 8 = 20 N
4 C The system is stationary (equilibrium), which means the tension and gravity forces are equal and opposite. The upper
rope is holding up both blocks with a total mass of 4 kg.
T = Fg = mg = (4)(10) = 40 N
5 D The forces of tension in the lower string are equal and opposite. The lower string is only holding up the bottom block.
T = Fg = mg = (2)(10) = 20 N
The tension acting on the upper block is directed downward.
6 D Most problems state the string or rope is massless. However, this rope has significant mass that must be included.
At very top of the rope the tension in the rope supporting the block and the rope itself with a total weight of 150 N.
The bottom of the rope is only holding up the block with a total weight of 100 N.
Tension is the rope varies between these values.
7 B The small mass m is being pushed to the right and accelerated by a normal force. Vertically the mass remains
stationary.
x-direction y-direction
SF = N f = Fg
ma = N µN = mg
N = mg / µ Sub this into the equation at the left
ma = mg / µ
a=g/µ
8 A Acceleration Sum Forces of entire system.
ΣFsys = Fg (2m) − Fg (m)
1
( m + 2m) a = 2mg − mg a=
3
g If given letters with no units. Answer in letter (leave g as g ) and no units.
AP Physics 1 Practice Multiple Choice Dynamics 2
9 B Acceleration Sum Forces of entire system.
ΣFsys = Fg (2m)
2mg 2
( m + 2m) a = 2mg a=
3m
= g
3
Variable problem: Given letters with no units. Answer with letter (leave g as g ) and no units.
10 B Tension: Sum forces for either the mass pulled by the string or the mass hanging from the string (not the system).
Pulled mass Hanging mass
ΣFm = T ΣF2m = Fg − T
ma = T 2ma = 2mg − T
Plug in a from #21 Plug in a from #21
⎛2 ⎞ ⎛2 ⎞
m ⎜ g⎟ = T 2m ⎜ g ⎟ = 2mg − T
⎝3 ⎠ ⎝3 ⎠
2mg 4mg
T= T = 2mg −
3 3
2mg
T=
3
11 C v = 0. Balanced forces.
f( 2m) = Fg (m)
µ 2mg = mg a = 0.5
Variable problem: Given letters with no units. Answer with letter (leave g as g ) and no units.
12 A Tension: Masses are not moving. Can use balanced forces for either mass.
Pulled mass Hanging mass
T = f( 2m) T = Fg (m)
T = µ 2mg T = mg
Plug in μ from #23 Done
( )
T = 0.5 2mg
T = mg
Either mass works. Sum forces for the mass with the fewest and easiest forces acting on it.
Variable problem: Given letters with no units. Answer with letter (leave g as g ) and no units.
13 D Centripetal means center seeking. The centripetal acceleration and force point to the center. Centrifugal means away
from the center. While you may have heard of centrifugal force, this is actually a FALSE force. It is actually a feeling of
force, NOT a force, that is caused by an object’s inertia.
14 C
( 2v )
2
v2
In the formula ac = if both v and r double, then a doubles 2ac = .
r ( 2r )
In the formula Fc = mac if a doubles, the Fc doubles ( )
2Fc = m 2ac
15 B If a mass at the top of a loop barely completes the loop, then there is no normal force (roller coasters) or tension (for
v2
masses swung on a string). The only force toward the center is gravity: Fc = Fg m = mg v = rg
r
AP Physics 1 Practice Multiple Choice Dynamics 2
16 B Weightless at the top of a loop is another trigger word for the scenario in the problem above. v = rg
17 D
(10) = 0.5
2
v2 v2
Fc = f m = µ mg µ= µ=
r rg ( 20) (10)
18 E In uniform circular motion (circular motion at a constant speed) the net force (centripetal force) is always directed
toward the center.
19 C v2
Tension is the only force vector acting along the radius, and it is equal to force centripetal. T = Fc = m
r
1. T =
mv 2
2.
1
T =
mv 2
3. 2T =
( 2m) v 2
4. T =
( 2m) v 2
r 2 ( )
2r r ( 2r )
20 B The bus is turning left, which means the center of the circle is to the left. Forces in circular motion are centripetial and
point toward the center of the circular motion.
21 D Mass m is being pushed toward the center of the circle by a normal force. Vertically the mass remains stationary.
x-direction y-direction
FC = N f = Fg
2
mv / R = N µN = mg
N = mg / µ Sub this into the equation at the left
mv2 / R = mg / µ
µ = Rg / v2
22 B Combine the two circular motion equations that contain tangential velocity and solve for the period T .
v2
ac = v = ac r
r
2π r 2π r 2π r 2π r 2π r 2π r
v= T= = = = T=
T v ac r ac r ac ac
Quadrupling the diameter also quadruples the radius
2T =
2π ( 4r )
ac
23 D The tension in the string splits into components. The tension is the horizontal direction is pointing in the direction of the
turn and is equal to the centripetal force. Vertically the object is stationary, so the vertical tension equals the force of
gravity.
x-direction y-direction
FC = Tx Ty = Fg
mv2 / r = T sin q T cos q = mg
T = mg / cos q Sub this into the equation at the left
mv2 / r = (mg / cos q ) sin q
v2 / r = g tan q
tan q = v2 / gr
