AP Physics Free Response Practice – Dynamics – ANSWERS
SECTION A – Linear Dynamics
1.
a.
b. F can be resolved into two components: F sin  acting into the incline and F cos  acting up
the incline.
The normal force is then calculated with F = 0; N – F sin  – mg cos  = 0 and f = N
Putting this together gives F = ma; F cos  – mg sin  – (F sin  + mg cos ) = ma, solve
for a
c. for constant velocity, a = 0 in the above equation becomes F cos  – mg sin  – (F sin  +
mg cos ) = 0
𝜇 cos 𝜃+sin 𝜃
solving for F gives 𝐹 = 𝑚𝑔 (cos 𝜃−𝜇 sin 𝜃) In order that F remain positive (acting to the
right), the denominator must remain positive. That is cos  >  sin , or tan  < 1/
2.
a. Combining the person and the platform into one object, held up by two sides of the rope we
have F = ma;
2T = (80 kg + 20 kg)g giving T = 500 N
b. Similarly, F = ma; 2T – 1000 N = (100 kg)(2 m/s2) giving T = 600 N
c. For the person only: F = ma; N + 600 N – mg = ma gives N = 360 N
3.
a.
i.
ii.
iii.
iv.
v.
b. The maximum friction force on the blocks on the table is f2max = s2N2 = s2(m1 + m2)g
which is balanced by the weight of the hanging mass: Mg = s2(m1 + m2)g giving M = s2(m1
+ m2 )
c.
For the hanging block: Mg – T = Ma; For the two blocks on the plane: T – f2 = (m1 +
m2)a
Combining these equations (by adding them to eliminate T) and solving for a gives 𝑎 =
𝑀−𝜇𝑘2 (𝑚1 +𝑚2 )
[ 𝑀+𝑚
]𝑔
+𝑚
1
d.
i.
2
f1 = k1m1g = m1a1 giving a1 = k1g
ii. For the two blocks: Mg – T = Ma2 and T – f1 – f2 = m2a2. Eliminating T and substituting
𝑀−𝜇𝑘1 𝑚1 −𝜇𝑘2 (𝑚1 +𝑚2 )
values for friction gives 𝑎2 = [
]𝑔
𝑀+𝑚
2
4.
a.
b. Fy = 0; N + F1 sin  – mg = 0 gives N = mg – F1 sin 
c. Fx = ma; F1 cos  – N = ma1. Substituting N from above gives  = (F1 cos  – ma1)/(mg –
F1 sin )
d.
e. The condition for the block losing contact is when the normal force goes to zero, which
means friction is zero as well. Fx = Fmax cos  = mamax and Fy = Fmax sin  – mg = 0
giving Fmax = mg/(sin ) and amax = (Fmax cos )/m which results in amax = g cot 
5.
a.
b. F = ma
Ma = Mg sin  – bv
M(dv/dt) = Mg sin  – bv
c. F = 0 so Mg sin  = bvT; vT = (Mgsin)/b
d.
𝑑𝑣
= (𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣)/𝑀
𝑑𝑡
𝑑𝑣
1
= 𝑑𝑡
𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣 𝑀
𝑣
∫
0
𝑑𝑣
1 𝑡
= ∫ 𝑑𝑡
𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣 𝑀 0
1
𝑡
𝑣
− 𝑙𝑛|𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣||0 =
𝑏
𝑀
𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣
𝑏𝑡
ln(𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣) − ln(𝑀𝑔𝑠𝑖𝑛𝜃) = ln (
)=−
𝑀𝑔𝑠𝑖𝑛𝜃
𝑀
𝑏𝑡
𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣
= 𝑒−𝑀
𝑀𝑔𝑠𝑖𝑛𝜃
𝑏𝑡
𝑀𝑔𝑠𝑖𝑛𝜃 − 𝑏𝑣 = 𝑀𝑔𝑠𝑖𝑛𝜃𝑒 − 𝑀
𝑣=
e.
𝑏𝑡
𝑀𝑔𝑠𝑖𝑛𝜃
(1 − 𝑒 − 𝑀 )
𝑏
6.
a. f = FN where FN is the radial force acting inward on the mass from the wall
FN = mac = mv2/R; f = mv2/R
b. the tangential acceleration comes from the frictional force along the wall
f = –mat (negative as it opposes the motion of the block)
mv2/R = –mat; at = dv/dt = –v2/R
c.
𝑑𝑣
𝜇𝑣 2
=−
𝑑𝑡
𝑅
𝑑𝑣
𝜇
= − 𝑑𝑡
2
𝑣
𝑅
𝑣0
3 𝑑𝑣
𝜇 𝑇
∫
= − ∫ 𝑑𝑡
2
𝑅 0
𝑣0 𝑣
𝑣0
1 3
𝜇𝑇
− | =−
𝑣 𝑣0
𝑅
1
1
2
𝜇𝑇
− 𝑣 − (− ) = − = −
0
𝑣0
𝑣0
𝑅
3
𝑇=
2𝑅
𝜇𝑣0