1984M3
a.
kmv
mg
b.
c.
d.
F = ma; mg – kmv = ma; a = g – kv; dv/dt = g – kv
F = 0, which gives 0 = g – kvT; vT = g/k
𝑑𝑣
= 𝑔 − 𝑘𝑣
𝑑𝑡
𝑑𝑣
= 𝑑𝑡
𝑔 − 𝑘𝑣
𝑣
∫
0
𝑡
𝑑𝑣
= ∫ 𝑑𝑡
𝑔 − 𝑘𝑣
0
1
𝑣
− 𝑙𝑛|𝑔 − 𝑘𝑣||0 = 𝑡
𝑘
ln(𝑔 − 𝑘𝑣) − ln(𝑔) = ln (
𝑔 − 𝑘𝑣
) = −𝑘𝑡
𝑔
𝑔 − 𝑘𝑣
= 𝑒 −𝑘𝑡
𝑔
𝑔 − 𝑘𝑣 = 𝑔𝑒 −𝑘𝑡
𝑣=
𝑔
(1 − 𝑒 −𝑘𝑡 )
𝑘
e.
1990M1
a. F = ma; F = –kv = ma; a0 = –kv0/m (negative sign indicates to the left)
b.
𝑑𝑣
𝑘𝑣
=−
𝑑𝑡
𝑚
𝑑𝑣
𝑘
= − 𝑑𝑡
𝑣
𝑚
𝑣
𝑡
𝑑𝑣
𝑘
= ∫ − 𝑑𝑡
𝑚
𝑣0 𝑣
0
∫
𝑣
𝑘
𝑙𝑛 | | = − 𝑡
𝑣0
𝑚
𝑘
𝑣
= 𝑒 −𝑚 𝑡
𝑣0
𝑘
𝑣 = 𝑣0 𝑒 −𝑚𝑡
2001B1
a.
b.
c.
The minimum speed occurs when gravity alone supplies the necessary centripetal force at the top of the circle
(i.e. tension is zero and is not required). Therefore we have Mg = Mvmin2/R which gives 𝑣𝑚𝑖𝑛 = √𝑅𝑔
At the bottom of the swing F = ma becomes T – Mg = Mv2/R which gives Tmax – Mg = Mvmax2/R and solving
𝑅
for vmax gives 𝑣𝑚𝑎𝑥 = √ (𝑇𝑚𝑎𝑥 − 𝑀𝑔)
𝑀
d.
At point P the ball is moving straight up. If the string breaks at that point, the ball would continue to move
straight up, slowing down until it reaches a maximum height and fall straight back to the ground.
1998B1
a. Fext = mtota gives mg = 2ma, or a = g/2
ℎ
b.
d = v0t + ½ at2; h = 0 + ½ (g/2)t2 gives 𝑡 = 2√
c.
d.
e.
Block A accelerates across the table with an acceleration equal to block B (g/2).
Block A is still in motion, but with no more applied force, Block A will move at constant speed across the table.
Since block B falls straight to the floor and stops, the distance between the landing points is equal to the
horizontal distance block A lands from the edge of the table. The speed with which block A leaves the tabletop
𝑔
𝑔
ℎ
2
𝑔
is the speed with which block B landed, which is found from v = v0 + at = (2√ ) = √ℎ𝑔 and the time for
ℎ
block A to reach the floor is found from 2h = ½ gt2, which gives 𝑡 = 2√ .
𝑔
ℎ
The distance is now d = vt = √ℎ𝑔 × 2√ = 2h
𝑔