vdv = ∫ 2 xdx
10
x
v ( x) = 2x
dx = 2dt x
dx v= = 2 x 得: dt
两边积分: 两边积分
t dx ∫10 x = ∫0 2dt x
得:
x = 10e
2t
1 解:
练习二
v0
v 与水平方向 成45度 v cos 45 = v 0 cos 60 v sin 45 = v 0 sin 60 − gt ∴t = 3 − 1 v0 ⋅ 2 g
FN
f
a
mg
f g tan θ + ω l cos θ ∴µ ≥ = 2 FN g − ω l sin θ
2
f − mg sin θ = mω 2 l cos θ cos θ ....(1) mg cos θ − FN = mω 2 l cos θ sin θ ....(2) f ≤ µ FN ..........................................(3) f g tan θ + ω 2 l cos θ ∴µ ≥ = 2 FN g − ω l sin θ
2 2
v2 = 7.80km / s R = 6.38 ×103km ∴v1 = 7.97km / s
2.动量不守恒, 2.动量不守恒,角动量守恒 动量不守恒
T
mV 人 R − mV 物 R = 0..(1) V 人 = U − V 物 ...(2) U ∴V 物 = 2
V物 V人 mg mg
U
练习六
v v v v v ∆r = r3 − r1 = 16i − 2 j
v v r1 = 2i
v v v r3 = 18i − 2 j
dv dv dx dv = =v = 2x 2.解：（ ） a = 解：（1） dt dx dt dx
得: 两边积分: 两边积分 得:
vdv = 2 xdx
∫
v
10 2
5 i ∆U = ν R∆T = 2 × × 8.31 × 100 = 4155( J ) 2 2
T2 − f − m2g sin 30 = m2a...(3) f = FN µ = µm2g cos30....(4) a = βr ...(5)
m2g
15 − 3 −2 a= = 2.65ms 5
练习九 1.解：角动量守恒 解 mvR sin θ = ( M + m ) R 2ω
mv sin θ ∴ω = ( M + m)R 2.解：角动量守恒 解
M =Jβ
0 − 1200 ⋅ 2π / 60 ωt − ω0 t= =− ≈ 4π ( s ) − 9.8 × 0.2 / 0.2 β
（2）由角动量定理
由 ωt
= ω 0 + βt
M gµ ∴β = =− J R
Mt = J ωt − J ω0
2.解： 解
FN f
T2 a T1 a m1g
m1g − T1 = m1a...(1) T1r − T2r = J β ....(2)
练习一
x = 2t 2 1.解：（ ）由运动方程 解：（1） y = 1− t 得质点的轨迹方程为
x = 2(1 − y ) 2
（2）1s和3s时的位置矢量分别是 ） 和 时的位置矢量分别是
v v v v v v ∆r 16i − 2 j v= = = 8i − j ∆t 3 −1 v v v v v (3) v = 4ti − j a = 4i
2.解: 解
3 kT = 5.6 ×10−21 J 2 m PV = M RT ρ=m V RT −2 8.31× 273 M =ρ = 1.43 × 10 = 32 ×10−3 kg / mol P 1.013 ×103 w=
氧气
练习十四 练习十四 1 2 5 m 1.解: 2 mv = 2 M R∆T 解 Mv 2 2 × 10−3 × 102 ∆T = = = 4.8 ×10−3 k 5R 5 × 8.31
3.
m0 5 m= = = m0 2 2 0 .6 3 1− v / c
m0
5 5 2 E = mc = m0 c = E0 3 3
2
练习十三 练习十三 1.解: 解
M mol 32 × 10−3 (1) : m0 = = = 5.3 × 10−26 kg N0 6.023 × 1023 (2) : I = 2m0 vx = 2mv cos 60 = 2.65 × 10−23 N ⋅ s I ⋅1023 ⋅ dt (3) : F = = 2.65 N dt 2.67 F (4) : P = = = 1.33 × 104 Pa s 2 × 10−4
a = β R...(3) 1 J = MR 2 ...(4) 2
kx
T1 mg
得:
2 m g − 2 kx a= ...(5) 2m + M
由:
dv dv a = = ⋅v dt dx
(4 mg − 2 kx ) x v=± ...(6) 2m + M
得:
机械能守恒: 解(2) 机械能守恒
1 2 1 2 1 2 0 = −mgx + mv + Jω + kx ...(1) 2 2 2
+
∫ MVdt + ∫ mvdt = 0
0 0
t
t
MS M + ms m = 0...(1)
Q sm = smM + S M ...( 2)
smM = l
SM
...(3)
MV + mv = 0Leabharlann m l =− m+M
练习五
1.以地心为固定点,卫星角动量守恒 1.以地心为固定点, 以地心为固定点
mv1 (R + 2.00 ×10 ) = mv2 (R + 3.47 ×10 )
v
v 与水平方向 成45度 v x = v 0 cos 60 v y = −v x = v 0 sin 60 − gt ∴t = 3 + 1 v0 ⋅ 2 g
2.
v aτ = a n = ....(1) R v t = v 0 + aτ t .....( 2) t = 5 / 3s
2
1 2 s = v 0 t + aτ t = 35 / 6( m ) 2 s 35 θ= = R 72
2 2
a0
a + a0 = 5.9m / s
1.
I = mv...(1)
练习四
v2 T − mg = m ...( 2) l ∴ I = (T − mg )ml = 0.86( kgm / s )
2.设劈尖、 2.设劈尖、球对地的水平速度 设劈尖 分别为V 分别为V、v，方向向右 水平方向动量守恒
由热力学第一定律
a d 若氢气体积不变， 1.解：(1)若氢气体积不变，热量变成气体的内能，则 V 若氢气体积不变 热量变成气体的内能，
5 Q = ν R∆T 2
0 2Q ∆T = = 12( K ) 5ν R
T2 = T1 + ∆T = 273 + 12 = 285( K )
T2 5 P2 = P = 1.0575 × 10 ( Pa ) 1 T1
m1 ∴x = b m1 + m2
2.解: 水平方向动量守恒 解 机械能守恒
2MgR v= M +m
0 = MV + mv
1 1 2 MV + mv 2 = mgR 2 2
练习八
1.解： 摩擦阻力矩 解
dM = − df ⋅ R = − Rµgdm ∴ M = − µmgR
(1)由转动定律 (1)由转动定律
m R ∆P = ∆ T = 1.99( Pa ) M V
2.解: 解
vp = v = v
2
2 RT = M mol 8 RT = π M mol = 3 RT = M mol
2 × 8 . 31 × 300 = 393 m / s −3 32 × 10 8 × 8 . 31 × 300 = 446 m / s −3 3 . 14 × 32 × 10 3 × 8 . 31 × 300 = 483 m / s −3 32 × 10
6mv 6mv0 ω= Ml + 3ml
棒上摆过程中机械能守恒
1 1 l 2 2 ( Ml )ω = Mg (1 − cos 60)...(3) 2 3 2
2 12m 2 v0 θ = arccos 1 − 2 ( M + 3m) gl
2. 解:(1) mg − T1 = ma ...(1) T1 R − kxR = J β ...(2)
3.
a = aτ + a = 3 2m / s
2 2 n
2
2
v雨地 = 2 2km / h
V人地 2 v雨地
练习三 1.
f cos θ − FN sin θ = mω 2 l cos θ ....(1) f sin θ + FN cos θ − mg = 0...........(2) f ≤ µ FN .......................................(3)
2.解:(1)对等体过程 解 对等体过程
i 5 ∆U = ν R∆T = 2 × × 8.31 × 100 = 4155( J ) 2 2
Q = ∆U + A = ∆U = 4155 J
A = P ∆V = ν R ∆T = 2 × 8.31 × 100 = 1662( J )
（2）对等压过程 ）
1.解: 解 2.解: 解
A = ∫ Fx dx = ∫ (3 x + 4)dx = 20(J )