动量部分习题参考答案.pdf

文档介绍：动量
2-1 2 2 −16 o .
PB = Pe + Pv = ×10 g ⋅cm / s. θ= 30
m Mm m2
2-2 (1)木块的速率 v = v0 和动量 p木= v0 ;子弹的动量 p v .
M + m M + m 子 M + m 0
Mm
(2)子弹施予木块的动量 I = v .
木 M + m 0
2-3 I = m(T0 − mg)l = ⋅m / s
ft1 ft1 ft2
2-4 v1 = , v2 = + .
m1 + m2 m1 + m2 m2
2-5 S船= . (对岸), S人= −S船+ S人对船= .(对岸).
v + v
2-6 0 乙
m乙= m货= 300kg .
v0 − v乙
m m
2-7 v = v + u , v = v , v = v − u
前 M + m 中后 M + m
Nm
2-8 (1) v = − u
车 M + Nm
1 1 1 1
(2) v = −m[ + + + + ]u
车N M + Nm M + (N −1)m L M + 2m M + m
(3) 比较(1)和(2),显然有 v车N > v车.
2 2 2 −1 v1 −1 v1 −1 2v0 cosθ 0
2-9 v2 = v1 + 4v0 cos θ0 , α= tg = sin = cos .
2v0 cosθ 0 v2 v2
240mv 240 ×10 ×10−3 × 900
2-10 F t = = = 36N
t 60
v0 M 0
2-11 (1) a0 = μ− g ; (2) μ= (a0 + g) = 735kg / s .
M 0 v0
m
2-12 (1) v = c ln 0 = 2500ln 3; v = 5000ln 3; v = 7500ln 3 = / s .
1 m 2 3
60
(2) v = 2500ln = 2500ln5 = / s .
60 − 48
dm
2-13 F = (v + u) 为向前的推力, 此式的 v、u 为绝对值.
dt
dm
2-14 (1) 水平总推力为 F = v (向前)
dt
(2) 以上问题的答案不改变
2-15 质点受力 f = ma = −mω 2 r , 恒指向原点.
2-16 F > μ(mA + mB )g
2-17
⎧ m g sinθ cosθ m g
a = − 2 = − 2
⎪ 1x 2
⎪ m2 + m1 sin θ(m1 + m2 )tgθ+ m2ctgθ
⎨
(m + m )g (m + m )gtgθ
⎪a = − 1 2 sin 2 θ= − 1 2
⎪ 1y 2
⎩ m2 + m1 sin θ(m1 + m2 )tgθ+ m2ctgθ
⎧ m1 g sinθ cosθ m1 g
⎪a2x = 2 =
⎨ m2 + m1 sin θ(m1 + m2 )tgθ+ m2ctgθ
⎪