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早稲田大学教育学部【物理】解答例
Ⅰ
問1 V1 = −e
√
2gh
::::::::
√
問2 V2 = −e 2gh +
L
1+e
√
√
g
L
,w2 = 2gh +
2h
1+e
::::::::::::::::::::::
L
1+e
問3 H1 =
{
e−
L
4(1 + e)h
√
g
2h
:::::::::::::::::::
}
問4 w3 =
:::::::::::::::::::::::
(m − M )w2 + 2M V2
m+M
::::::::::::::::::::
94
h
25
:::
問5 H2 =
問6 小球 A:M aA = M g − kx ,小球 B:maB = kx + mg
(M + m)k
x
Mm
:::::::::::::
問7 aG = g ,aBA = −
:
√ (
)
2
Mg
h−
問8 T1 =
g
k
√
問9 T2 = 2π
:::::::::::::::
Mm
(M + m)k
::::::::::::::
Ⅱ
問1 右図 問2 E =
問3 I =
eV
V
,u =
L
kL
::
:::
Sne2
V kL
:::::::
問6 IR =
問4 P =
V2
V 2T
[W] ,Q =
[J] R
R
:::::::
::::::::
RV
V
,VR =
r:::::
+R
r:::::
+R
問7 金属棒の消費電力は,
IR VR =
R
V2
(r + R)2
= (
V2
√
r
√ + R
R
)2 5
V2
4r
等号成立のとき,消費電力は最大となり,このとき
√
r
√ = R ゆえに R = :r
R
4RV 2
4RV 2
問8 Pa =
,
P
=
b
(2r + R)2
(2R + r)2
:::::::::
:::::::::
問5 6
倍
: