(1) 依题意画出如图所示:
∵
t1 =
=
=" 0.4" s (1分)
∴
s1 =
vo t1 = 5×0.4 =" 2" m (2分)
球与地碰前瞬间竖直方向的速度:
v1 =
=" "
=" 4" m/s (1分)
碰后瞬间竖直方向的速度
v2 =
v1 =
×4 =" 3" m/s(1分)
∴
t2 =
v2 /
g =" 3/10" =" 0.3" s (1分)
与地碰后水平方向的速度
vo’ =
vo =
×5 =" 4" m/s(1分)
∴
s2 =
vo’ t2 = 4×0.3 =" 1.2" m(1分)
而球与地面接触时的位移为
s3 =
(
vo +
vo’)
t =
×(5 + 4)×0.02 =" 0.09" m(2分)
∴
s =
s1 +
s2 +
s3 =" 2" + 1.2 + 0.09 =" 3.29" m(1分)
(2) 球与地面碰撞时的弹力为:
(
FN -
mg)
t =
m v2 -
m(-
v1)(3分)
解得
FN = [
m(
v2 +
v1)/
t ] +
mg= [0.5(3 + 4)/0.02 ] + 0.5×10 =" 180" N (1分)
球与地面碰撞时的摩擦力为
f ,由动量定理,有:
-
f t =
m vo’ -
m vo (2分)
解得:
∴
μ =
f /
FN =" 25/180" =" 5/36" ≈ 0.14 (1分)