急物理動量功課

Let the angle subtended by ball b be @ and the mass of each ball be m.
By conservation of energy,
PE lost by ball a = KE gain by ball a
m(10)(0.1)=0.5mu^2, where u is the speed of ball a just before collision
u=sqrt 2 m/s
By conservation of momentum,
total momentum before collsion = total momentum after collision
m(sqrt 2) + m(0) = m(0) + mv, where v is the speed of ball a just after collision
v=sqrt 2 m/s
By conservation of energy,
KE lost by ball b = PE gained by ball b
0.5m(sqrt 2)^2=m(10)h, where h is the magnitude vertical displacement of ball b after collision
h = 0.1m
Hence cos @ = (0.5-0.1)/0.5
@ = 36.9 degree

這題非常簡單，碰撞為彈性碰撞，所以碰撞的情況不需考慮，只消考慮最初與最終的情況就可以了。

用能量守恆定律

A球失去的勢能 = B球得到的勢能

mgh1 = mgh2

h2 = h1

所以，B球得到最高的位移也是10 cm

而最大的角度則為cos^-1(50 - 10)/50 = 36.9*