A下滑过程机械能守恒:
(1/2)m1v1^2=m1gR+(1/2)m1v0^2,
解得v1=5m/s.
碰撞后A上滑过程中机械能守恒:
m1gR=(1/2)m1v1’^2,
解得v1'=4m/s.
碰撞过程中系统动量守恒、机械能守恒:
m1v1=m2v2-m1v1'
(1/2)m1v1^2=(1/2)m2v2^2+(1/2)m1v1'^2
代入数值,解得v2=1m/s,
m2:m1=9:1.得解.
A下滑过程机械能守恒:
(1/2)m1v1^2=m1gR+(1/2)m1v0^2,
解得v1=5m/s.
碰撞后A上滑过程中机械能守恒:
m1gR=(1/2)m1v1’^2,
解得v1'=4m/s.
碰撞过程中系统动量守恒、机械能守恒:
m1v1=m2v2-m1v1'
(1/2)m1v1^2=(1/2)m2v2^2+(1/2)m1v1'^2
代入数值,解得v2=1m/s,
m2:m1=9:1.得解.