弯弓射虎

big bird · Nov 28th, 2012, 21:25

Acceleration (a) is not important here, velocity at the time of contact and the duration of the contact are! The law to use is not Newton's 2nd law of motion, but rather the law of conservation of momentum and conservation of mass-energy. The formulas to use include m1 * v1 + m2 * v2 = constant, f * t = m * v, 0.5 * m * v * v = m * g * h, etc. If elasticity is included (as in all realistic situation), it becomes more complicated.

Your pushing of a mass off the table is misleading. f = m * a means that if there is a force, there is acceleration. But when a mass is pushed off the table, the force is not there anymore. So the mass will fall way short of the 1 meter that you suggested.

How did you pass 高考?

Nov 28th, 2012, 21:25	#1
big bird Senior Member 注册日期: Sep 2009 帖子: 543	Acceleration (a) is not important here, velocity at the time of contact and the duration of the contact are! The law to use is not Newton's 2nd law of motion, but rather the law of conservation of momentum and conservation of mass-energy. The formulas to use include m1 * v1 + m2 * v2 = constant, f * t = m * v, 0.5 * m * v * v = m * g * h, etc. If elasticity is included (as in all realistic situation), it becomes more complicated. Your pushing of a mass off the table is misleading. f = m * a means that if there is a force, there is acceleration. But when a mass is pushed off the table, the force is not there anymore. So the mass will fall way short of the 1 meter that you suggested. How did you pass 高考?
big bird Senior Member 注册日期: Sep 2009 帖子: 543	老實談太極,認真打太極,追求真太極,嚴防假大空.