Physics Question

Now, there is no external force acting on the system,

So, by the law of conservation of momentum

Initial momentum = Final momentum

0 = mv + Mv'

0 = 10(0.02)(1500) + 60v'

Velocity of the man, v' = -5 ms-1

This is the take-off speed of the man, in which it is in horizontal direction.

So, consider the vertical motion of the man,

by s = ut + 1/2 gt2

100 = 0 + 1/2 (10)t2

Time of flight, t = 4.47 s

Therefore, the horizontal distance required

= v't

= (5)(4.47)

= 22.4 m



2009-05-08 07:46:17 補充：
The WEIGHT of the bullets is 0.02g

Since it is WEIGHT, so g is the gravitational field strength (Acceleration due to gravity), but not the unit gram.

So, the mass of the bullets should be 0.02 kg, not 0.02 X 10^-3 kg

In this question, the terms "weighs" and "weight" both mean "mass". Otherwise, it is not reasonable to amend "a weight of 0.02 g" but not to amend "he weighs 60 kg".

2009-05-08 09:53:01 補充：
Besides, 0.02 g is too small and not reasonable (If the density is 5 g/cm^-3, the volume will be 0.004 cm^3. It is very small). I think it is 0.02 kg and mistyped.