Answer to Question #90913 in Mechanics | Relativity for layla

Question

An object of mass m = 100g falls freely during 3sec:

a) Find the received impulse,

b) Deduce the change of the speed

Answer 1

a) Impulse is the change of momentum $p=mv$ of an object when the object is acted upon by a force $F$  for an interval of time $\Delta t$ . The formula for impulse is:

$$Impulse=Force\times time=F\cdot \Delta t$$

Object falls under the force of gravity, which is

$$F=mg$$

where m = 100g = 0.1kg,   g=9.8m/s² is the acceleration of gravity. Then we get

$$Impulse=0.1\,kg\cdot 9.8\,m/{{s}^{2}}\cdot 3\,s=\text{2}\text{.94}\,kg\cdot \,m/s$$

b) Because impulse is a measure of how much the momentum changes as a result of force acting on it for a period of time, an alternative formula for impulse is:

$$Impulse=\Delta p={{p}_{final}}-{{p}_{initial}}$$ $$=m{{v}_{final}}-m{{v}_{initial}}=m\Delta v$$

Then change of the speed is

$$\Delta v=\frac{Impulse}{m}=\frac{\text{2}\text{,94}\,kg\cdot \,m/s}{0.1\,kg}=29.4\,m/s$$