Monday, July 14, 2014

If a hammer weighs `0.5` kg and travels at `8` m/s and impacts a nail at a constant force of `4*10^5` N and bounces off at the same velocity, how...

Hello!


Let's start from Newton's Second law, `F=ma.` Here `m` is the mass of a hammer, `a` is its acceleration and `F` is the force which acts on it. By Newton's Third law this force is the same in magnitude as the force which acts on a nail, and it is given that it is constant.


Next, recall that acceleration `a` is the derivative of velocity `V,`  `a=V'(t),`  `t` is for time. Then `F=ma=(mV)'` and we can integrate this equality over time interval in question.


At the left side we obtain `F*T` where `T` is the time we have to determine, at the right side we obtain the change of `(mV).`


Denote the initial speed of a hammer as `V_0,` then the change of `(mV)` is equal to `2mV_0,` because the velocity was `V_0` in one direction and becomes `V_0` in the opposite direction, or `-V_0,` and `V_0-(-V_0)=2V_0.`


This way we have  `T=(2mV_0)/F.`


In numbers it is `(2*0.5*8)/(4*10^5)=2*10^(-5) (s).` This is the answer.



That said, it is absolutely impossible that the force was constant. Actually the given value for force means average force during the collision.

No comments:

Post a Comment

How does the choice of details set the tone of the sermon?

Edwards is remembered for his choice of details, particularly in this classic sermon. His goal was not to tell people about his beliefs; he ...