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...
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.
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