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...

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.