I get that the force of a collision is m(Δv/Δt). What I do not get is why the change in velocity for a straight-line collision m(2v/Δt). Why is this?


Firstly I think you might be a tad confused about what's going on. It's not the change in velocity that equals $\frac{2mv}{t}$ but that is the force.

Suppose we have a perfectly elastic collision of a ball hitting a wall. The force the ball exerts on the wall is indeed $$F=ma=m\frac{dv}{dt}$$ However we can also express this force as $$F=\frac{d}{dt}(mv)$$ As $p=mv$ it can be said that $$F=\frac{dp}{dt}$$In fact this is the original formulation of $F=ma$ and it states that the force is equal to the change in momentum over time.

For an elastic collision in a straight line where the object rebounds in the same direction the velocity after the collision will be of the same magnitude but in the opposite direction. So if an object starts out with velocity $2v$ after the collision it will have velocity $-2v$.

This means that the force, F is $$F=\frac{m(v_0)-m(-v_0))}{t}$$And therefore $$F=\frac{2mv}{t}$$I hope this is a clear explanation for you :).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.