Physics question....it's easy I think but I'm confused on where to begin.?

Question

An arrow, starting from rest, leaves the bow with a speed of 30.0 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

Answer 1

F = 1/2mv^2

So, doubling the force would equate to an increase in the velocity by sqrt(2). Final velocity would be 30*sqrt(2) m/s or about 42.43 m/s.

Answer 2

The momentum of an object is its mass times its velocity. You can figure out the change in momentum of an object by multiplying the average force applied by the time it was applied.

p_change = F_average * t

If you apply a force that's twice as large for the same amount of time, the momentum will be twice as large. Since momentum is just m*v, and m doesn't change, the velocity will be twice as large.

However, if the bow exerts twice as much force, the arrow probably won't be in contact with the string for the as much time. If we assume the string exerts a constant force F over a set distance (say, 0.5 meters), then stops exerting any force at all, then doubling the force will not produce twice the velocity, since the arrow will cover the 0.5 meters in a shorter amount of time. In this case, twice the force will mean the arrow is in contact with the string for only 1/sqrt(2) as much time.

Using our change-of-momentum equation, we can see that
F * t will only be 2/sqrt(2) = sqrt(2) times as fast. In this case (which is more realistic than a constant amount of time), doubling the force will producing 1.414 times the velocity.

Answer 3

you have already answered your question the arrow will leave the bow at a speed of 30m/s