A 3.0-g bullet traveling at 350 m/s hits a tree and slows down uniformly to a stop while penetrating...?

Question

a distance of 12 cm into the tree's trunk. What was the force exerted on the bullet in bringing it to rest? If you could show me how you got the answer so I can understand it for a similar problem next time I'd appreciate it. Thanks

Answer 1

Force is a change in momentum over time. Thus, delp/delt = m*delv/delt = m(v1 - v0)/(t1 - t0) = F the force to stop the bullet of mass m = .003 kg going v0 = 350 mps and stopping v1 = 0 mps. So we have everything we need to know except delt, the amount of time the bullent took to stop in the tree.

We can find delt from v1^2 = v0^2 - 2 (delv/delt) S; where S = .12 m, the stopping distance in the tree. [NB: This is just the SUVAT equation V^2 = U^2 + 2 AS.) Solve for delt; v0^2 = 2 delv/delt S; so that delt = 2 (v1 - v0)/(S*v0^2); where S = .12 m, v0 = 350 mps, and v1 = 0. You can do the math.

Plug delt back into m*delv/delt = m(v1 - v0)/delt = F = -mv0/delt and solve for F. Note the minus sign, which connotes a decelerating force. You can do the math.

Key to solving this kind of problems is to recognize that the stopping force F = m delv/delt, which is just F = ma. And couple that with the given SUVAT equation to find delt, the stoppage time, and you can find the force.

Answer 2

Kinetic potential (additionally complete) E of the Bullet formerly hitting the tree is: E= (a million/2) m v^2, the place m= 3g =3*10^(-3) kg and v = 350m/s. So potential E =(a million/2) 3*10^(-3) (350)^2 Joules =183.seventy 5 Joules. Now if F Newtons is the (consistent) tension which the tree exerts on the bullet mutually because it extremely is penetrating in it.Then, the paintings W finished via this tension is: W = F*s , the place s = 12 cm = 0.12m. So: W = 0.12 F Joules.This paintings is comparable to the kinetic potential of the bullet.(paintings required to convey the bullet to sit down down back or its kinetic potential to 0). So: 0.12 F =183.seventy 5 Or F=1531.25 N