In a ballistics test, a 20.0kg bullet traveling horizontally at 1100m/s goes thr
ID: 2186020 • Letter: I
Question
In a ballistics test, a 20.0kg bullet traveling horizontally at 1100m/s goes through a 40.0cm -thick 450kg stationary target and emerges with a speed of 900m/s . The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? so i saw on other solutiions it was ((900)^2-(1100)^2)/(2*.4) then that's -500,000 (900-1100)/(500,000) = .000625 but that's wrong also What average force does the bullet exert on the target? What is the target's speed just after the bullet emerges?Explanation / Answer
Let's first calculate the speed of the target when the bullet emerges using conservation of momentum 20*1100=450*v+20*900 solve for v, this is the speed of the target v=20*200/450 =8.89 m/s Now consider the bullet w/r/t the target from the moment of impact to the moment the bullet emerges from the target. This is an accelerating reference frame, so I will assume the acceleration is constant. In this reference frame, the starting speed of the bullet is 1100 m/s, and the emerging speed of the bullet is 900-8.89m/s or 891.11 m/s the equations of motion of the bullet are x(t)=v0*t-.5*a*t^2 and v(t)=v0-a*t where t is in seconds while the bullet is in the target, and a is the acceleration magnitude. solve for t, the time the bullet is the target. With the values, the two equations are: 0.40=1100*t-.5*a*t^2 891.11=1100-a*t find a,t from above equations (two eqns-two unknows) now that we have t, consider the target in the Earth's frame of reference v=a*t And F=m*a, where F is the average force the bullet exerts on the target