posted by Rob on .
A bullet of mass m=0.060 kg hits a 5.0 kg block with an initial speed of 225 m/s. The block is connected to a spring that is attached to a wall. The friction between the block and the table is negligible. Upon impact, the bullet bounces back from the box with a speed of 75 m/s.
A) Calculate the speed of the block right after the collision.
B) As a result of the collision, the spring compresses to a mazimum of 0.20m. Find the spring constant
--The equation to this is 1/2mv^2 = 1/2kx^2 right? I can't get the answer to this question without the answer to part A.
C) Find the inelastic energy loss during the collision.
a. Momentum conserved. m*225=(5+.06)V
solve for V
I got 2.67 m/s from your equation. But the correct answer is supposed to be 3.6 m/s.
Also, for part C, do I do KE final minus KE initial?
So 1/2(5.06)(75^2) - 1/2(0.06)(225^2) ?
the correct equation is this: (.06)(225) = (5)(v) - (.06)(75)
momentum is conserved. total momentum is equal to the momentum of the block minus the momentum of the bullet. you subtract because the two objects move in opposite directions.