a. Momentum conserved. m*225=(5+.06)V
solve for V
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.
4 answers
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) ?
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.
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.