Question

Ballistic and Springs-- An 8g bullet is shot into a 2.2-kg block of wood thatis initially at rest. The bullet embeds in the block, and together they slide on a horizontal frictionless surface at 1.2 m/s towards on unstrained ideal spring that is fixed to a wall. The spring is compressed by the block as it comes to a momentary stop. Ignore the effects of air resistance. (a) What is the velocity of the bullet immediately before it hits the block? (b) Calculate the total kinetic energy of the bullet and block before and after the bullet hits the block. (c) If the spring constant is 400 N/m, by how much is the spring compressed when the block comes to a momentary stop? (d) How long does it take for the block (+ embedded bullet) to stop from the moment it touches the spring?

Answer 1

(A) Applying momentum conservation for block-bullet system,

(8 x 10^-3 v0) + (2.2 x 0) = ((8 x 10^-3) + 2.2 )(1.2)

v0 = 331.2 m/s .......Ans

(b) Ki = (8 x 10^-3)(331.2^2) / 2 = 438.8 J

Kf = (2.208)(1.2^2) / 2 = 1.59 J

(c) Applying energy conservation,

1.59 = 400x^2 /2

x = 0.089 m Or 8.9 cm

(d) T = 2 pi sqrt( m/ k) = 2 pi sqrt(2.208 / 400)

T = 0.467 sec

t = T/4 = 0.117 sec