A 0.880 kg block is attached to a horizontal spring with spring constant 3650 N/

Question

A 0.880 kg block is attached to a horizontal spring with spring constant 3650 N/m . The block is at rest on a frictionless surface. A 8.60 g bullet is fired into the block, in the face opposite the spring, and sticks. What was the bullet's speed if the subsequent oscillations have an amplitude of 16.2 cm?

I tried following similar problems, but can't get the right answer. I made the spring energy=kinetic energy and I'm still having issues. I converted the amplitude to .162 m, and the bullet to .0086 kg.

Explanation / Answer

given, m = 0.88 kg
k = 3650 Nm ( spring constant)
fricitonless table
bullet mass m' = 8.6*10^-3 kg
amplitude, A = 0.162 m

now, initial speed of bullet = u
then final speed of the mass = v
from conservation of momentum
m'*u = (m' + m)v
v = m'*u/(m' + m) = 0.009678145*u

now, from the amplitude
0.5*k*A^2 = 0.5(m + m')v^2
3650*0.162^2 = (0.8886)v^2 = 0.8886*(0.009678145*u)^2
hence
u = 1072.793 m/s

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