Question

A block of mass M = 5.60 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5890 N/m. A bullet of mass m = 9.30 g and velocity v of magnitude 650 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Answer 1

(a) By conservation of momentum,

Just after the impact, momentum of bullet and block is equal to the momentum of bullet

So, bullet momentum = 0.0093kg*650m/s = 6.045kg*m/s

Final momentum = (0.0093+5.60)*v = 5.6093kg*v

Now, by applying Conservation of momentum

6.045kg*m/s = 5.6093kg*v

v = 6.045/5.6093m/s = 1.078m/s

(b) When the block stops, all of it's kinetic energy has been transferred to the spring as potential energy

So by applying Conservation of energy

Initial K.E of bullet and block = P.E stored in spring

0.5*m*v^2 = 0.5*k*x^2

0.5*(0.0093+5.60)*(1.078)^2 = 0.5*5890*x^2

x = 0.033m