Question

A spring with spring constant k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block, get embedded in it, and the spring's maximum compression d is measured. The block and bullet system then oscillates.

Find an expression for the bullet's speed vB in terms of m,    M, k and d.

    Find the frequency ? of the oscillation.

Find the amplitude A of the oscillation.

      

Answer 1

from the law of conservation of energy...

(1/2)*m*VB^2 = (1/2)*(m+M)*V^2+(1/2)*k*d^2...(1)

here V is teh common velocity of bullet and block after firing...

m*VB = (m+M)*V..
V = m*VB/(m+M)....(2)...

from equation (1)....
m*VB^2 = [(m+M)*m^2*VB^2/(m+M)^2] + k*d^2....
m*VB^2[1-(m/m+M)] = k*d^2...
VB^2 = k*d^2/[(m*M)/(m+M)]...

VB = sqrt[k(m+M)/(m*M)]*d

Time period of the oscillation is T = 2*pi*sqrt[(m+M)/k]....

frequency f = 1/T = (1/2*pi)*sqrt[k/(m+M)]...

Answer 2