A19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a horizontal spring with a force constant of 500 n/m. the maximum compression of the spring when the block stops is 21 cm. what is the speed of the bullet just before it hits the block? a) 900 m/s b) 600 m/s c) 500 m/s d) 700 m/s e) 800 m/s 2.

d) 700 m/s

Explanation:

if k is the force constant and x is the maximum compression distance, then:

the potential energy the spring can acquire is given by:

U = 1/2×k×(x^2)

and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy  = final energy

            Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

                  Ki = Uf

   1/2×m×(v^2) = 1/2×k×(x^2)

         m×(v^2) = k×(x^2)

                v^2 = k×(x^2)/m

                       = (500)×((21×10^-2)^2)/(19×10^-3 + 8)

                       = 2.75

                    v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

          m1×v1 = v×(m1 + M)

                v1 =  v×(m + M)/(m1)

                    =  (1.66)×(19×10^-3 + 8)/(19×10^-3)

                    = 699.86 m/s

                    ≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.  

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: