Question

Not A 5.00-g bullet moving with an initial speed of v 410 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 940 N/m. The block moves d-5.60 cm to the right after impact before being brought to rest by the spring (a) Find the speed at which the bullet emerges from the block 146 your-esponse dffers from the correct answer by more than 100%, m/s (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet-block system during the collision Noed Help? Ls

Answer 1

a) From energy conservation, we get

From the linear momentum conservation, we have

$mv_i = MV_i +mv \\ \\ \Rightarrow v = \frac{mv_i-MV_i}{m} = \frac{(5.0\times 10^{-3})(410)- (1.00)(1.72)}{(5.0\times 10^{-3}~kg)} \\ \\ \\ \Rightarrow v = 0.066\times 10^3 = 66~m/s$

b) The initial enrgy is only the kinetic enrgy of the bullet, while as the final energy is the sum of the kinetic enrgy and the potential energy

$E_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(5.00\times 10^{-3})(410)^2 = 420.25~J$

also

$E_f= \frac{1}{2}mv^2+\frac{1}{2}kx^2 = \frac{1}{2}(5.00\times 10^{-3})(66)^2 +\frac{1}{2}(940)(5.60\times 10^{-2})^2 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~10.89+1.47 = 12.36~J$

hence the amount of energy converted is

$\Delta E = E_f - E_i = 12.36-420.25 = -407.89~J$