Answer
(a) The initial velocity of the bullet was 920 m/s.
(b) The fraction of the bullet's initial kinetic energy that was dissipated in the collision is 0.999
Work Step by Step
(a) Note that the total mass of the block and the bullet is:
0.999 kg + 0.0010 kg = 1.000 kg.
We can use conservation of energy to find the speed of the block when it started to slide. The initial kinetic energy of the block is equal to the elastic potential energy stored in the spring plus the energy taken from the system by friction.
$KE = EPE + mg~\mu_k\cdot d$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2 + mg~\mu_k\cdot d$
$v^2 = \frac{kx^2 + 2mg~\mu_k\cdot d}{m}$
$v = \sqrt{\frac{kx^2 + 2mg~\mu_k\cdot d}{m}}$
$v = \sqrt{\frac{(140~N/m)(0.050~m)^2 + (2)(1.000~kg)(9.80~m/s^2)(0.50)(0.050~m)}{1.000~kg}}$
$v = 0.92~m/s$
We can use conservation of momentum to find the initial speed $v_1$ of the bullet.
$m~v_1 = (m+M)~v$
$v_1 = \frac{(m+M)(v)}{m}$
$v_1 = \frac{(1.000~kg)(0.92~m/s)}{0.0010~kg}$
$v_1 = 920~m/s$
The initial velocity of the bullet was 920 m/s.
(b) We can find the initial kinetic energy $KE_1$ of the bullet:
$KE_1 = \frac{1}{2}mv_1^2$
$KE_1 = \frac{1}{2}(0.0010~kg)(920~m/s)^2$
$KE_1 = 423.2~J$
We can find the kinetic energy $KE_2$ of the block after the collision:
$KE_2 = \frac{1}{2}Mv^2$
$KE_2 = \frac{1}{2}(1.000~kg)(0.92~m/s)^2$
$KE_2 = 0.4232~J$
We can find the fraction of the bullet's initial kinetic energy that was dissipated in the collision:
$\frac{\Delta KE}{KE_1} = \frac{423.2~J - 0.4232~J}{423.2~J} = 0.999$
The fraction of the bullet's initial kinetic energy that was dissipated in the collision is 0.999
