Answer
a) KE decreases
b) $v_{\circ}=0.240Km/s$
c) $t=0.433s$
Work Step by Step
(a) As the collision between the bullet and the bob is an inelastic collision, the kinetic energy immediately after the collision will be less than the kinetic energy before the collision. In other words, the kinetic energy in an inelastic collision is not conserved (it is converted into sound energy and "deforming" the bullet and bob).
(b) We can find the initial speed of the bullet as follows:
$v=\sqrt{2gh}=\sqrt{2(9.81)(0.124)}=1.56m/s$
According to the law of conservation of momentum
$mv_{\circ}=(M+m)v$
This can be rearranged as:
$v_{\circ}=\frac{(M+m)}{m}v$
We plug in the known values to obtain:
$v_{\circ}=\frac{1.45Kg+0.00950Kg}{0.00950}(1.56m/s)=0.240Km/s$
(c) We can find the required time as
$t=\frac{1}{4}T$
$\implies t=\frac{1}{4}(2\pi)\sqrt{\frac{l}{g}}$
We plug in the known values to obtain:
$t=\frac{\pi}{2}\sqrt{\frac{0.745}{9.81}}$
$t=0.433s$
