Help with Phys. Question! : A 10 g, bullet is fired into a 10 kg wood block that is at rest on a wood table.?

Question

A 10 g, bullet is fired into a 10 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5 cm across the table. What was the speed of the bullet?

( coefficients of static and kinetic friction are not given but from a table in the book wood on wood coefficient are:
static= 0.5
kinetic= 0.2
I thought that might help!)

This is question 38, from chapter 9 of book:
Physics for Scientist and Engineers, By Knight

Answer 1

first, the energy lost by the slide is
.05*(10+10/1000)*9.81*0.2 J
0.982 J

This is related to the KE of the combined block and bullet just after collision

.5*(10+10/1000)*v^2=0.982
v=0.443 m/s

Using conservation of momentum
10*vi/1000=(10+10/1000)*0.443

vi=443 m/s

j

Answer 2

The block has a mass of 1kg and the bullet has a mass of 10g (or 0.010kg). Using the conservation of momentum equation for completely inelastic collisions, (M1 x V1) + (M2 x V2) = (M1 + M2) xV' M1 lets say is the mass of the bullet in kilograms M2 lets say is the mass of the wood block V1 is the velocity at which the bullet is traveling V2 is the velocity at which the wood block is traveling V' is the velocity at which the bullet and the wood block are traveling combined In this case the term (M2 x V2) can be eliminated because V2 = zero. (The wood block isn't moving.) Therefore, we have 0.010kg x V1 = (1.010kg) x V' Unfortunately, we aren't given the blocks final speed, but instead the distance it travels with UK of .20. So, the repelling force of friction = 1kg x 9.8m/s/s x 0.20 and this equals 1.96 N. (Ffriction = Fnormal x Coefficient of Kinetic Friction) Since the mass of the block w/ the bullet is 1.010, we take 1.96/1.010 to get the acceleration of the pair due to friction. The answer to this is 1.94 m/s/s. Now, we take this answer and plug it into the final velocity equation. Vf^2 = Vi^2 + 2ad Vf^2 - stands for the final velocity of the block w/ bullet squared. Since we know that at the end, the block comes to rest, this value is zero. Vi^2 - this stands for the initial velocity of the bullet/block after the collision. This is the unknown we need to solve for. a - acceleration due to friction (1.94 m/s/s we found earlier) d - the 4 meters traveled Anyway, plug in the values as shown below. 0 = Vi"^2 + 2 x 1.94 x 4 this simplifies to... 0 = Vi^2 + 15.52 So now we know that Vi^2 = the absolute value of 15.52, so square root this number to get 3.94m/s Now we can plug this value into the first equation @ the top. 0.010kg x V1 = (1.010kg) x 3.94m/s This simplifies to... 0.010kg x V1 = 3.98 Now, 3.98/0.010 = V1 which is the speed of the bullet. Your final answer is 397.94 m/s