kinetic energy problem?

Question

A 10-g bullet hits a 200-g block of wood and sticks inside. The block is on the frictionless horizontal surface and initially at rest. What fraction of the bullet's kinetic energy is converted into the internal energy?

if only they gave me velocities =[

Answer 1

conservation of momentum
10*vi=210*vo
or vi=21*vo

and
the starting KE is
5*vi^2
the ending KE is
105*vo^2

the internal energy is

5*vi^2-105*vo^2

divide
(5*vi^2-105*vo^2)/(5*vi^2)

substitute vi from the momentum

(2205*vo^2-105*vo^2)/(2205*vo^2)
divide out vo and solve for the fraction
0.952

j

Answer 2

Velocities aren't necessary:
If v1=speed of bullet and
v2=speed of block after impact
Then by conservation of momentum v2=10/210v1
Kinetic energy of block+bullet = 1/2*210*100/210^2*(v1)^2, which simplifies to 50/210(v1)^2
The kinetic energy of the bullet was simply 1/2*10*(v1)^2
Therefore, the fraction of the bullet's kinetic energy transformed into the kinetic energy of block+bullet = (50/210(v1)^2)/(5(v1)^2)=1/21.
So 20/21 of the initial kinetic energy is converted into internal energy.

Answer 3

You don't need velocities for this, find v1 as a ratio of v2.

m1 = mass of the bullet
v1 = velocity of the bullet
m2 = mass of the block
vf = final velocity

Using momentum
m1*v1 + m2*0 = (m1+m2)*vf
vf = m1/(m1+m2) v1 = 10/210 *v1 = 0.048 v1

Initial energy of the bullet
KE = 1/2*m1*v1^2

Final Energy of the bullet
KE = 1/2*m1*vf^2

Fraction = (KEinitial - KEfinal)/KEinitial = 1 - KEfinal/KEintial

= 1 - (1/2*m1*vf^2)/(1/2*m1*v1^2)

= 1 - vf^2/vi^2 = 1- .048^2 = 0.997

So, 99.7% of the bullets energy is transferred into the block.