Conservation of momentum with friction

[Thepoint]

Homework Statement

A bullet of mass 1.9×10^−3 kg embeds itself in a wooden block with mass 0.991kg
which then compresses a spring (k= 100N/m ) by a distance 4.5×10^−2m before coming to rest. The coefficient of kinetic friction between the block and table is 0.55.

a) What is the initial speed of the bullet?

b) What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

Homework Equations

.5kx^2,
normal force x Mu,
.5mv^2,
mgy
m1*v1=(m1+m2)*v2

The Attempt at a Solution

.5kx^2= .5mv^2- (Mu x F(g))
2.25J= .5(.0019kg)v^2 - (.55)1.036kg(9.8m/s^2)
v=91 m/s

I don't know what I did wrong. Also I can answer the second part if I know the first part so just help me out with the first part.

 

[LowlyPion]

Homework Statement

A bullet of mass 1.9×10^−3 kg embeds itself in a wooden block with mass 0.991kg
which then compresses a spring (k= 100N/m ) by a distance 4.5×10^−2m before coming to rest. The coefficient of kinetic friction between the block and table is 0.55.

a) What is the initial speed of the bullet?

b) What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? ]2. Homework Equations
.5kx^2,
normal force x Mu,
.5mv^2,
mgy
m1*v1=(m1+m2)*v2

3. The Attempt at a Solution
.5kx^2= .5mv^2- (Mu x F(g))
2.25J= .5(.0019kg)v^2 - (.55)1.036kg(9.8m/s^2)
v=91 m/s

I don't know what I did wrong. Also I can answer the second part if I know the first part so just help me out with the first part.

Calculate the PE put in the spring:
1/2*100(.045)² = .10 not the 2.25 J you used. (You apparently used 1/2*k*x not x² )

Also m1 + m2 is .9929. Where did the 1.036kg come from?

 

[thomate1]

Your attempt at a solution is incorrect because you did not take into account the fact that the bullet embeds itself in the wooden block, meaning that the mass of the system changes after the collision. Here's a corrected solution:

a) To find the initial speed of the bullet, we can use the conservation of momentum equation: m1v1 = (m1+m2)v2. We know the masses of the bullet and block, so we can plug those in: (0.0019 kg)v1 = (0.0019 kg + 0.991 kg)v2. We also know that after the collision, the system comes to rest, so v2 = 0. Solving for v1, we get v1 = 0.991 kg / 0.0019 kg * 0 = 0 m/s. This means that the initial speed of the bullet was 0 m/s.

b) To find the fraction of the bullet's initial kinetic energy that is dissipated, we can use the equation for kinetic energy: KE = 0.5mv^2. We know the initial speed of the bullet is 0 m/s, so the initial kinetic energy is 0. After the collision, the bullet is embedded in the block and the system comes to rest, meaning that all of the bullet's kinetic energy has been dissipated. Therefore, the fraction of the bullet's initial kinetic energy that is dissipated is 1.