Solving Momentum Problem w/ Friction: Find Bullet Speed

[pringless]

A 6.56g bullet is fired horizontally into a 94.2 g wooden block that is initially at rest on a rough horizontal surface and connected to a massless spring of constant 70.4 N/m. If the bullet-block system compresses the spring by 1.16 m, what was the speed of the bullet just as it enters the block? Assume the coefficient of kinetic friction between the block surface is .485. Answer in units of m/s.

how would you solve this with friction?

is it still m1v1 = (m1+m2)v2? but I am not sure how to incorporate the friction

 

[Doc Al]

Originally posted by pringless
is it still m1v1 = (m1+m2)v2? but I am not sure how to incorporate the friction

Yes, momentum is conserved during the collision.

After the collision, consider what happens to the mechanical energy. The bullet+block starts off with purely KE. The friction force does work against the block, dissipating some energy. Whatever's left transforms into PE as the spring is compressed.

 

[M98Ranger]

in this equation

To solve this problem with friction, we need to take into account the work done by friction on the bullet-block system. The work done by friction can be calculated by multiplying the coefficient of kinetic friction (μ) by the normal force (N) and the displacement (d). In this case, the normal force is equal to the weight of the block (mg) and the displacement is equal to the compression of the spring (1.16 m).

So, the work done by friction can be calculated as W = μmgd.

Now, let's look at the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. In this case, the net work done on the bullet-block system is equal to the work done by the bullet (m1v1) minus the work done by friction (μmgd).

Therefore, we can set up the following equation:

m1v1 - μmgd = (m1+m2)v2

Substituting the given values, we get:

(6.56g)v1 - (0.485)(94.2g)(9.8m/s^2)(1.16m) = (6.56g+94.2g)v2

Solving for v1, we get:

v1 = [(6.56g+94.2g)v2 + (0.485)(94.2g)(9.8m/s^2)(1.16m)] / 6.56g

Plugging in the given values for v2 and solving, we get:

v1 = 11.6 m/s

Therefore, the speed of the bullet just as it enters the block is 11.6 m/s.