A 12.0-g bullet is fired horizontally into a 100-g wooden block initially at rest on a horizonta

A 12.0-g bullet is fired horizontally into a 100-g wooden...

A $12.0-\mathrm{g}$ bullet is fired horizontally into a $100-\mathrm{g}$ wooden block initially at rest on a horizontal surface. After impact, the block slides $7.5 \mathrm{~m}$ before coming to rest. If the coefficient of kinetic friction between block and surface is $0.650$, what was the speed of the bullet immediately before impact?

Key Concepts

Conservation of Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant during a collision. In collision problems, even if energy is lost, the momentum before and after the collision is conserved, which is a key tool for relating the velocities and masses of the colliding bodies.

Inelastic Collision

In an inelastic collision, the colliding bodies stick together or deform, and while momentum is conserved, some of the kinetic energy is converted into other forms of energy. This concept is crucial when analyzing systems where the collision results in a combined mass moving with a common velocity immediately after impact.

Work-Energy Principle

This principle connects the work done by forces acting on a body to the change in its kinetic energy. In scenarios where friction is present, the work done by the frictional force (which is equal to the friction force multiplied by the distance over which it acts) causes the kinetic energy of the moving system to be dissipated, eventually bringing it to rest.

Kinetic Friction

Kinetic friction is the resistive force that opposes the motion of a moving object and is proportional to the normal force. Its coefficient quantifies the amount of friction between two surfaces. In energy problems, the work done by kinetic friction over a distance is used to calculate the loss in kinetic energy and determine deceleration or stopping distance.

Transcript

00:01 Okay, so in this situation, we have a bullet being fired into a block initially at rest, and together the bullet stays in the block, and it travels a 7 .5 meters along a table with the coefficient of friction of 0 .650.

00:17 And we're asked to find the speed of the bullet right before it goes in the block.

00:21 So in order to find that, we can work backwards in this problem.

00:25 So we can find what the initial speed of the bullet and the block right, it enters in event number one.

00:34 So let's do that.

00:36 So we have initially, right as the bullets fighting the block, there's a kinetic energy inside.

00:44 They have a certain speed and there's a kinetic energy.

00:47 Now, as the travels along the table, there's an external work being applied to this, taking energy out of the system, and at the end of its travel, there's no more energy left in the bullet and the bullet in the block come to rest.

00:59 There's no gravitational potential energy gained or anything so it's zero so we have one over two the bullet in the block are together so we'll say m a plus mb and there's v uh we'll call it v1 that's the that's the speed of the bullet in the block right after the bullet goes into it now the work external is a friction force times the delta d we're not going to do the cosine part because it's already negative equaling zero let's add the numbers up m a 0 .0 and mb is 0 .1 .1.

01:39 We're trying to find v1 squared.

01:41 Ff friction is equal to mu times fn and our delta d is 7 .5.

01:51 And let's continue this out.

01:58 We're going to get mu is 0 .650.

02:04 Fn is equal to the mass times gravity and because the masses are combined again, we'll just add those right away.

02:10 It's the mass of the bullet plus the mass of the block, 0 .012 plus .0 .2 plus.

02:13 0 .100.

02:14 We'll get 0 .112.

02:17 And that's mass and gravity is 9 .81 times the 7 .5.

02:23 And now we can isolate for v1 and solve...