A 4.0-kg wooden block is at rest on a tabletop. A 20-g bullet enters the block going at 800 m /

A 4.0-kg wooden block is at rest on a tabletop. A 20-g...

A $ 4.0-\mathrm{kg}$ wooden block is at rest on a tabletop. $\mathrm{A} 20-\mathrm{g}$ bullet enters the block going at $800 \mathrm{~m} / \mathrm{s}$. It passes through the $20-\mathrm{cm}-$ thick block, emerging at $425 \mathrm{~m} / \mathrm{s}$. (a) How much time was the bullet in the block? (b) What average force did the bullet exert on the block? (c) What was the block's initial velocity when the bullet emerged? (d) If the block slides $81 \mathrm{~cm}$ before stopping, what's the coefficient of friction between block and table?

A $ 4.0-\mathrm{kg}$ wooden block is at rest on a tabletop. $\mathrm{A} 20-\mathrm{g}$ bullet enters the block going at $800 \mathrm{~m} / \mathrm{s}$. It passes through the $20-\mathrm{cm}-$ thick block, emerging at $425 \mathrm{~m} / \mathrm{s}$.
(a) How much time was the bullet in the block? (b) What average force did the bullet exert on the block? (c) What was the block's initial velocity when the bullet emerged? (d) If the block slides $81 \mathrm{~cm}$ before stopping, what's the coefficient of friction between block and table?

Key Concepts

Coefficient of Kinetic Friction

This is a dimensionless scalar that represents the ratio between the frictional force resisting the motion of two surfaces in contact and the normal force pressing them together. Determining this coefficient is essential when analyzing how friction affects the speed and stopping distance of moving objects.

Impulse-Momentum Theorem

This concept relates the impulse applied to an object to its change in momentum. It states that the product of the average force and the time interval over which it acts is equal to the change in momentum of the object, providing a way to determine unknown forces or times in collision problems.

Conservation of Momentum

This principle holds that in a closed system with no external forces, the total momentum remains constant. It is particularly important in collision problems, including cases where one object passes through another, as it allows one to relate the momenta before and after the interaction.

Kinematic Equations for Uniform Motion

These equations describe the relationships between displacement, velocity, acceleration, and time for objects under constant acceleration or uniform motion. They are used to calculate the duration of events or distances traveled when some of these quantities are known.

Work-Energy Principle

This principle states that the work done by all forces acting on an object results in a change in the object’s kinetic energy. It is useful for relating forces such as friction, which dissipate energy, to the motion and eventual stopping distance of an object.

Transcript

00:01 Alright, and this problem will be talking about momentum and collisions.

00:04 So this first part, we're looking for the time that the bullet was inside or passing through the block.

00:14 So we're going to use kinematics equation.

00:18 We're going to use this one, the x finals equal to x initial plus initial velocity times time, plus one 1⁄2, a t squared.

00:32 For our velocity, or for our acceleration, we're going to use the two velocities we know from the bullet and the time we don't know because we're solving for time.

00:45 So we're going to write our acceleration as 425 meters per second minus 800 meters per second over t.

00:54 So we're going to plug that in right here.

00:58 So when i do that, i'm just going to show this term.

01:07 Have 425 minus 800 over t squared.

01:13 This t is going to cancel with one of these ts.

01:17 So i'm just going to cancel that exponent and cancel this.

01:20 So we have one t left.

01:23 So then i'm going to simplify that, plug in all of the other numbers i know.

01:27 So that is 0 .2 meters is equal to 800t because that's the initial velocity, and then minus 187 .5t.

01:40 Then once you solve for t with that equation, you get t is equal to 3 .27 times 10 to the negative 4 seconds, which is very fast, but it makes sense because it's a bullet going through a very small space.

02:00 Alright, part b.

02:02 We are looking for, what force the bullet exerted.

02:11 So that equation is f equals m .a.

02:17 We know that the mass of the bullet is 0 .02 kilograms and we just had that equation for acceleration.

02:25 And now we have the actual time.

02:27 So i'm just going to rewrite that like this with the time that it took.

02:37 And once you solve that, you get 2 .30 times 10 to the 4th newton's of what the bullet exerted onto the block.

02:51 All right, then for part c, we want to know the initial velocity of the block.

02:59 So we're going to set up a conservation of momentum equation.

03:04 So we initially only have the bullet...