. A A 5.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal su

. A A 5.00 g bullet is fired horizontally into a 1.20 kg...

. A $\mathrm{A} 5.00$ g bullet is fired horizontally into a 1.20 $\mathrm{kg}$ wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is $0.20 .$ The bullet remains embedded in the block, which is observed to slide 0.230 $\mathrm{m}$ along the surface before stopping. What was the initial speed of the bullet?

. A $\mathrm{A} 5.00$ g bullet is fired horizontally into a 1.20 $\mathrm{kg}$ wooden
block resting on a horizontal surface. The coefficient of kinetic
friction between block and surface is $0.20 .$ The bullet remains embedded in the block, which is observed to slide 0.230 $\mathrm{m}$
along the surface before stopping. What was the initial speed
of the bullet?

Key Concepts

Work-Energy Principle

This principle relates the work done by forces (such as friction) to the change in kinetic energy of an object. When the block (with the embedded bullet) slides and eventually stops due to friction, the work done by friction equals the loss in kinetic energy, allowing us to relate frictional force, displacement, and initial kinetic energy.

Inelastic Collisions

In an inelastic collision, the colliding objects stick together after impact and kinetic energy is not conserved, even though momentum is. This concept is important for understanding how the bullet embeds into the block and how the subsequent motion is governed by the combined mass.

Conservation of Linear Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant. In collision problems, especially in inelastic collisions where objects might stick together, the momentum before collision is equal to the momentum after collision, providing a way to relate the velocities of the colliding bodies.

Transcript

00:01 In this problem, we have a bullet with a mass of 0 .005 kilograms, which hits a block, which is initially at rest.

00:09 The block has a mass of 1 .2 kilograms, and it sits on a surface with a coefficient of kinetic friction of 0 .2.

00:16 After the collision, the block moves a displacement of 0 .23 from its initial location, and from this information, we have to determine the initial speed of the bullet.

00:28 So there's two parts to this question.

00:30 The first part is determining the speed of the bullet and the block combined because the bullet embeds itself in the block, using the displacement that it travels after the collision as well as conservation of energy.

00:48 So right after the collision, the systems energy is all kinetic.

00:55 And over 0 .23 meters, that gets dissipated through work done by friction.

01:01 So, mathematically, one -half times the two masses together, times their velocity squared.

01:15 I'm going to call this v2, and the reason why we'll become apparent later, that's equal to the work done by friction, which is the coefficient of kinetic energy, times acceleration due to gravity, times the masses, times the distance traveled by the block with the blood embedded in it.

01:44 So we can cross out this factor here, and solving for the velocity, it is the square root of two times the coefficient of friction, times acceleration due to gravity, times the distance, traveled.

02:04 Now we want to plug this velocity into conservation of momentum to find the initial speed of just the bullet...

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