A bullet of mass m is fired into a block of mass M that is...

Key Concepts

Projectile Motion

After the collision, the combined mass moves as a projectile. Projectile motion involves both horizontal and vertical components of motion, where the time of flight is determined by the vertical free-fall under gravity. This time of flight and the horizontal displacement provide a method to calculate the initial horizontal velocity of the system.

Free-Fall Kinematics

Free-fall kinematics explains the vertical motion of an object under the influence of gravity. In the problem, the time it takes for the combined block and bullet to fall from a given height is found using the equations for uniformly accelerated motion, which then ties into the horizontal motion analysis to derive the initial conditions.

Conservation of Momentum

This principle states that the total momentum of a closed system remains constant in the absence of external forces. In the context of collisions, especially inelastic ones where objects stick together, the momentum before the collision (from the bullet) equals the momentum after the collision (of the combined mass), allowing us to relate their velocities.

Inelastic Collision Dynamics

In a perfectly inelastic collision, the colliding bodies stick together after impact, meaning that while momentum is conserved, kinetic energy is not. This concept is crucial in problems involving a bullet embedding in a block, as it impacts the analysis by necessitating the use of momentum conservation rather than energy conservation.

Transcript

00:01 Here, we're going to use both momentum conservation as well as some kinematic equations to generally figure out what the velocity of the bullet has to be.

00:11 If it collides, if it is shot at a block, which is at the edge of a frictionless table, the bullet gets embedded in the block, they fly off the table and they land a distance d from the table that has a height h.

00:26 And first the first thing we're going to do is use momentum conservation of this collision since it will happen first the momentum conservation says that the initial momentum of the system which is the bullet and the block will be the addition of the two momenta since this is this collision is happening in one dimension the bullet is moving in one direction which will be the momentum of the mass of the bullet which is denoted as lowercase m times the velocity of the bullet plus the initial momentum of the block.

01:04 The block is initially stationary, so that'll be zero.

01:07 So this has to be equal to the final momentum, which again has to point in the same direction, which means that the magnitudes of the two momentum have to be the same.

01:18 And this means, and since the bullet is going to embed in the block, the two will move together with the same velocity.

01:28 So this will be little m the mass of the bullet plus the mass of the block times which we're calling v -0, which is the velocity of both of them moving after the collision.

01:40 We want to solve for the velocity of the bullet.

01:43 So we're going to divide both sides by little m.

01:47 So this will be massive bullet plus massive block divided by mass of the bullet times the velocity of the two moving together after the collision.

01:58 So now we just need to find this velocity, for which we know that the two will move in a projectile motion down until the two hit a distance d from the table of height h.

02:17 So we know in general to move in a projectile motion at assuming constant acceleration, the equation for the y coordinate will be one -half.

02:31 Acceleration of y direction times t squared plus initial velocity in the y direction times t plus the initial height same thing with the x coordinate except everything is in the x direction initial velocity in the x direction times time plus the initial the initial position the initial x coordinate we can set the origin such that this last the initial x position is zero we're given that the initial y position is just h since the this bullet is moving horizontally into the block.

03:09 The velocity of the block plus the bullet after the collision also has to be horizontal.

03:17 So there's going to be no y component to the initial velocity.

03:20 So this term is zero.

03:23 And likewise, there's while the block is in the air, there's no force in the x direction...

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