A 6.26-g bullet is moving horizontally with a velocity of +344 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1205 g, and its velocity is +0.673 m/s after the bullet passes through it. The mass of the second block is 1578 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.
initial momentum = .00626*344 = 2.15
2.15 = .00626 Vb1 + 1.205*.673
Vb1 = 214 m/s exiting first block
bullet momentum entering second = .00626*214 = 1.34
1.34 = 1.584 v
v = .846 m/s (a)
(b)final = (1/2)(1.584)(.846^2)
= .567 Joules
initial = (1/2)(.00626)(344)^2
=370
ratio = .567/370 = .0015
To solve this problem, we need to use the principles of conservation of momentum and conservation of kinetic energy. Let's tackle each part of the question step-by-step.
(a) What is the velocity of the second block after the bullet embeds itself?
We can start by finding the initial momentum of the system before the collision. The momentum of an object is calculated as the product of its mass and velocity.
Initial momentum (before the collision) = (mass of the bullet) × (velocity of the bullet) + (mass of the first block) × (velocity of the first block)
Given:
Mass of the bullet (m1) = 6.26 g = 0.00626 kg (converted from grams to kg)
Velocity of the bullet (v1) = +344 m/s (to the right)
Mass of the first block (m2) = 1205 g = 1.205 kg (converted from grams to kg)
Velocity of the first block (v2) = +0.673 m/s (to the right)
Initial momentum (before the collision) = (0.00626 kg) × (+344 m/s) + (1.205 kg) × (+0.673 m/s)
Next, let's find the final momentum of the system after the collision. Since the bullet embeds itself in the second block, both blocks move together as a single unit.
Final momentum (after the collision) = Total mass × Final velocity
Total mass = mass of the first block + mass of the second block
Total mass = 1.205 kg + 1.578 kg
Now, we can calculate the final velocity using the conservation of momentum principle:
Final momentum (after the collision) = (total mass) × (final velocity)
(total mass) × (final velocity) = (0.00626 kg) × (+344 m/s) + (1.205 kg + 1.578 kg) × final velocity
Solving for the final velocity, we get:
final velocity = [(0.00626 kg) × (+344 m/s)] / [1.205 kg + 1.578 kg]
b) Find the ratio of the total kinetic energy after the collision to that before the collision.
The kinetic energy of an object is calculated as (1/2) × mass × velocity^2. We need to find the ratio of the total kinetic energy after the collision to that before the collision.
To calculate the total kinetic energy before the collision, we need to find the kinetic energy of the bullet and the first block separately, and then sum them together.
Kinetic energy before the collision = (kinetic energy of the bullet) + (kinetic energy of the first block)
The kinetic energy of an object is given by (1/2) × mass × velocity^2.
Kinetic energy of the bullet = (1/2) × (mass of the bullet) × (velocity of the bullet)^2
Kinetic energy of the first block = (1/2) × (mass of the first block) × (velocity of the first block)^2
After finding the kinetic energies, we can calculate the total kinetic energy after the collision by finding the kinetic energy of the combined blocks.
Total kinetic energy after the collision = (1/2) × (total mass) × (final velocity)^2
Finally, we can calculate the ratio of the total kinetic energy after the collision to that before the collision:
Ratio = (Total kinetic energy after the collision) / (Total kinetic energy before the collision)
To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy. Let's break down the steps to find the answers to both parts (a) and (b) of the question.
First, let's calculate the bullet's initial momentum before the collision (P_initial_bullet):
P_initial_bullet = m_bullet * v_bullet
Given:
m_bullet = 6.26 g = 0.00626 kg
v_bullet = +344 m/s (since the bullet is moving to the right)
Calculating the initial momentum:
P_initial_bullet = 0.00626 kg * 344 m/s
Next, let's calculate the momentum of the first block after the bullet passes through it (P_first_block):
P_first_block = m_first_block * v_first_block
Given:
m_first_block = 1205 g = 1.205 kg
v_first_block = +0.673 m/s
Calculating the momentum of the first block:
P_first_block = 1.205 kg * 0.673 m/s
Since momentum is conserved during an inelastic collision, the momentum of the bullet before the collision should be equal to the sum of the momenta of the first and second blocks after the collision:
P_initial_bullet = P_first_block + P_second_block
Now, let's determine the velocity of the second block after the bullet embeds itself (v_second_block) in part (a) of the question:
P_second_block = P_initial_bullet - P_first_block
Knowing the value of P_initial_bullet and P_first_block, we can now calculate the momentum of the second block:
P_second_block = (0.00626 kg * 344 m/s) - (1.205 kg * 0.673 m/s)
Finally, we can find the velocity of the second block by dividing the momentum (P_second_block) by its mass (m_second_block):
v_second_block = P_second_block / m_second_block
Given:
m_second_block = 1578 g = 1.578 kg
Calculating the velocity of the second block:
v_second_block = P_second_block / 1.578 kg
To find the ratio of the total kinetic energy after the collision to that before the collision in part (b) of the question, we can use the formulas for kinetic energy:
Kinetic energy (KE) = 1/2 * m * v^2
To find the total kinetic energy after the collision, we need to calculate the kinetic energy of both blocks:
KE_first_block = 1/2 * m_first_block * v_first_block^2
KE_second_block = 1/2 * m_second_block * v_second_block^2
The total kinetic energy after the collision is the sum of the kinetic energies of both blocks:
Total KE_after = KE_first_block + KE_second_block
To calculate the ratio of KE_after to KE_before, we need to calculate the initial kinetic energy of the bullet:
KE_bullet = 1/2 * m_bullet * v_bullet^2
Finally, we can find the desired ratio:
Ratio = (Total KE_after + KE_bullet) / KE_bullet
By following these steps and performing the calculations, you should be able to find the answers to both part (a) and part (b) of the question.