A 4.15-g bullet is moving horizontally with a velocity of +369 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 1192 g, and its velocity is +0.635 m/s after the bullet passes through it. The mass of the second block is 1554 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.
To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy.
(a) To find the velocity of the second block after the bullet embeds itself, we can use the principle of conservation of momentum.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) is equal to the product of mass (m) and velocity (v).
Before the collision:
Momentum of the bullet:
m_bullet * v_bullet = (0.00415 kg) * (369 m/s) = 1.52735 kg·m/s
No momentum for the blocks, as they are at rest.
After the collision:
Since the bullet embeds itself in the second block, we consider the momentum of the bullet and the two blocks together.
Total momentum after the collision:
(m_bullet + m_block1 + m_block2) * v_after = 1.52735 kg·m/s
Mass of the first block (m_block1) = 1192 g = 1.192 kg
Mass of the second block (m_block2) = 1554 g = 1.554 kg
Therefore,
(1.52735 kg·m/s) = (0.00415 kg + 1.192 kg + 1.554 kg) * v_after
Simplifying the equation,
v_after = (1.52735 kg·m/s) / (2.75015 kg)
v_after ≈ 0.555 m/s
So, the velocity of the second block after the bullet embeds itself is approximately +0.555 m/s.
(b) To find the ratio of the total kinetic energy after the collision to that before the collision, we can use the principle of conservation of kinetic energy.
The total kinetic energy before the collision is equal to the sum of the individual kinetic energies of the bullet and the two blocks.
Before the collision:
Kinetic energy of the bullet:
KE_bullet = (1/2) * m_bullet * v_bullet^2 = (1/2) * (0.00415 kg) * (369 m/s)^2
Since the blocks are at rest, their kinetic energy is zero.
Total kinetic energy before the collision:
KE_total_before = KE_bullet + KE_block1 + KE_block2
KE_total_before = (1/2) * (0.00415 kg) * (369 m/s)^2
After the collision:
Since the bullet embeds itself in the second block, the total kinetic energy after the collision is the kinetic energy of the second block.
Kinetic energy of the second block:
KE_block2_after = (1/2) * m_block2 * v_after^2 = (1/2) * (1.554 kg) * (0.555 m/s)^2
Therefore,
Ratio of total kinetic energy after the collision to that before the collision:
KE_total_after / KE_total_before = KE_block2_after / KE_total_before
Substituting the values,
Ratio = [(1/2) * (1.554 kg) * (0.555 m/s)^2] / [(1/2) * (0.00415 kg) * (369 m/s)^2]
Simplifying the equation,
Ratio ≈ 0.334
So, the ratio of the total kinetic energy after the collision to that before the collision is approximately 0.334.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
Part (a):
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of the bullet before the collision can be calculated as the product of its mass and velocity:
Momentum of bullet before collision = mass of bullet × velocity of bullet
= 4.15 g × 369 m/s
= 4.15 × 10^(-3) kg × 369 m/s
= 1.5285 kg·m/s
The momentum of the first block after the collision can be calculated as the product of its mass and velocity:
Momentum of first block after collision = mass of first block × velocity of first block
= 1192 g × 0.635 m/s
= 1192 × 10^(-3) kg × 0.635 m/s
= 0.75748 kg·m/s
Since the bullet embeds itself in the second block, the momentum of the second block after the collision is the same as the momentum of the bullet before the collision:
Momentum of second block after collision = Momentum of bullet before collision
= 1.5285 kg·m/s
Now, we can use the principle of conservation of momentum to find the velocity of the second block after the bullet embeds itself:
Total momentum before collision = Total momentum after collision
(1.5285 kg·m/s) = (Mass of first block × Velocity of first block) + (Mass of second block × Velocity of second block)
Solve for the unknown velocity of the second block:
Velocity of second block = (1.5285 kg·m/s - Mass of first block × Velocity of first block) / Mass of second block
= (1.5285 kg·m/s - 0.75748 kg·m/s) / 1554 g
= (1.5285 kg·m/s - 0.75748 kg·m/s) / 1554 × 10^(-3) kg
= 0.77102 kg·m/s / 1554 × 10^(-3) kg
= 0.496 m/s (rounded to three decimal places)
So, the velocity of the second block after the bullet embeds itself is +0.496 m/s (towards the right).
Part (b):
To find the ratio of the total kinetic energy after the collision to that before the collision, we need to calculate the total kinetic energy before and after the collision, and then take their ratio.
The total kinetic energy before the collision is the sum of the kinetic energy of the bullet and the kinetic energy of the first block.
Kinetic energy before collision = 1/2 × mass of bullet × (velocity of bullet)^2 + 1/2 × mass of first block × (velocity of first block)^2
= 1/2 × 4.15 g × (369 m/s)^2 + 1/2 × 1192 g × (0.635 m/s)^2
= 1/2 × 4.15 × 10^(-3) kg × (369 m/s)^2 + 1/2 × 1192 × 10^(-3) kg × (0.635 m/s)^2
= 0.293606725 J + 0.15166608 J
= 0.445272805 J (rounded to nine decimal places)
The total kinetic energy after the collision is the sum of the kinetic energy of the first block and the kinetic energy of the second block.
Kinetic energy after collision = 1/2 × mass of first block × (velocity of first block)^2 + 1/2 × mass of second block × (velocity of second block)^2
= 1/2 × 1192 g × (0.635 m/s)^2 + 1/2 × 1554 g × (0.496 m/s)^2
= 1/2 × 1192 × 10^(-3) kg × (0.635 m/s)^2 + 1/2 × 1554 × 10^(-3) kg × (0.496 m/s)^2
= 0.19047918 J + 0.153678136 J
= 0.344157316 J (rounded to nine decimal places)
Therefore, the ratio of total kinetic energy after the collision to that before the collision is:
Ratio = (Total kinetic energy after collision) / (Total kinetic energy before collision)
= 0.344157316 J / 0.445272805 J
= 0.772 (rounded to three decimal places)
So, the ratio of the total kinetic energy after the collision to that before the collision is approximately 0.772.