A 0.0200 kg bullet moving horizontally at 500 m/s embeds itself into an initially stationary 0.500 kg block.
(a) What is their velocity just after the collision?
(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity?
(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping?
Well, well, well, it seems like we've got some physics problems here! Let's dive in and sprinkle some humor into these equations.
(a) To find the velocity just after the collision, we need to apply the law of conservation of momentum. The momentum before the collision equals the momentum after the collision. So we have:
(mass of bullet x velocity of bullet) + (mass of block x initial velocity of block) = (total mass of bullet and block x final velocity)
Plugging in the values, we get:
(0.0200 kg x 500 m/s) + (0.500 kg x 0 m/s) = (0.5200 kg x final velocity)
Solving the equation, you'll find the final velocity. But be warned, the answer might shoot right into your face!
(b) Now we have a bullet-embedded block sliding on a surface with friction. We need to take into account the kinetic friction force, which opposes the motion. But don't worry, friction can also be quite funny!
To find the velocity after sliding 8.0 m, we'll first calculate the work done by friction, which is force × distance. The work done by friction will change the block's kinetic energy. So:
Work done by friction = change in kinetic energy
Substituting the values, we have:
Friction force x 8.0 m = (0.500 kg x final velocity^2) / 2
Now, solve for the final velocity and see if it comes sliding hilariously!
(c) Now we have a total circus going on! The bullet-embedded block collides and sticks to a stationary block. We need to determine the distance traveled before stopping.
To find the distance, we'll use the concept of work and energy again. The initial kinetic energy of the bullet-embedded block will convert to work done by the frictional force. We'll set this equal to the final kinetic energy of the whole combination of blocks.
So, (0.5200 kg x initial velocity^2) / 2 = (total mass x final velocity^2) / 2
Solving this equation will give us the final velocity, and once we have that, we can calculate the distance traveled. But remember, this journey might end with a boisterous finale or a silent halt!
Keep in mind that these calculations assume an idealized scenario without external factors. Physics is full of twists and turns, making it as unpredictable as a clown show!
To answer these questions, we need to use the principles of conservation of momentum and work done by friction.
a) To find the velocity of the bullet and block system just after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the bullet is given by:
Initial momentum of the bullet = mass of the bullet × velocity of the bullet
= 0.0200 kg × 500 m/s
= 10 kg·m/s
Since the block is initially stationary, its initial momentum is zero.
Total momentum before collision = Initial momentum of the bullet + Initial momentum of the block
= 10 kg·m/s + 0 kg·m/s
= 10 kg·m/s
After the collision, the bullet and block become a single system and move together with a common velocity.
Total momentum after collision = Final momentum of the combined system
Let's assume the common velocity of the bullet and block system after the collision is V.
Total momentum after collision = (mass of the bullet + mass of the block) × V
= (0.0200 kg + 0.500 kg) × V
= 0.5200 kg × V
According to the principle of conservation of momentum:
Total momentum before collision = Total momentum after collision
10 kg·m/s = 0.5200 kg × V
Now, we can solve for V:
V = 10 kg·m/s / 0.5200 kg
V ≈ 19.23 m/s
Therefore, the velocity of the bullet and block system just after the collision is approximately 19.23 m/s.
b) To find the velocity of the bullet-embedded block after sliding a certain distance, we need to consider the force of friction acting on the block. The work done by friction will cause a change in the kinetic energy of the system.
The work done by friction (W) can be calculated using the formula:
W = force of friction × displacement × cosine(180°)
Since the frictional force opposes the motion, the angle between the force and displacement is 180°.
Let's calculate the normal force (N) acting on the block first:
N = mass of the block × acceleration due to gravity
= 0.500 kg × 9.8 m/s²
N = 4.90 N
The force of friction (f) can be calculated using the formula:
f = coefficient of friction × N
= 0.30 × 4.90 N
f = 1.47 N
The work done by friction can be calculated using the formula:
W = f × displacement × cosine(180°)
W = 1.47 N × 8.0 m × cos(180°)
W = -11.76 J
The negative sign indicates that the work done by friction is in the opposite direction of motion.
Change in kinetic energy (ΔKE) of the system is equal to the work done by friction:
ΔKE = -11.76 J
ΔKE = Final kinetic energy - Initial kinetic energy
Since the block initially starts from rest, its initial kinetic energy is 0.
Final kinetic energy is given by:
Final kinetic energy = 0.5 × (mass of the bullet-embedded block) × (velocity)^2
= 0.5 × 0.5200 kg × (velocity)^2
To find the final velocity (V), we can equate the change in kinetic energy to the final kinetic energy:
-11.76 J = 0.5 × 0.5200 kg × (V)^2 - 0 J
(V)^2 ≈ (11.76 J / (0.5 × 0.5200 kg))
V ≈ √45.3 m²/s²
V ≈ 6.73 m/s
Therefore, the velocity of the bullet-embedded block after sliding 8.0 m is approximately 6.73 m/s.
c) To find the distance the combination of the bullet-embedded block and the stationary block travels before stopping, we will again use the principles of conservation of momentum and work done by friction.
The momentum before the collision is equal to the momentum after the collision.
The momentum before the collision is given by:
Initial momentum of the bullet-embedded block = mass of the bullet-embedded block × velocity of the bullet-embedded block
= 0.5200 kg × 6.73 m/s
= 3.50 kg·m/s
The momentum after the collision is given by:
Final momentum of the combined system = mass of the combined system × final velocity
Let's assume the final velocity of the combined system is V'.
The mass of the combined system is the sum of the masses of the bullet-embedded block and the stationary block:
Mass of the combined system = mass of the bullet-embedded block + mass of the stationary block
= 0.5200 kg + 2.00 kg
= 2.52 kg
According to the principle of conservation of momentum:
Initial momentum of the bullet-embedded block = Final momentum of the combined system
3.50 kg·m/s = 2.52 kg × V'
Now, we can solve for V':
V' = 3.50 kg·m/s / 2.52 kg
V' ≈ 1.39 m/s
Now, let's consider the work done by friction to calculate the distance traveled before stopping.
The frictional force can be calculated using the same formula as before:
f = coefficient of friction × N
= 0.30 × (mass of the combined system × acceleration due to gravity)
= 0.30 × 2.52 kg × 9.8 m/s²
f = 7.41 N
The work done by friction can be calculated using the formula:
W = f × displacement × cosine(180°)
Since the displacement is not given, let's assume it to be D.
W = 7.41 N × D × cos(180°)
W = -7.41 D J
The work done by friction is equal to the change in kinetic energy of the system:
-7.41 D J = ΔKE
Since the system comes to a stop, the final kinetic energy is 0.
Initial kinetic energy of the system is given by:
Initial kinetic energy = 0.5 × (mass of the combined system) × (V')^2
= 0.5 × 2.52 kg × (1.39 m/s)^2
To find the distance (D), we can equate the change in kinetic energy to the initial kinetic energy:
-7.41 D J = 0.5 × 2.52 kg × (1.39 m/s)^2 - 0 J
7.41 D J = 0.5 × 2.52 kg × (1.39 m/s)^2
D ≈ (0.5 × 2.52 kg × (1.39 m/s)^2) / 7.41 J
D ≈ 0.186 m
Therefore, the combination of the bullet-embedded block and the stationary block travels approximately 0.186 m before stopping.