a rifle bullet of mass 8.0g strikes and embeds itself in a block of mass 0.992kg that rests on a frictionless horizontal surface and is attached to a coil spring. the impact compresses the spring 15.0cm. calibration of the spring shows that a force of 0.750N is required to compress the spring 0.250cm.
a). find the magnitude of the velocity of the block just after impact.
b). what was the original speed of the bullet?
a)1/2mv^2 = 1/2kx^2
k=F/x = .75/.0025
so
1/2(.992)v^2 = 1/2(.75/.0025)(.15^2)
Solve for v
b)mv(bullet) = mv(block+bullet)
.008v = 1*v (you found in part a)
Sure, let's tackle this question with a touch of humor!
a) To find the magnitude of the velocity of the block just after impact, we need to use some good ol' physics. Since momentum is conserved during the collision, we can equate the initial momentum of the bullet to the final momentum of the bullet-block system.
Now, imagine the bullet saying to the block, "Hey there, I'm going to join you on this adventure!" And the block replies, "Great, but remember, I'm just going to lay here and take all the momentum you give me."
So, the bullet's initial momentum is given by mass of the bullet (8.0g) multiplied by its initial velocity (let's call it v_i). The final momentum of the bullet-block system is equal to the combined mass of the bullet and the block (since they stick together) multiplied by their final velocity (let's call it v_f).
Hence, we have (8.0g)(v_i) = (0.992kg + 8.0g)(v_f).
Now, let's convert grams to kilograms, so we don't have to deal with mass confusion. Voila! (0.008kg)(v_i) = (0.992kg + 0.008kg)(v_f).
Since embedded bullets are not too good at multitasking, we'll assume the spring did not affect the bullet's velocity. Therefore, the velocity of the block just after impact (v_f) is the same as the magnitude of the bullet's velocity just before impact (v_i).
Now, let's move on to b) and then we'll come back to solve for v_i.
b) To find the original speed of the bullet, we need to figure out how much the spring was compressed by the bullet-block system. The question tells us that compressing the spring by 0.250cm requires a force of 0.750N.
So we can say, "Hey spring, I know you're not a fan of bullet hugs, but would you be so kind as to tell us how much you got compressed?"
And the spring says, "Alright, the force you need to compress me by 0.250cm is 0.750N. But remember, I'm only doing this because I'm a spring with a nice sense of humor."
Now, we know that spring constant (k) is defined as force divided by compression. So we have k = 0.750N / 0.250cm.
But hold on a second! We need to be consistent with units here. We don't want any mix-ups between meters and centimeters. So, let's convert the compression to meters, since the SI unit of k is N/m.
0.250cm is equivalent to 0.0025m. Now we have k = 0.750N / 0.0025m.
Now, let's go back to solving for v_i.
Remember, we have (0.008kg)(v_i) = (0.992kg + 0.008kg)(v_f).
Since v_i = v_f, we can simplify it to (0.008kg)(v_i) = (1kg)(v_i).
Now, we'll substitute the value for v_i and solve the equation.
(0.008kg)(v_i) = (1kg)(v_i).
Finally, we have v_i = 0 (zero).
So, my friend, the original speed of the bullet was 0 m/s!
Now, you might be thinking, "Wait, what? Was this a trick question?" Well, it turns out that the bullet was already at rest just before the collision, giving it an original speed of zero. But hey, at least the block got to make a friend, right?
Hope this answer tickled your physics funny bone!
To solve this problem, we can apply the principles of conservation of momentum and energy.
a) Finding the magnitude of the velocity of the block just after impact:
According to the conservation of momentum, the total momentum before the impact is equal to the total momentum after the impact. The momentum before the impact is given by the equation:
m_bullet * v_bullet = (m_block + m_bullet) * v_block
where:
m_bullet = mass of the bullet = 8.0 g = 0.008 kg
v_bullet = velocity of the bullet before impact (unknown)
m_block = mass of the block = 0.992 kg
v_block = velocity of the block after impact (unknown)
Rearranging the equation, we get:
v_block = (m_bullet * v_bullet) / (m_block + m_bullet)
Now, let's calculate the velocity of the block after impact:
v_block = (0.008 kg * v_bullet) / (0.992 kg + 0.008 kg)
v_block = (0.008 * v_bullet) / 1.0
v_block = 0.008 * v_bullet
Next, we need to use the information about the compression of the spring to find the velocity of the block. According to the principle of conservation of energy, the mechanical energy before the impact is equal to the mechanical energy after the impact.
The mechanical energy before the impact is given by:
E_before = 0.5 * m_bullet * v_bullet^2
The mechanical energy after the impact is given by the potential energy stored in the compressed spring:
E_after = 0.5 * k * x^2
where:
k = spring constant (unknown)
x = compression of the spring = 15.0 cm = 0.15 m
We are given the force required to compress the spring by a certain amount. Using Hooke's Law, we can find the spring constant:
F = k * x
Rearranging the equation, we get:
k = F / x
Now, let's calculate the spring constant:
k = 0.75 N / 0.0025 m
k = 300 N/m
Now that we have the value of the spring constant, we can equate the mechanical energies before and after the impact:
0.5 * m_bullet * v_bullet^2 = 0.5 * k * x^2
Rearranging the equation, we get:
v_bullet^2 = (k * x^2) / m_bullet
v_bullet = √((k * x^2) / m_bullet)
Substituting the given values, we can calculate the magnitude of the velocity of the bullet before impact:
v_bullet = √((300 N/m * (0.15 m)^2) / 0.008 kg)
v_bullet = √(6.75 m^2/s^2 / 0.008 kg)
v_bullet = √843.75 m^2/s^2
v_bullet = 29.06 m/s
Therefore, the magnitude of the velocity of the block just after impact is 29.06 m/s.
b) To find the original speed of the bullet, we need to consider the momentum conservation equation:
m_bullet * v_bullet = (m_block + m_bullet) * v_block
Rearranging the equation, we get:
v_bullet = (m_block + m_bullet) * v_block / m_bullet
Let's substitute the given values to calculate the original speed of the bullet:
v_bullet = (0.992 kg + 0.008 kg) * 29.06 m/s / 0.008 kg
v_bullet = 1.0 kg * 29.06 m/s / 0.008 kg
v_bullet = 1.0 * 29.06 / 0.008 m/s
v_bullet = 3632.5 m/s
Therefore, the original speed of the bullet is 3,632.5 m/s.
To find the magnitude of the velocity of the block just after impact, we can use the principle of conservation of momentum. According to this principle, the total momentum before the impact is equal to the total momentum after the impact.
a) Let's calculate the momentum before the impact:
The momentum of the bullet before the impact can be calculated using the equation:
p_bullet = m_bullet * v_bullet
Given:
Mass of the bullet, m_bullet = 8.0 g = 0.008 kg
Initial velocity of the bullet is unknown, let it be v_bullet.
So, the momentum of the bullet before the impact is:
p_bullet = 0.008 kg * v_bullet
The momentum of the block before the impact is zero since it is at rest.
Now, let's calculate the momentum after the impact:
The bullet is embedded in the block, so the total mass after the impact is the sum of the bullet and block masses:
m_total = m_bullet + m_block
= 0.008 kg + 0.992 kg
= 1 kg
Let the velocity of the block and bullet together after the impact be v_total.
The total momentum after the impact is:
p_total = m_total * v_total
Since there is no external force acting horizontally on the system, the total momentum before and after the impact remains the same. Therefore, we can write:
p_bullet = p_total
Equating the two expressions for momentum, we get:
0.008 kg * v_bullet = 1 kg * v_total
Simplifying the equation, we find:
v_total = (0.008 kg * v_bullet) / (1 kg)
Now, to calculate the magnitude of the velocity of the block just after impact, we need to determine v_total.
b) To calculate the original speed of the bullet, we need to use the principle of conservation of mechanical energy.
The energy stored in the spring is given by:
E_spring = (1/2) * k * x^2
where k is the spring constant and x is the compression of the spring.
The force required to compress the spring is given by:
F = k * x
Given:
Force required to compress the spring, F = 0.75 N
Compression of the spring, x = 0.15 m
We are also given that when the spring is compressed by 0.25 cm (0.0025 m), the force required is 0.75 N. We can use this information to find the spring constant, k.
0.75 N = k * 0.0025 m
Solving for k, we find:
k = 0.75 N / 0.0025 m
Now we have the spring constant, k.
To calculate the original speed of the bullet, we need to consider the conservation of mechanical energy between the bullet and the block-spring system.
The mechanical energy of the bullet before the impact is given by:
E_bullet = (1/2) * m_bullet * v_bullet^2
The mechanical energy stored in the compressed spring is:
E_spring = (1/2) * k * x^2
Since mechanical energy is conserved, we can write:
E_bullet = E_spring
Substituting the values we found earlier, we have:
(1/2) * m_bullet * v_bullet^2 = (1/2) * k * x^2
Solving for v_bullet, we find:
v_bullet = sqrt((k * x^2) / m_bullet)
Now we have the original speed of the bullet.
To summarize:
a) Calculate the magnitude of the velocity of the block just after impact using the equation: v_total = (0.008 kg * v_bullet) / (1 kg).
b) Calculate the original speed of the bullet using the equation: v_bullet = sqrt((k * x^2) / m_bullet), where k is calculated as k = 0.75 N / 0.0025 m.