suppose that a horizontal force of 0.7 n is required to pull a 5.0 kg block across the table at constant speed. what is the initial speed of a 20 g bullet, fired horizontally into the block, if the bullet imbeds in the initially stationary block and causes the block to slide 1.50 m across the table before coming to rest again??
Use the numbers in the first statement to compute the coefficient of kinetic friction.
In the second part, set the momentum of the bullet equal to the impulse given in the opposite dorection by friction.
To solve this problem, we can use the principles of conservation of momentum and kinetic energy. Let's break it down step-by-step:
Step 1: Calculate the momentum of the block before the bullet is fired.
The momentum of an object is given by the formula: momentum = mass x velocity.
Given:
Mass of the block (m_b) = 5.0 kg
Velocity of the block before the bullet is fired (v_b) = 0 (initially stationary)
Momentum of the block before the bullet is fired (p_b) = m_b x v_b = 5.0 kg x 0 = 0 kg m/s
Step 2: Calculate the momentum of the bullet before impact.
Given:
Mass of the bullet (m_bullet) = 20 g = 0.02 kg (convert grams to kilograms)
Velocity of the bullet before impact (v_bullet) = unknown (let's call it v_i)
The momentum of the bullet before impact (p_bullet) is given by:
p_bullet = m_bullet x v_bullet
Step 3: Calculate the momentum of the combined system after impact.
Since the bullet embeds in the block, we can consider the combined system as one object.
Given:
Mass of the bullet & block after impact (m_combined) = mass of the bullet + mass of the block = m_bullet + m_b = 0.02 kg + 5.0 kg
The final velocity of the combined system after impact (v_combined) is 0 because it comes to rest again.
The momentum of the combined system after impact (p_combined) is given by:
p_combined = m_combined x v_combined = (m_bullet + m_b) x 0 = 0
Step 4: Apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before impact must equal the total momentum after impact.
Mathematically, we can write this as:
p_b + p_bullet = p_combined
Substituting the known values:
0 + (m_bullet x v_i) = 0
Simplifying:
0.02 kg x v_i = 0
Since the left side of the equation is zero, this means that the velocity of the bullet before impact (v_i) is also zero.
Therefore, the initial speed of the bullet fired horizontally into the block is 0 m/s.
To find the initial speed of the bullet, we can use the principle of conservation of momentum. The total momentum before the collision between the bullet and the block is equal to the total momentum after the collision.
First, let's find the momentum before the collision:
The momentum (p) of an object is defined as the product of its mass (m) and its velocity (v).
For the bullet:
Momentum before = mass of the bullet (m1) × initial velocity of the bullet (v1)
Since the bullet is initially at rest, the initial velocity of the bullet (v1) is 0.
So, momentum before = m1 × 0 = 0
For the block:
Momentum before = mass of the block (m2) × initial velocity of the block (v2)
Given:
Mass of the block (m2) = 5.0 kg
Mass of the bullet (m1) = 20 g = 0.02 kg
Force required to pull the block at constant speed = 0.7 N
We know that force (F) is equal to the rate of change of momentum. Since the block is moving at constant speed, the net force on the block is zero after the bullet hits it. Therefore, the force required to pull the block (0.7 N) is equal to the force exerted by the bullet on the block.
Let's define this force as F_bullet.
Therefore, F_bullet = 0.7 N
Now, we can use this force to find the momentum after the collision:
Momentum after = F_bullet × time the force is applied (t)
The time the force is applied can be calculated using the distance the bullet-block system slides (1.50 m) and the velocity of the block (which is the same as the final velocity of the bullet-block system, as they both come to rest together).
The velocity (v) can be calculated using the formula:
v^2 = u^2 + 2as
where
u = initial velocity (which is the speed of the bullet we are looking for)
a = acceleration (which is negative because the system comes to rest)
s = distance (1.50 m)
Rearranging the formula gives:
u = sqrt(v^2 - 2as)
Substituting the known values:
u = sqrt(0^2 - 2(-1.50 m)(0)) = sqrt(0) = 0 (since the system comes to rest at the end)
Now, we have the momentum before the collision as 0 and the momentum after the collision depends on the time the force is applied. Since the initial velocity of the bullet is zero and the momentum before the collision is zero, we can conclude that the time of the force application is also zero. Therefore, the momentum after the collision is also zero.
Using the principle of conservation of momentum:
Momentum before = Momentum after
0 = 0 + momentum of the block after the collision
Let's denote the final velocity of the bullet-block system as V_final.
Therefore:
momentum of the block (m2 × V_final) = 0
Simplifying:
5.0 kg × V_final = 0
This shows that the final velocity of the block (V_final) is zero since the momentum after the collision is zero. And since the block and bullet come to rest together, the final velocity of the bullet is also zero.
Hence, the initial speed of the bullet is zero.