A bullet with a velocity of 200 m/s is shot at a 2-kg block resting on a horizontal frictionless surface. The bullet imbeds in the block and the bullet-block system slides across the surface with a speed of 5 m/s.
A) what is the mass of the bullet?
B)The coefficient of friction between the rough surface and the block,
C) the time it takes the bullet-block combination to come to rest (after it comes in contact with the rough surface)
A) To find the mass of the bullet, we can use the principle of conservation of linear momentum. The initial momentum of the bullet is equal to the final momentum of the bullet-block system. The initial momentum is calculated as the product of the bullet's mass (m) and velocity (v).
Initial momentum = mass of bullet × initial velocity
Final momentum = (mass of bullet + mass of block) × final velocity
Assuming the bullet remains embedded in the block, the mass of the block does not change. Therefore:
Initial momentum = Final momentum
m × 200 m/s = (m + 2 kg) × 5 m/s
Expanding this equation gives:
200m = 5m + 10
195m = 10
m = 10/195
m ≈ 0.051 kg
Therefore, the mass of the bullet is approximately 0.051 kg.
B) To find the coefficient of friction between the rough surface and the block, we can use the concept of friction force. The friction force is equal to the coefficient of friction (μ) multiplied by the normal force (N).
Normal force (N) = mass of the block × acceleration due to gravity
N = 2 kg × 9.8 m/s^2
N = 19.6 N (rounded to one decimal place)
Friction force = coefficient of friction × normal force
Since the block comes to rest, the friction force will be equal to the force exerted by the block and bullet combination at 5 m/s.
Friction force = (mass of bullet + mass of block) × final velocity
Friction force = (0.051 kg + 2 kg) × 5 m/s
Friction force = 2.051 kg × 5 m/s
Friction force = 10.255 N (rounded to three decimal places)
Now, we can find the coefficient of friction (μ) using the friction force and normal force:
Friction force = coefficient of friction × normal force
10.255 N = μ × 19.6 N
μ ≈ 0.523 (rounded to three decimal places)
Therefore, the coefficient of friction between the rough surface and the block is approximately 0.523.
C) To find the time it takes for the bullet-block combination to come to rest after coming in contact with the rough surface, we can use the concept of deceleration. The deceleration is given by the equation:
Deceleration = (final velocity - initial velocity) / time
Here, the final velocity is 0 m/s (as the combination comes to rest), the initial velocity is 5 m/s, and we need to find the time.
0 = (5 m/s - 0 m/s) / time
0 = 5 m/s / time
Solving for time:
time = 5 m/s / 0
time is undefined
The time it takes for the bullet-block combination to come to rest is undefined. This could be due to an error in the problem statement or other factors not accounted for.
To solve this problem, we will need to use two important principles of physics: conservation of momentum and work-energy theorem.
A) To find the mass of the bullet, we can start by applying the principle of conservation of momentum. According to this principle, the total momentum before and after the collision should be equal.
The initial momentum of the bullet-block system can be calculated as the sum of the bullet's momentum and the block's momentum. The bullet's momentum can be found by multiplying its mass (m_bullet) with its velocity (v_bullet). The block is initially at rest, so its momentum is zero.
Initial momentum = m_bullet * v_bullet + 0 = m_bullet * v_bullet
After the collision, the bullet gets embedded in the block, so the system becomes a combined bullet-block system. The total momentum of the system can be calculated by multiplying the combined mass (m_system) with the final velocity (v_system) using the same equation as above.
Total momentum after collision = m_system * v_system
Since momentum is conserved, we can set the initial momentum equal to the final momentum and solve for the mass of the bullet (m_bullet):
m_bullet * v_bullet = m_system * v_system
Given:
v_bullet = 200 m/s
v_system = 5 m/s
m_system = 2 kg (mass of the block)
Substituting the given values into the equation:
200 m/s * m_bullet = 5 m/s * (m_bullet + 2 kg)
Solving for m_bullet:
200 m_bullet = 5 m_bullet + 10 kg
195 m_bullet = 10 kg
m_bullet = 10 kg / 195
m_bullet ≈ 0.051 kg (approximately)
So, the mass of the bullet is approximately 0.051 kg.
B) To calculate the coefficient of friction between the rough surface and the block, we can use the work-energy theorem. According to this theorem, the work done on an object is equal to the change in its kinetic energy.
The kinetic energy of the bullet-block system before coming in contact with the rough surface is given by:
Initial kinetic energy = (1/2) * m_system * v_system^2
The work done by friction on the bullet-block system (which eventually brings it to rest) can be calculated by:
Work done by friction = frictional force * distance
The frictional force can be expressed as the coefficient of friction (μ) multiplied by the normal force (N). In this case, the normal force is equal to the weight of the block, which is given by the formula:
Weight of the block = m_system * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the bullet-block system is initially moving, the work done by friction will be negative and equal to the change in kinetic energy:
Initial kinetic energy - Work done by friction = 0
Using the above information, we can solve for the coefficient of friction (μ):
(1/2) * m_system * v_system^2 - μ * m_system * g * d = 0
where d is the distance over which the bullet-block system comes to rest.
Given:
m_system = 2 kg
v_system = 5 m/s
g = 9.8 m/s^2
Substituting the given values into the equation:
(1/2) * 2 kg * (5 m/s)^2 - μ * 2 kg * 9.8 m/s^2 * d = 0
Simplifying the equation:
25 J - 19.6 μ d = 0
19.6 μ d = 25 J
μ = 25 J / (19.6 J/m)
μ ≈ 1.27 (approximately)
So, the coefficient of friction between the rough surface and the block is approximately 1.27.
C) To find the time it takes for the bullet-block combination to come to rest after coming in contact with the rough surface, we need to use the equation of motion.
The equation for the arrested motion of an object is:
v = u + a * t
where:
v is the final velocity (0 m/s in this case),
u is the initial velocity (5 m/s),
a is the acceleration due to friction,
t is the time taken to stop.
Since the object comes to rest, v = 0 m/s. Substituting the given values into the equation:
0 m/s = 5 m/s + a * t
Solving for t:
a * t = -5 m/s
Given that:
a = μ * g
μ = 1.27 (as calculated in part B)
g = 9.8 m/s^2
Substituting the values:
1.27 * 9.8 m/s^2 * t = - 5 m/s
t = -5 m/s / (1.27 * 9.8 m/s^2)
t ≈ -0.407 seconds (approximately)
However, we cannot have a negative time in this context. Therefore, the approximation above is incorrect.
Upon closer inspection, it is evident that the bullet-block combination will not lose all its velocity in going zero - it will, in fact, decelerate to a certain value.