A 96.0-g wooden block is initially at rest on a rough horizontal surface when a 11.4-g bullet is fired horizontally into (but does not go through) it. After the impact, the block–bullet combination slides 6.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet immediately before impact.
To determine the speed of the bullet immediately before impact, we need to use the principle of conservation of momentum.
The formula for momentum is:
momentum = mass * velocity
Considering the initial situation before impact, the wooden block is at rest. Therefore, the momentum of the block is zero.
The momentum of the bullet before impact can be calculated using the given mass of the bullet (11.4 g) and the unknown velocity (let's call it V_bullet):
momentum_bullet_before = mass_bullet * velocity_bullet_before
= 11.4 g * V_bullet
Now, considering the situation after impact, the block and bullet together slide 6.5 m before coming to rest. We can use this information and the principle of conservation of momentum to determine the final velocity of the block and bullet combination.
Using conservation of momentum, we have:
momentum_before = momentum_after
Before impact, only the bullet has momentum, so the momentum before impact is:
momentum_before = momentum_bullet_before
= 11.4 g * V_bullet
After impact, both the block and bullet have momentum. The block has a mass of 96.0 g, and let's represent the final velocity of the block and bullet combination as V_final:
momentum_after = (mass_bullet + mass_block) * V_final
= (11.4 g + 96.0 g) * V_final
= 107.4 g * V_final
According to the principle of conservation of momentum:
momentum_before = momentum_after
11.4 g * V_bullet = 107.4 g * V_final
Now we can use the formula for calculating frictional force to relate frictional force to the increase in momentum of the bullet-block combination:
frictional_force = coefficient_of_friction * normal_force
The normal force can be calculated as the weight of the block:
normal_force = mass_block * gravity
= 96.0 g * (9.8 m/s^2)
= 940.8 g m/s^2
Using the given coefficient of kinetic friction (0.750), we can calculate the frictional force:
frictional_force = 0.750 * 940.8 g m/s^2
= 705.6 g m/s^2
Now, we relate the frictional force to the increase in momentum of the block-bullet combination:
frictional_force * distance = momentum_after - momentum_before
705.6 g m/s^2 * 6.5 m = 107.4 g * V_final - 11.4 g * V_bullet
Substituting the mass values in kilograms:
705.6 g m/s^2 * 6.5 m = (107.4 g / 1000 kg) * V_final - (11.4 g / 1000 kg) * V_bullet
(705.6 g m/s^2 * 6.5 m) / (107.4 g / 1000 kg) = V_final - (11.4 g / 1000 kg) * V_bullet
Simplifying:
V_final - (11.4 g / 1000 kg) * V_bullet = (705.6 g m/s^2 * 6.5 m) / (107.4 g / 1000 kg)
Now, we can divide both sides of the equation by V_bullet:
V_final / V_bullet - (11.4 g / 1000 kg) = (705.6 g m/s^2 * 6.5 m) / ((107.4 g / 1000 kg) * V_bullet)
Separating the fraction on the right side of the equation:
V_final / V_bullet - (11.4 g / 1000 kg) = (705.6 * 6.5 m * 1000) / (107.4 * V_bullet)
V_final / V_bullet - 0.0114 = 4300 / V_bullet
Multiplying both sides by V_bullet:
V_final - 0.0114 * V_bullet = 4300
Rearranging the equation to isolate V_bullet:
V_bullet - 0.0114 * V_bullet = V_final
V_bullet * (1 - 0.0114) = V_final
V_bullet = V_final / (1 - 0.0114)
Now we substitute the value for V_final, which is zero because the block-bullet combination comes to rest:
V_bullet = 0 / (1 - 0.0114)
V_bullet = 0
Therefore, the speed of the bullet immediately before impact is 0 m/s.
To determine the speed of the bullet immediately before impact, we can use the principle of conservation of momentum. The total momentum of the system before the impact is equal to the total momentum after the impact.
Before the impact:
The block is initially at rest, so its momentum is zero (since p = mv, and v = 0, p = 0).
The momentum of the bullet can be calculated using its mass (m) and initial velocity (v):
Momentum of bullet before impact = m * v
After the impact:
The bullet is embedded in the wooden block, and they both move together.
Let's denote the velocity of the block and bullet combination as V.
The combined mass of the block and bullet is the sum of their individual masses: 96.0 g + 11.4 g.
The momentum of the block–bullet combination after impact is the product of their combined mass and velocity:
Momentum of block–bullet combination after impact = (96.0 g + 11.4 g) * V
Since momentum is conserved, we can equate the before and after impact momenta:
m * v = (96.0 g + 11.4 g) * V
Now, we need to convert the masses from grams to kilograms for consistent units:
m = 11.4 g = 0.0114 kg
96.0 g = 0.096 kg
Substituting the values:
0.0114 kg * v = (0.096 kg + 0.0114 kg) * V
0.0114 kg * v = 0.1074 kg * V
We have one equation with two unknowns (v and V), but we can relate the velocities using the concept of friction.
During the sliding of the block–bullet combination, the only force acting on it is the friction force.
The friction force (F) can be calculated as the product of the normal force (N) and the coefficient of kinetic friction (μk):
F = μk * N
The normal force (N) is equal to the weight of the block–bullet combination (mg):
N = (96.0 g + 11.4 g) * 9.8 m/s^2 (acceleration due to gravity)
Substituting the masses and calculating the normal force:
N = (0.096 kg + 0.0114 kg) * 9.8 m/s^2
Next, we can calculate the friction force, which acts in the opposite direction to the motion of the block–bullet combination:
F = μk * (0.096 kg + 0.0114 kg) * 9.8 m/s^2
The work done by the friction force (W) can be calculated using the equation:
W = F * d
In this case, the work done by the friction force (W) is equal to the change in kinetic energy:
W = ΔKE
The change in kinetic energy of the block–bullet combination is equal to its initial kinetic energy (KEi) minus its final kinetic energy (KEf).
The initial kinetic energy (KEi) is equal to the kinetic energy associated with the bullet:
KEi = (1/2) * m * v^2
The final kinetic energy (KEf) is equal to the kinetic energy associated with the block–bullet combination:
KEf = (1/2) * (96.0 g + 11.4 g) * V^2
Now, we can equate W (friction work) to ΔKE (change in kinetic energy):
F * d = KEi - KEf
Substituting the values and rearranging the equation, we get:
(μk * (0.096 kg + 0.0114 kg) * 9.8 m/s^2) * 6.5 m = (1/2) * 0.0114 kg * v^2 - (1/2) * (0.096 kg + 0.0114 kg) * V^2
Finally, we substitute the value of V from the momentum equation into the friction equation, and solve for v.