A 10.5 g bullet is fired into a 104 g wooden block initially at rest on a horizontal surface.The acceleration of gravity is 9.8 m/s2. After impact, the block slides 6.66 m before coming to rest.If the coefficient of friction between block and surface is 0.658 , what was the speed of the bullet immediately before impact?
Answer in units of m/s.
Vbullet = Vblock-bullet * (Mblock-bullet) / Mbullet
1/2 Mbb Vbb^2 = Mbb g * .658 * 6.66
Vbb = √(2 g * .658 * 6.66)
To find the speed of the bullet immediately before impact, we can use the principle of conservation of momentum. The momentum before the impact is equal to the momentum after the impact.
The momentum of an object is calculated by multiplying its mass by its velocity. Given that the bullet has a mass of 10.5 g (or 0.0105 kg), and the wooden block has a mass of 104 g (or 0.104 kg), let's denote the initial speed of the bullet as v.
Before the impact, the bullet has a momentum of 0.0105 kg * v.
After the impact, both the bullet and the wooden block move together. The momentum of the system after the impact is equal to the momentum of the block, which slides a distance of 6.66 m before coming to rest. To find the momentum of the block, we need to know its final velocity.
We can use the kinematic equation for linear motion to find the final velocity of the block. The equation is:
vf^2 = vi^2 + 2aΔx
Where:
vf = final velocity
vi = initial velocity (0 in this case)
a = acceleration
Δx = displacement
The acceleration of the block can be calculated using Newton's second law:
Fnet = m * a
μ * mg = m * a
Where:
Fnet = net force
μ = coefficient of friction
m = mass of the block
g = acceleration due to gravity
Simplifying the equation:
μ * mg = ma
μ * g = a
Now we can substitute the values in the kinematic equation:
vf^2 = 0 + 2 (μ * g) Δx
vf^2 = 2μgΔx
Solving for vf:
vf = √(2μgΔx)
Now, the momentum of the block after the impact is equal to the momentum of the system:
0.104 kg * vf = 0.0105 kg * v
Solving for v:
v = (0.104 kg * vf) / 0.0105 kg
Substituting the value of vf into the equation:
v = (0.104 kg * √(2μgΔx)) / 0.0105 kg
Now we can plug in the given values:
v = (0.104 kg * √(2 * 0.658 * 9.8 m/s^2 * 6.66 m)) / 0.0105 kg
Simplifying the equation:
v = 25.6 m/s
Therefore, the speed of the bullet immediately before impact is 25.6 m/s.