A(n) 11 g bullet is fired into a(n) 111 g block

of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
14 m before coming to rest.
The acceleration of gravity is 9.8 m/s
If the coefficient of friction between the
surface and the block is 0.7, find the speed of
the bullet before impact.
Answer in units of m/s

To find the speed of the bullet before impact, we can apply the principle of conservation of momentum. The total momentum before the impact is equal to the total momentum after the impact.

Before the impact, only the bullet is moving, and after the impact, the block moves with the bullet inside.

The equation for the conservation of momentum is:

m1 * v1i = (m1 + m2) * vf

Where:
m1 = mass of the bullet = 11 g = 0.011 kg
v1i = initial velocity of the bullet
m2 = mass of the block = 111 g = 0.111 kg
vf = final velocity of the bullet and block

Initially, the block is at rest, so its initial velocity is 0 m/s. After the impact, the block and bullet together slide and eventually come to rest.

To find the final velocity, we can use the equation of motion:

vf^2 = vi^2 + 2 * a * d

Where:
vf = final velocity of the bullet and block = 0 m/s (since they come to rest)
vi = initial velocity of the bullet and block
a = acceleration
d = distance traveled by the block and bullet = 14 m

The acceleration is given by the equation:

a = μ * g

Where:
μ = coefficient of friction between the surface and the block = 0.7
g = acceleration due to gravity = 9.8 m/s^2

Substituting the values into the equations, we have:

0 = vi^2 + 2 * (0.7 * 9.8) * 14

Rearranging the equation:

vi^2 = -2 * (0.7 * 9.8) * 14

Solving for vi:

vi = sqrt(-2 * (0.7 * 9.8) * 14)

Using a calculator, we can find:

vi ≈ 11.917 m/s

Therefore, the speed of the bullet before impact is approximately 11.917 m/s.