A 20.0 g bullet is fired horizontally into a 85 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 156 N/m. The bullet becomes embedded in the block. The bullet block system compresses the spring by a maximum of 1.30 cm. (a) What is the velocity of the block once the bullet is imbedded? (b)What was the speed of the bullet at impact with the block?

Law of conservation of energy

KE =PE
(m1+m2) •u²/2 =kx²/2
u=sqrt{kx²/(m1+m2)}=.....
Law of conservation og linear momentum
m1v + 0= (m1+m2)u
v=(m1+m2)u/m1 =....

Work backwards, because you cannot solve it forwards since there are two unknowns.

(a) Start with the energy transferred to the compressed spring. The block transfers energy to the spring, that's why it compresses.

Ek = Ee
mv^2 = kx^2 (halves cancel)
v = sqrt[(kx^2)/m]
v = .310 m/s

(b) Momentum is conserved. The momentum before is the momentum of the bullet. After, it is the combined mass of bullet and block.
Let u be the initial velocity of bullet
mu = (m1+m2)v
u = 1.63 m/s

RICHMOND- you are sir should stop trying to show people how to do physics when you are completely wrong.

To find the velocity of the block once the bullet is embedded, we can use the principle of conservation of momentum.

(a) Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.

Before the collision, only the bullet is moving, and the wooden block is at rest. After the collision, the bullet becomes embedded in the block, so the final system includes both the bullet and the block moving together.

Let's define the variables:
m1 = mass of the bullet = 20.0 g = 0.020 kg
v1 = velocity of the bullet before collision
m2 = mass of the block = 85 g = 0.085 kg
v2 = velocity of the block after collision (what we want to find)

Using the conservation of momentum equation, we have:

(m1 * v1) + (m2 * 0) = (m1 + m2) * v2

Since the wooden block starts at rest, the initial velocity of the bullet is equal to the speed of the bullet at impact:

v1 = speed of bullet at impact (what we need to find)

Now we need to find the speed of the bullet at impact.

(b) To find the speed of the bullet at impact, we can use the principle of conservation of kinetic energy.

The initial kinetic energy of the bullet is equal to its final kinetic energy after embedding in the wooden block. Assume all the energy is conserved and no external work is done.

The initial kinetic energy is given by KE1 = (1/2) * m1 * (speed of bullet at impact)^2

The final kinetic energy is given by KE2 = (1/2) * (m1 + m2) * V^2, where V is the final velocity of both the bullet and the block after the collision.

Since we don't have the value of the final velocity yet, we need to find it.

Using the conservation of energy equation, we have:

KE1 = KE2
(1/2) * m1 * (speed of bullet at impact)^2 = (1/2) * (m1 + m2) * V^2

Now we have two different equations we can solve simultaneously to find both the velocity of the block after the bullet is embedded (a) and the speed of the bullet at impact (b).

Let's calculate the values.

(a) To find the velocity of the block once the bullet is imbedded:

0.020 kg * speed of bullet at impact + 0.085 kg * 0 m/s = (0.020 kg + 0.085 kg) * v2

v2 = (0.020 kg * speed of bullet at impact) / (0.020 kg + 0.085 kg)

(b) To find the speed of the bullet at impact:

(1/2) * 0.020 kg * (speed of bullet at impact)^2 = (1/2) * (0.020 kg + 0.085 kg) * v2^2

Solve for speed of bullet at impact:

(speed of bullet at impact) = sqrt((0.105 kg * v2^2) / 0.020 kg)

Now we can calculate both the velocity of the block once the bullet is embedded and the speed of the bullet at impact using the above equations.