A 0.330-kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 29.5-N/m force constant that is at its equilibrium length. A 0.0600-kg wad of Play-Doh is thrown horizontally at the block with a speed of 3.00 m/s and sticks to it. Determine the amount by which the Play-Doh–block system compresses the spring?

i used conservation of energy and got 13.53 cm and its wrong.

Before applying conservation of energy law, you have to evaluate the speed of the block+wad system by applying conservation of impulse law:

m(block+wad)*V = m(wad)*V(wad)

So, find V and then use it in conservation of energy law (keep in mind that the the kinetic energy of block+wad system should be calculated using total mass (block+wad) and above evaluated speed V.

were hella gunna fail lmao

To determine the amount by which the Play-Doh-block system compresses the spring, let's analyze the problem using the law of conservation of momentum.

The momentum of an object is given by the formula p = mv, where p is the momentum, m is the mass of the object, and v is its velocity.

Initially, we have two separate objects: the block of wood and the wad of Play-Doh. Since the wad of Play-Doh is thrown horizontally and sticks to the block, their combined momentum before the collision is equal to the momentum after the collision.

Let's denote the velocity of the block of wood after the collision as V_block and the velocity of the wad of Play-Doh after the collision as V_Play-Doh. The mass of the block of wood is 0.330 kg, the mass of the wad of Play-Doh is 0.0600 kg, and their initial velocity is 3.00 m/s.

Using conservation of momentum, we can write:

(m_block * V_block) + (m_Play-Doh * V_Play-Doh) = (m_block + m_Play-Doh) * V_final

(0.330 kg * V_block) + (0.0600 kg * V_Play-Doh) = (0.330 kg + 0.0600 kg) * V_final

(0.330 kg * V_block) + (0.0600 kg * V_Play-Doh) = (0.3900 kg) * V_final

However, we need to consider that the combined Play-Doh–block system compresses the spring, indicating that the final velocity after the collision is zero (since the system comes to rest). Therefore, V_final = 0.

Now we can solve for V_block:

(0.330 kg * V_block) + (0.0600 kg * V_Play-Doh) = 0

Since the wad of Play-Doh sticks to the block, they move as a combined system. Thus, their velocities are the same after the collision, so we can write V_block = V_Play-Doh = V.

Simplifying the equation:

0.330 kg * V + 0.0600 kg * V = 0

0.390 kg * V = 0

Therefore, we find that V = 0 m/s. This means that the final velocity of the Play-Doh-block system after the collision is zero.

Next, we need to use the concept of the work-energy principle to determine the compression of the spring.

When the spring is compressed by an external force, it stores potential energy, given by the formula:

PE_spring = (1/2) * k * x^2

Where PE_spring is the potential energy stored in the spring, k is the force constant of the spring, and x is the distance compressed.

Since the velocity becomes zero, all of the initial kinetic energy of the Play-Doh-block system is converted into potential energy stored in the spring. Therefore, we can equate the initial kinetic energy to the potential energy stored in the spring:

(1/2) * m_system * V^2 = (1/2) * k * x^2

Considering that m_system = m_block + m_Play-Doh and V = 0, the equation simplifies to:

0 = (1/2) * k * x^2

This means that x = 0, indicating that there is no compression of the spring in this scenario.

Therefore, the correct answer is that the Play-Doh–block system does not compress the spring.