A 12.0-g piece of clay is launched horizontally at a 102-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 52 N/m. The piece of clay sticks to the side of the block. If the clay-block system compresses the spring by a maximum of 10.0 cm, what was the speed of the piece of clay at impact with the block?

Question

A 12.0-g piece of clay is launched horizontally at a 102-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 52 N/m. The piece of clay sticks to the side of the block. If the clay-block system compresses the spring by a maximum of 10.0 cm, what was the speed of the piece of clay at impact with the block?

m/s

Answer 1

Use 1/2kx^2 = 1/2 (m1+m2) vf.

Use that v to kind
(m1+m2)vf = m1v1i
And watch units.

Answer 2

To find the speed of the piece of clay at impact with the block, we can use the principle of conservation of momentum.

The momentum before the collision is equal to the momentum after the collision, since no external forces act on the system. The momentum of an object is given by the product of its mass and velocity.

Let's assume the initial velocity of the clay is v clay, the final velocity of the clay and block combined after the collision is v final, and the mass of the clay is m clay.

The clay is launched horizontally, so its initial vertical velocity is 0 m/s. Since no external force acts on the system, the vertical component of momentum is conserved. This means that m clay * 0 = (m clay + m block ) * v final, since the clay and block stick together.

From this equation, we can determine that v final = 0 m/s.

Now, let's consider the horizontal component of momentum. The initial horizontal momentum is m clay * v clay, and the final horizontal momentum is (m clay + m block ) * v final.

Since the clay and block stick together, their masses add up. The mass of the block is given as 102 g, or 0.102 kg, and the mass of the clay is given as 12 g, or 0.012 kg.

Using the conservation of momentum, we can set up the following equation:

m clay * v clay = (m clay + m block ) * v final

Substituting the values, we get:

0.012 kg * v clay = (0.012 kg + 0.102 kg) * 0 m/s

Simplifying, we find:

0.012 kg * v clay = 0.114 kg * 0 m/s

Dividing both sides of the equation by 0.012 kg, we get:

v clay = (0.114 kg * 0 m/s) / 0.012 kg

v clay = 0 m/s

Therefore, the speed of the piece of clay at impact with the block is 0 m/s.