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

See prev post

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 equation for conservation of momentum is:

m1 * v1 = (m1 + m2) * vf

Where:
m1 and m2 are the masses of the objects involved (the clay and the block)
v1 is the initial velocity of the clay
vf is the final velocity of the combined clay-block system

In this case, the clay and the block stick together after the collision, so their final velocity will be the same. We can denote this final velocity as v.

First, let's find the initial velocity of the clay by rearranging the equation:

v1 = (m1 + m2) * vf / m1

Given:
m1 = 12.0 g = 0.012 kg (converted from grams to kilograms)
m2 = 102 g = 0.102 kg (converted from grams to kilograms)

Now, we need to determine the final velocity vf. Since the clay and block system compresses the spring by a maximum of 10.0 cm, we can use the potential energy stored in the spring to find the final velocity.

The potential energy stored in a spring is given by:

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

Where:
PE is the potential energy stored in the spring
k is the spring constant (52 N/m)
x is the compression of the spring (10.0 cm = 0.10 m)

The potential energy stored in the spring will be equal to the initial kinetic energy of the clay, which is given by:

KE = (1/2) * m1 * v1^2

Setting the two equations equal and solving for vf, we get:

(1/2) * k * x^2 = (1/2) * m1 * v1^2

Simplifying the equation, we have:

k * x^2 = m1 * v1^2

vf = sqrt((k * x^2) / m1)

Now, we can substitute the known values to find vf:

vf = sqrt((52 N/m * (0.10 m)^2) / 0.012 kg)

vf ≈ sqrt((52 N * 0.01 m^2) / 0.012 kg)

vf ≈ sqrt(43.33 m^2/s^2)

vf ≈ 6.58 m/s

Therefore, the speed of the piece of clay at impact with the block is approximately 6.58 m/s.