Question

A blob of clay with mass 18 grams and velocity 110 m/s collides with a wooden block of mass 2.6 kg. The wooden block is initially at rest and is connected to a spring with k = 700 N/m. What is the final velocity of the clay and wooden block system after the collision?

1) 10 m/s
2) 20 m/s
3) 30 m/s
4) 40 m/s

Answer 1

The final velocity of the clay and wooden block system after the collision is approximately 755.79 m/s.

Step-by-step explanation:

To find the final velocity of the clay and wooden block system after the collision, we can use the principle of conservation of momentum. Before the collision, the momentum of the clay is given by its mass (18 grams) multiplied by its velocity (110 m/s), which is equal to 1,980 grams*m/s. The wooden block is initially at rest, so its momentum is zero.

After the collision, the clay and wooden block stick together and move as one. Let's assume their final velocity is v. Since momentum is conserved, we have:

(mass of clay + mass of wooden block) * v = 1,980 grams*m/s

Converting the masses to kilograms: (0.018 kg + 2.6 kg) * v = 1,980 grams*m/s

Simplifying the equation: 2.618 kg * v = 1,980 kg*m/s

Dividing both sides of the equation by 2.618 kg, we get: v = 1,980 kg*m/s / 2.618 kg = 755.79 m/s

Therefore, the final velocity of the clay and wooden block system after the collision is approximately 755.79 m/s.