Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
To perform an experiment to determine the force constant of a spring, you will need a stand with a boss and clamp, a spiral spring, a meter rule and different weights.
The setup is arranged as shown in the image attached. The natural length of the spring is first recorded. Different weights are added to the spring one after the other and the extension is recorded.
The weight is now plotted on the vertical axis and the extension is plotted on the horizontal axis. The slope of the graph is the force constant of the spring.
Learn more: brainly.com/question/10991960
<span>Transformed into potential energy</span>