A 5.20 g pellet is shot horizontally from a BB gun at a speed of 29.1 m/s into a 20.8 g wooden block. The wooden block is attached to a spring and lies on a frictionless table. If the collision is inelastic and the spring constant k = 22.0 N/m, what is the maximum compression of the spring?
momentum is conserved with the pellet/block
find the velocity of the pellet/block
the KE of the p/b is the work that compresses the spring
1/2 m v^2 = 1/2 k x^2
To find the maximum compression of the spring, we need to analyze the conservation of momentum and conservation of energy during the collision.
First, let's analyze the conservation of momentum. According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.
The momentum of the pellet before the collision is given by:
Initial momentum of the pellet = mass of the pellet * velocity of the pellet
p_pellet = m_pellet * v_pellet
= 5.20 g * 29.1 m/s (converting grams to kilograms)
= 0.00520 kg * 29.1 m/s
≈ 0.15132 kg·m/s
The momentum of the block before the collision is zero since it's at rest initially, so:
Initial momentum of the block = 0
After the collision, the pellet and the block move together with a common velocity. Let's call this common velocity "vf".
The total mass of the pellet and the block together after the collision is:
m_total = m_pellet + m_block
= 5.20 g + 20.8 g (converting grams to kilograms)
= 0.00520 kg + 0.0208 kg
= 0.026 kg
Using the conservation of momentum:
Final momentum of the pellet and block = m_total * vf
0.15132 kg·m/s + 0 = 0.026 kg * vf
Simplifying the equation:
0.026 kg * vf = 0.15132 kg·m/s
vf = (0.15132 kg·m/s) / (0.026 kg)
vf ≈ 5.825 m/s
Now that we have the final velocity after the collision, we can analyze the conservation of mechanical energy during the collision.
The mechanical energy before the collision is zero since both the pellet and block are initially at rest.
The mechanical energy after the collision is the potential energy stored in the compressed spring. Assuming no energy is lost in the collision, the kinetic energy of the pellet and block is fully converted into potential energy of the spring.
The potential energy stored in a spring is given by:
Potential energy = (1/2) * k * x^2
Where:
k is the spring constant (given as 22.0 N/m)
x is the compression of the spring (what we want to find)
The total kinetic energy after the collision is given by:
Kinetic energy = (1/2) * m_total * vf^2
Setting the kinetic energy equal to the potential energy, we get:
(1/2) * m_total * vf^2 = (1/2) * k * x^2
Substituting the values we have:
(1/2) * 0.026 kg * (5.825 m/s)^2 = (1/2) * 22.0 N/m * x^2
Simplifying the equation:
0.026 kg * 5.825 m^2/s^2 = 22.0 N/m * x^2
0.0760545 kg·m^2/s^2 = 22.0 N/m * x^2
Now, solve for x, the maximum compression of the spring:
x^2 = (0.0760545 kg·m^2/s^2) / (22.0 N/m)
x^2 ≈ 0.003457 kg·m^2/N
x ≈ sqrt(0.003457 kg·m^2/N)
x ≈ 0.0589 m or 5.89 cm
So, the maximum compression of the spring is approximately 5.89 cm.