Answer:
The maximum distance that the block will compress the spring after the collision is 0.365 m
Explanation:
Given;
mass of block, m₁ = 15.0kg
spring constant, k = 450 N/m
mass of stone, m₂ = 3 kg
initial velocity of the stone, u₂ = 8.00m/s
final velocity of the stone, v₂ = 2.00m/s
initial velocity of the block, u₁ = 0 (since it is resting on a smooth horizontal table)
final velocity of the block after collision, v₁ = ?
First, apply the principle of conservation of linear momentum, to determine the velocity of the block after collision;
let right direction be positive and left direction be negative
m₁u₁ + m₂u₂ = m₁v₁ - m₂v₂
15(0) + 3(8) = 15(v₁) - 3(2)
24 = 15v₁ - 6
24 + 6 = 15v₁
30 = 15v₁
v₁ = 30 / 15
v₁ = 2 m/s to the right
Finally, apply the principle of conservation energy to determine the maximum distance the block will compress the spring;
KE₁ + PE₁ = KE₂ + PE₂
¹/₂m₁u₁² + ¹/₂Kx₁² = ¹/₂m₁v₁² + ¹/₂Kx₂²
where;
u₁ is the initial velocity of the block after collision before compressing the spring, = 2 m/s
x₁ is the initial compression of the spring
v₁ is the final velocity of the block after compressing the spring
x₂ is the maximum distance that the block will compress the spring after the collision
¹/₂(15)(2)² + ¹/₂(450)(0)² = ¹/₂(15)(0)² + ¹/₂(450)x₂²
30 = 225x₂²
x₂² = 30 / 225
x₂² = 0.13333
x₂ = √0.13333
x₂ = 0.365 m
Therefore, the maximum distance that the block will compress the spring after the collision is 0.365 m
