A 15.0kg block is attached to a very light horizontal spring of force constant 450N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00kg stone traveling horizontally at 8.00m/s to the right, where upon the stone rebounds at 2.00m/s horizontally to the left. Required:
Find the maximum distance that the block will compress the spring after the collision.

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

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