Question

A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s (figure 9-E12) towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compression of the spring.

Answer 1

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ANSWER::

Mass of each block M₁ and M₂ = 2 kg

Initial velocity of 1st block , V₁ = 1 m/s

Initial velocity of 1st block , V₂ = 0 m/s

Spring constant of the spring = 100 N/m

Block 1 strikes spring with a velocity 1 m/s

After collision , velocity starts to decrease continuously and at instant whole system i.e. Block 1 + Compound Spring + Block 2 move together with common velocity .

Let that velocity be V

Using conservation of energy,

(1/2)M₁V₁² + (1/2)M₂V₂² = (1/2)M₁V² + (1/2)M₂V² + (1/2)kx²

(1/2) x 2(1)² + 0 = (1/2) x 2 x V² + (1/2) x 2 x V² + (1/2)(x)² x 100

{x = maximum compression of spring}

1 = 2v² + 50x² {Equation 1}

As there is no external force in horizontal direction , the momentum should be conserved.

M₁V₁ + M₂V₂ = (M₁ + M₂)V

2 x 1 = 4 x V

V = 1/2 m/s {Equation 2}

Putting in Equation 1

1 = 2 x (1/4) + 50x²

1/2 = 50x²

x² = 1/100 m²

x = (1/10)m = 0.1 m = 10 cm

Hope it helps!

Answer 2

Mass of each block MA and MB = 2kg.
Initial velocity of the 1st block, (V) = 1m/s
VA = 1 m/s, VB = 0m/s
Spring constant of the spring = 100 N/m.
The block A strikes the spring with a velocity 1m/s/
After the collision, it’s velocity decreases continuously and at a instant the whole system (Block A + the compound spring + Block B) move together with a common velocity.
Let that velocity be V.

Using conservation of energy,

(1/2) MAVA 2 + (1/2)MBVB 2 = (1/2)MAv2 + (1/2)MBv2 + (1/2)kx2

(1/2) × 2(1)2 + 0 = (1/2) × 2× v2 + (1/2) × 2 × v2 + (1/2) x2 × 100

(Where x = max. compression of spring)

⇒1 = 2v2 + 50x2 …(1)

As there is no external force in the horizontal direction, the momentum should be conserved.