Question

A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. The other end of the spring is attached to a fixed wall. A lump of putty travels horizontally with a speed of 8.0 m/s towards the block from the side directly opposite the spring. The putty strikes and sticks to the block. What is the maximum distance the spring is compressed after the impact

Answer 1

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

$(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m$

Answer 2

The maximum distance the spring is compressed after the impact is 0.27 m

Assumption

Let the mass of the putty be 3 Kg

How to determine the speed after the impact

• Mass of putty (m₁) = 3 Kg
• Velocity of putty (v₁) = 8 m/s
• Mass of block (m₂) = 15 Kg
• Velocity of block (v₂) = 0 m/s
• Common velocity (v) =?

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Divide both side by (m₁ + m₂)

v = (m₁v₁ + m₂v₂) / (m₁ + m₂)

v = [(3×8) + (15×0)] / (3 + 15)

v = [24 + 0] / 18

v = 24 / 18

v = 1.33 m/s

How to determine the energy

• Velocity (v) = 1.33 m/s
• Mass of putty + block (m)= 3 + 15 = 18 Kg
• Energy (E) =?

E = ½mv²

E = ½ × 18 × 1.33²

E = 15.92 J

How to determine the compression

• Energy (E) = 15.92 J
• Spring constant (K) = 450 N/m

• Compression (e) =?

E = ½Ke²

15.92 = ½ × 450 × e²

15.92 = 225 × e²

Divide both side by 225

e² = 15.92 / 225

Take the square root of both side

e = √15.92 / 225)

e = 0.27 m

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