Question

the pan spiring balance has mass of 300g. when suspended from the coil spring it is found to stretch the spring 10cm. a lump of putty of mass 200g si dropped from rest onto the pan from the height of 30cm. find the maximum distance the pan moves downward​

Answer 1

Answer:

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Answer 2

Given:

Mass of pan (m₁) = 300 g = 0.3 kg

Mass of putty (m₂) = 200 g = 0.2 kg

Spring extended (x₀) = 10 cm = 0.1 m

Height from which putty was dropped (h) = 30 cm = 0.3 m

To find:

The maximum distance the pan moves downward​.

Solution:

Previously when pan was suspended, then mass spring system was balanced.

So, Spring force Fs = kx₀

Gravity force = mg

kx₀ = m₁g

k = m₁g/x₀

k = (0.3)(9.8)/0.1

k = 29.4 N/m

Now when putty was dropped, then when it falls on pan, its initial velocity = v₀

From equation: v₀² = u₀² + 2gh

v₀² = 0 + 2(9.8)(0.3)

v₀ = 2.425 m/s

Initial velocity of pan = 0

Now putty and pan move together with final velocity = vf

From conservation of momentum

pi = pf

m₂v₀ = (m₁ + m₂)vf

vf = m₂v₀/(m₁ + m₂)

vf = (0.2)(2.425)/(0.3 + 0.2)

vf = 0.97 m/s

So, at x = x₀

Kinetic energy of system Ki = (1/2)(m₁ + m₂)vf²

Potential energy of the system Vg = (m₁ + m₂)g(x_max - x₀)

at x = x₀, potential energy in the spring

Vsi = (1/2)kx₀²

Now at x = x_max, potential energy of the spring

Vsf = (1/2)kx_max²

Gravitational potential energy, Ugf = 0

Final kinetic energy, Kf = 0

So, from energy conservation

Ki + Ug + Usi = Kf + Ugf + Usf

1/2(m₁ + m₂)vf² + (m₁ + m₂)g(x_max - x₀) + (1/2)kx₀² = (1/2)kx_max²

(0.5 × 0.97²) + (1 × 9.8)(x_max - 0.1) + (29.4 × 0.1²) = 29.4 × x_max²

0.47045 + 9.8x_max - 0.98 + 0.294 = 29.4 × x_max²

-0.21555 + 9.8 x_max = 29.4 x_max²

29.4 x_max² - 9.8 x_max + 0.21555 = 0

x_max = 0.31 , 0.023

So, x_max = 0.31 m

If, consider 0.21555 ≈ 0

Then, 29.4 x_max² - 9.8 x_max = 0

x_max(29.4 x_max - 9.8) = 0

x_max = 0

Or, 29.4 x_max - 9.8 = 0

29.4 x_max = 9.8

x_max = 9.8 /  29.4

So, x_max = 3.33 m