A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.63kg steel block at rest on a horizontal frictionless surface. On collision, one-half the kinetic energy is converted to internal energy. Find the final speeds.

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

priyankatutor priyankatutor · Answer 2 · 2021-12-16T08:04:36+01:00

The final speed of the given ball law of conservation of energy. The final speed of the given ball is 1.21 m/s.

The law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where

U₁ = initial potential energy

K₁ = initial kinetic energy

E₁ = initial internal energy

U₂ = final potential energy

K₁ = final kinetic energy

E₁ = final internal energy

So,

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy,

ΔE = 1/2ΔK

Where

ΔK = change in kinetic energy.

So,

ΔE = 1/2ΔK = 1/2(K₂ - K₁)

ΔE = K₂/2

ΔE = 1/2(m + M)v²/2

ΔE = (m + M)v²/4

Thus,

mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

v² = 4mgh/3(m + M)

Take square root of both sides,

v = √[4mgh/3(m + M)

put the values in the formula,

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/3(0.514 kg + 2.63 kg)

v = 1.21 m/s

Therefore, the final speed of the given ball is 1.21 m/s.

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