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28 November, 15:31

A 20 kg wood ball hangs from a 2.0-mlong wire. The maximum tension the wire can withstand without breaking is 400 N. A 1.0 kg projectile traveling horizontally hits and embeds itself in the wood ball.

1. What is the greatest speed this projectile can have without causing the wire to break?

2. What is the greatest speed the wood ball can have after the collision without breaking?

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  1. 28 November, 16:51
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    1) the greatest speed the projectile can have without causing the wire to break is v₂ = 94.836 m/s

    2) the greatest speed the wood ball can have after the collision without breaking is v = 4.516 m/s

    Explanation

    Since the wood ball generates an circular motion when is hit by the projectile, then from Newton's second law:

    F = m*a, where a = radial acceleration = v²/R (circular motion), F = force, m = mass

    since F=T - m*g cos θ

    T - m*g cos θ = m*v²/R

    where T = tension of the wire, R = radius of the circle = wire's length, θ = angle with respect to the vertical axis

    In the moment that the projectile hits the wood block, the force is

    T₀ - m*g = m*v₀²/R (θ=0)

    then by energy conservation:

    kinetic energy at the bottom + potencial energy at the bottom = kinetic energy at height h + potencial energy at height h

    1/2*m*v₀² = 1/2*m*v² + m*g*R (1 - cos θ)

    v₀²/R = v²/R + 2*g (1 - cos θ)

    v²/R = v₀²/R - 2*g (1 - cos θ)

    therefore

    T - m*g cos θ = m*v²/R = m*v₀²/R - 2*m*g + 2*m*g*cos θ

    since T₀ - m*g = m*v₀²/R

    T - m*g cos θ = T₀ - m*g - 2*m*g + 2*m*g*cos θ

    thus

    T = T₀ - 3*m*g + 3*m*g*cos θ = T₀ - 3*m*g * (1 - cos θ)

    T = T₀ - 3*m*g * (1 - cos θ)

    since T increases with increasing cos θ, the maximum tension is for cos θ=1 (θ=0)

    thus

    T max = T₀

    T₀ - m*g = m*v₀²/R

    v max = v₀ = √[ (T₀/m - g) * R] = √[ (400 N/20kg - 9.8 m/s²) * 2 m] = 4.516 m/s

    then if the projectile embeds itself in the wood ball, we can assume an inelastic collision. Then by conservation of momentum

    m₁*v₁ + m₂*v₂ = (m₁+m₂) * v₀

    where 1 represents the wood ball, and 2 the projectile

    since 1 is at rest initially, v₁=0. Therefore

    m₂*v₂ = (m₁+m₂) * v₀

    v₂ = (m₁+m₂) / m₂ * v₀

    replacing values

    v₂ = (m₁+m₂) / m₂ * v₀ = (20 kg+1 kg) / 1 kg * 4.516 m/s = 94.836 m/s
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