Respuesta :
Answer:
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
