Answer to Question #164625 in Physics for Johnny Tim

Question

a) The average force F is 667 N and the time the force is applied to the 0.023 kg arrow is Δt1 = 0.0038 s. Calculate the “muzzle velocity” or speed v with which the arrow leaves the crossbow.

b) The arrow has initial speed v and a mass 0.023 kg and the block (initially at rest) has a mass of 0.1 kg. After the collision, block and arrow stick together, calculate the final velocity vf for the moment after this collision.

c) The arrow has an initial speed v and a mass m1 of 0.023kg and the block (initially at rest) has a mass of 1.4 kg. The collision between the arrow and block lasts Δt2 = 0.0075 s. Calculate the magnitude of average force experienced by the arrow during the collision. Compare this average force to the force felt by the block during the collision.

d) The arrow-block system then slides with an initial speed v (vf from part b) a distance d meters over a rough surface where the coefficient of friction between the arrow-block system and the surface is 𝛍k = 0.63. Calculate the distance d, the arrow-block system travels before coming to a stop.

Answer 1

a)

$$F\Delta t_1=m_1v\\667\cdot0.0038=0.023v\\v=110.2\frac{m}{s}$$

b)

$$(m_1+m_2)u=F\Delta t_1\\(0.023+1.4)u=667\cdot0.0038\\u=1.78\frac{m}{s}$$

c)

$$(0.023+1.4)1.78=F\cdot0.0075\\F_1=337.7\ N=0.506F$$

d)

$$F\Delta t_1=m_1v$$

$$m_1v=(m_1+m_2)u=F\Delta t_1$$

$$0.5(m_1+m_2)u^2=\mu (m_1+m_2)gd$$

$$d=\frac{F^2\Delta t_1^2}{2\mu g (m_1+m_2)^2}$$

$$d=\frac{(667\cdot0.0038)^2}{2(0.63)(9.8)(0.023+1.4)^2}=0.27\ m$$