Answer:
Bounce 1 , pass 3, emb2
Explanation:
(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle. So it is Bounce1, pass3, emb2.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ
A is the answer for the problem
