A projectile of mass m moves to the right with a speed v; (see figure below). The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle perpendicular to the page through O. We wish to find the fractional change of kinetic energy in the system due to the collision. The moment of inertia of a rod is I = moment of inertia of a particle is I = mr². Md² and the 12 a. What is w, the angular speed of the system after the collision. Answer: W= mdvi Md²+md² b. What is the kinetic energy before the collision? Answer: KE₁ = m(v₁) ². c. What is the kinetic energy after the collision? m²d² Answer: KEy = 2 ( M² +4 m²) (v₁) ² 1/ 12

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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KE = mv², KE = Iw², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef
L = mrv, 1 =1w
Transcribed Image Text:KE = mv², KE = Iw², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef L = mrv, 1 =1w
Problem 5:
A projectile of mass m moves to the right with a speed v; (see figure below). The projectile
strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless
axle perpendicular to the page through O. We wish to find the fractional change of kinetic
energy in the system due to the collision. The moment of inertia of a rod is I =
moment of inertia of a particle is I = mr².
Md² and the
12
a. What is w, the angular speed of the system after the collision.
Answer: W=
mdvi
Md²+1md²
4
2
12
b. What is the kinetic energy before the collision? Answer: KE¡ = m(v¡)².
½
c. What is the kinetic energy after the collision?
¹m²d²
4
Answer: KEf
1
Ma²+¹md² (v₁) ²
m
12
O
M
d
O
3
Transcribed Image Text:Problem 5: A projectile of mass m moves to the right with a speed v; (see figure below). The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle perpendicular to the page through O. We wish to find the fractional change of kinetic energy in the system due to the collision. The moment of inertia of a rod is I = moment of inertia of a particle is I = mr². Md² and the 12 a. What is w, the angular speed of the system after the collision. Answer: W= mdvi Md²+1md² 4 2 12 b. What is the kinetic energy before the collision? Answer: KE¡ = m(v¡)². ½ c. What is the kinetic energy after the collision? ¹m²d² 4 Answer: KEf 1 Ma²+¹md² (v₁) ² m 12 O M d O 3
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