Question
+ hotmail . Bing Grades: View fh Main Menu User Login w 75% @D. + 3: e Chegg Study FINAL EX ? Mail-B-1.3-i ??? ? https://www.webassign.net/web/Student/Assignment-Responses/last2dep-18907201 O> Assignment Submission For this assignment, you subrnit answers by question parts The number of submissions remaininch question part anly changes ifyou submit or change the answer Assignment Scoring Yaur last submission is used for your score 8. -125 points SerF3ET9 9.P.083 My Notes Ask Your Teac A 0.500 kg sphere moving with a velocity given by (2.001 3.10j1.00k) mys strikes another sphere of mass 1.50 kg moving with an initial velocity of (-1.00f 2.00j - 2.80k) m/s. (a) The velocity of the 0.500 kg sphere after the collision is (-0.5oi 3j-.ook) m/s. Find the final velocity of the 1.50 kg sphere. m/s Identify the kind of collision (elastic, inelastic, or perfectly inelastic). O elastic O inelastic O perfectly inelastic (b) Now assume the velocity af the 0.500 kg sphere after the collision is (-0 250? + 0.725 1.85k m/s. Find the final velocity of the 1.50 kg sphere. m/s Identify the kind af collision O elastic O inelastic O perfectly inelastic (c) Take the velocity of the 0.500 kg sphere after the collision as (.00l 2.90) +ak) m/s. Find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a positlve value and a negative value. Report each with their corresponding final velocities.) a (positlve value) a (negative value] m/s Need Help? Read it Submit Answer Save Progress Practice Another Version c> Home Miy Assignmenls Exlenin Reques Webkesign3 4.0 1007-2013 Advaneed Inatruesonal Syclema. Ina. All righta reserved. AM 124 6/30/2018
Explanation / Answer
8. m = 0.5 kg
u = 2i - 3.1j + k
M = 1.5 kg
U = -i + 2j - 2.8k
a. after collision, velocities be v and V
then from conservation of momentum
mu + MU = mv + MV
i - 1.55j + 0.5k - 1.5i + 3j - 4.2k = 0.5v + 1.5V
v =-0.5i + 3j - 8k
hnce
-0.5i + 1.45j - 3.7k = -0.25i + 1.5j - 4k + 1.5V
V = (-0.25i - 0.05j + 0.3k)/1.5 = -0.1666666i - 0.0333333333j + 0.2k
|V| = 0.262424177430 m/s
initial KE = 0.5*0.5*(2^2 + 3.1^2 + 1^2) + 0.5*1.5*(1 + 4 + 2.8^2) = 13.2825J
Final KE = 0.5*0.5*(0.5^2 + 9 + 64) + 0.5*1.5*(0.26242417)^2 = 18.363983 J
so the collision is inelastic
b) v = -0.25i + 0.725j - 1.85k
hence
-0.5i + 1.45j - 3.7k = -0.125i + 0.3625j - 0.925k + 1.5*V
V = -0.25i + 0.725j -1.85k
|V| = 2.0026 m/s
in this case
final KE = 4.01046132 J
as v = V, so it is perfectly inelastic collision
c) v = -i + 2.9j + ak
for elastic collision
-0.5i + 1.45j - 3.7k = -0.5i + 1.45j + 0.5ak + 1.5V
V = -(3.7 + 0.5a)k/1.5
for elastic collision
13.2825 = 0.5*0.5*(1 + 2.9^2 + a^2) + 0.5*1.5*(3.7 + 0.5a)^2/1.5^2
10.93 = 0.25a^2*1.5^2 + 0.75(3.7 + 0.5a)^2
0.6625 = 0.75a^2 + 2.775a
a^2 + 3.7a + 0.8833333333 = 0
a = -3.44347628 or 0.2565237163
hence
|V| = 1.31884124 m/s or -2.552166666 m/s
