Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Q Serch ignment.asp. I Help ssign 10 Begin Date: 4/9/2018 11:00:00 AM- -Due Date

ID: 2041157 • Letter: Q

Question

Q Serch ignment.asp. I Help ssign 10 Begin Date: 4/9/2018 11:00:00 AM- -Due Date: 4/16/2018 9:00:00 AM End Date: 4/16/2018 9:00:00 AM (25%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1400 kg and is approaching at 5.5 m/s due south. The second car has a mass of 750 kg and is approaching at 17.5 m/s due west. 33% Part(a) Calculate the magnitude of the final velocity, in meters per second of the cars. ee 33% Part (b) Calculate the direction of the final velocity, in degrees south of west, of the cars. 33% Part (c) what is the change in kinetic energy in oules, for the collision? This energy goes into deformation of the cars Grade Summa Deductions Potential AKE I Attempts remai (34 per attemp detailed view sin0 cotanO asin0acos0 atanO acotan0 sinhO cosh0 | tanh() | cotanh() | | 0 O Degrees Radians Hint I give up!

Explanation / Answer

a) apply conservation of momentum in x-direction

(1400 + 750)*vx = 750*17.5

==> vx = 750*17.5/(1400 + 750)

= 6.10 m/s


apply conservation of momentum in y-direction

(1400 + 750)*vy = 1400*5.5

==> vy = 1400*5.5/(1400 + 750)

= 3.58 m/s

so, v = sqrt(vx^2 +vy^2)

= sqrt(6.1^2 + 3.58^2)

= 7.07 m/s

c) change in kinetic energy = KEf - KEi

delta_KE = (1/2)*(m1 + m2)*v^2 - ( (1/2)*m1*v1^2 + (1/2)*m2*v2^2)

= (1/2)*(1400 + 750)*7.07^2 - ( (1/2)*1400*5.5^2 + (1/2)*750*17.5^2)

= -8.23*10^4 J <<<<<<<-----answer

negative represents loss in kinetic energy