Collision at an Angle Two cars, both of mass m, collide and stick together. Prio
ID: 1337441 • Letter: C
Question
Collision at an Angle Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle Phi south of east (as indicated in the figure). After the collision, the two-car system travels at speed Ufinal at an angle Theta east of north. (Figure1) Part A Find the speed Ufjnal of the joined cars after the collision. Express your answer in terms of v and phi . vfinal = v/2. suareroot 5-2sinphicosphi Part B This question will be shown after you complete previous question(s).Explanation / Answer
let +ve x axis be along east and positive y axis be along north.
unit vector along +ve x axis is i
unit vector along +ve y axis is j.
veloicty of first car=2*v j
veloicty of second car=v*(cos(phi) i -sin(phi) j)
hence total initial momentum=2*m*v j +m*v*cos(phi) i -m*v*sin(phi) j
=m*v*cos(phi) i +m*v*(2-sin(phi)) j
after collision, they move with a speed v_final and at an angle theta with +ve x axis.
then final momentum=(m+m)*v_final*(sin(theta) i +cos(theta) j)
=2*m*v_final*sin(theta) i + 2*m*v_final*cos(theta) j
using conservation of momentum principle,
initial momentum=final momentum
==>m*v*cos(phi) i +m*v*(2-sin(phi)) j = 2*m*v_final*sin(theta) i + 2*m*v_final*cos(theta) j
equating components along x and y axis respectively,
m*v*cos(phi)=2*m*v_final*sin(theta)
==>v_final*sin(theta)=0.5*v*cos(phi) .....(1)
and
m*v*(2-sin(phi))=2*m*v_final*cos(theta)
==>v_final*cos(theta)=0.5*v*(2-sin(phi))....(2)
squaring both equation 1 and equation 2 and adding them:
v_final^2=0.25*v^2*(cos^2(phi)+4+sin^2(phi)-4*sin(phi))
==>v_final^2=0.25*v^2*(5-4*sin(phi))
==>v_final^2=v^2*(1.25-sin(phi))
==>v_final=v*sqrt(1.25-sin(phi))