For the situation in the figure below, use momentum conservation to determine (a

Question

For the situation in the figure below, use momentum conservation to determine (a) the magnitude and (b) the direction of the final velocity of ball 1 after the collision. The angle = 54.0o.

link to image a: https://www.flickr.com/photos/134122729@N02/19153574892/in/dateposted-public/

link to image b: https://www.flickr.com/photos/134122729@N02/18538746023/in/dateposted-public/

(a) Top view of two balls colliding on a horizontal surface.
(b) This part of the drawing shows the x and y components of the velocity of ball 1 after the collision.

Explanation / Answer

m1 = 0.15

vo1x = vo1*sin54

vo1y = -vo1*cos54

m2 = 0.26

vo2x = vo2

vo2y = 0

after collision

vf1x = vf1*costheta

vf1y = vf1*sintheta

vf2x = vf2*cos35

vf2y = -vf2*sin35

total initial momentum of the balls before collision

Pox = m1*vo1*sin54 + m2*vo2

Poy = -m1*v01*co54 + 0

after collision

pfx = m1*vf1costheta + m2*vf2*cos35

Pfy = m1*vf1*sintheta - m2*vf2*sin35

from momentum conservation

Pfy = Poy

m1*vf1*sintheta - m2*vf2*sin35 = -m1*v01*cos54

0.15*vf1*sintheta = (0.26*0.7*sin35)-(0.15*0.9*cos54)

vf1*sintheta = 0.167 m/s

Pfx = Pox     ,

m1*vf1costheta + m2*vf2*cos35 = m1*vo1*sin54 + m2*vo2

0.15*vf1*costheta = (0.15*0.9*sin54)+(0.26*0.54)-(0.26*0.9*sin54)

vf1*costheta = 0.402

part(a)

vf1 = sqrt(0.167^2+0.402^2) = 0.435 m/s

part(b)

direction

tan theta = 0.167/0.402

theta = 22.56 degrees   <-------answer

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