One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is

Question

One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is shot west at 1.1 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 and sending it north at 1.49 m/s .

a) What is the speed of the first ball after the collision? Express your answer to two significant figures and include the appropriate units.

b) What is the direction of the first ball after the collision? Give the direction as an angle south of east. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

m1 = m                m2 = m

before collision

speeds

u1x = 1.8 i                    u2x = -1.1 i

u1y = 0                           u2y = 0

after collision

v1x = v1*costheta                      v2x = 0

v1y = v1*sintheta                     v2y = 1.49j

from momentum conservation

along x axis

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m*1.8 - m*1.1 = m*v1x + m*0

v1x = 0.7 m/s

along y

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

0 = m*v1y + m*1.49

v1y = -1.49 m/s

(a)

speed v1 = sqrt(v1x^2+v1y^2) = sqrt(0.7^2+1.49^2) = 1.65 m/s

(b)

direction = tan^-1(v1y/v1x) = 64.8 degrees south of east

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