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

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Physics

One billiard ball is shot east at 1.8 m/s . A second, identicalbilliard ball is shot west at 1.1 m/s . The balls have a glancingcollision, not a head-on collision, deflecting the second ball by90∘ 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 theappropriate units.

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

Answer & Explanation Solved by verified expert
3.7 Ratings (332 Votes)

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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