Question

A ball with a mass of 5.8 kg is moving to the right at 10.3 m/s and collides with a stationary ball of mass 10.9 kg. After the collision, the 5.8 kg ball moves to the right, but at an angle of 20° below the horizontal, at a speed of 4.3 m/s. Find the speed of the 10.9 kg ball if it travels to the right, but at an angle above the horizontal.

Answer 1

m1•v1+0 = m1•u(1x) +m2•u(2x)
0=-m1•u(1y) + m2•u(2y)

m1•v1 = m1•u1•cos20⁰+m1•u2•cosα
0= - m1•u1•sin20⁰ +m2•u2•sinα .

m1•(v1 -u1•cos20⁰)=m1•u2•cosα, .....(1)
m1•u1•sin20⁰ = m2•u2•sinα ...............(2)
Divide (2) by (1)
tan α = m1•u1•sin20⁰/m1•(v1 - u1•cos20⁰)=
=u1•sin20⁰/(v1- u1•cos20⁰)=
=4.3•0.34/(10.3-4.3•cos20⁰)=0.234.
α = 13.16⁰
u2=m1•v1•sin20/m2•sinα=
=5.8•10.3•0.34/10.9•0.227 = 8.21 m/s
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