Question

A small ball of mass m, 0.24kg and initial speed vy-11.2m/s makes a collision with another ball of mass m2 -1.08kg which is initially at rest on a frictionless horizontal table. After the collision the balls move as shown in the figure. Given 8-51°, determine the speed of ma just after the collision, that is, v2'? Express your answer using one decimal place. "L. mie mi m2 m2

Answer 1

Sol" :- mi VO m. m₂ m = 0.24 kg 1.08 kg 11.2 mis m₂ = ₂ = 0 51° x-axis from conservation of momentum along miv + m₂ V = m rincoso + ma vi cos(go) Gr mu, to : m, v Los (0) + o IVO + Cosgo=0 mu = mu' Cos (0) ☺ Now again from momenturos conservation along along y-axis

mulsino a msingo mu sino = ma u Singo'z / or V's m ₂ m, sino 2 By putting the value 06 Vi an egn mv, m, mv, Loso m sino or my m₂ų loto

VO mv, m₂ loto V' - my tamo m₂ m 0.24 kg 11.2 mis m₂ = ko8 kg V 51° So ve' 0.24X112 tamisu 1.08 V2 - 2,488 % 1.234 VA 11 3.070 OY 3.1 mis Ang So speed of m₂ after collision V- 3. I mis