Collision Question: (URGENT) Exam tomorrow Hi all, I have an exam on collision and projectile motion tomorrow, and was wondering if anyone could answer this question: I get an answer, but my friend gets a completely different answer. I was wondering if anyone could just clarify and show us the correct answer. All help greatly appreciated. Thank you for your time. 1. The problem statement, all variables and given/known data A black billiard ball travels up the centre line at 1.5 m/s. It's mass is 250g, it strikes a white ball of the same mass which is propelled at an angle of 35 degrees to the left of the original direction of the black ball. The white ball final speed is 0.9 m/s. What is the final velocity of the black ball? 2. Relevant equations m1u1 + m2u2 = m1v1 + m2v2 tanθ = sinθ/cosθ 3. The attempt at a solution This is what I did: Find x and y components of the unknown final velocity's vector's magnitude. Let Vb = Velocity of black ball θb = angle the black ball is moved in after collision x-component: m1u1 + m2u2 = m1v1 + m2v2 (0.25 * 0) + (0.25 * 1.5 cos 90) = (0.25 * 0.9 cos 125) + (0.25 Vb cos θb) 0 = -0.13 + 0.25 Vb cos θb 0.13 = 0.25 Vb cos θb Vb cos θb = 0.13 / 0.25 = 0.52 m/s y-component m1u1 + m2u2 = m1v1 + m2v2 (0.25 * 0) + (0.25 * 1.5 sin 90) = (0.25 * 0.9 sin 125) + (0.25 * Vb sin θb) 0 + 0.375 = 0.18 + 0.25 Vb sin θb 0.375 - 0.18 = 0.25 Vb sin θb 0.195 = 0.25 Vb sin θb Vb sin θb = 0.195/ 0.25 = 0.78 m/s So from this, using the magnitudes a right angle triangle can be formed: positive 0.78 (north direction) being the vertical side positive 0.52 (east direction) being the horizontal side to find the resultant vector r = sqroot (0.52^2 + 0.78^2) r = 0.87 m/s to find the direction: tanθ = sin θ / cosθ tanθb = vb sin θb / vb cos θb tanθb = 0.78 / 0.52 θb = tan^-1 (0.78 / 0.52) θb = 56.30 Therefore: the black ball after collision is moving at 0.87 m/s in a direction of 56.30 degrees north of east.