A 5 g particle moving 60 m/s collides with a 2 g particleintially at rest. After
ID: 1741211 • Letter: A
Question
A 5 g particle moving 60 m/s collides with a 2 g particleintially at rest. After the collision each of the particles has avelocity that is directed 30 degrees from the original direction ofmotion of the 5 g particle. What is the speed of the 2 gparticle after the collision? Thanks! A 5 g particle moving 60 m/s collides with a 2 g particleintially at rest. After the collision each of the particles has avelocity that is directed 30 degrees from the original direction ofmotion of the 5 g particle. What is the speed of the 2 gparticle after the collision? Thanks!Explanation / Answer
Given masses m = 5 g M = 2 g u = 60 m / s U = 0 m / s angle = 30 degrees In original direction : --------------------- from law of conservation of momentum , m u + M U = m v cos30 + M V cos 30 5 * 60 + 0 = 5 * v * 0.866 + 2 V * 0.866 from this 5 v + 2 V =346.41 ------( 1) Perpendicular to original direction : ---------------------------------- from law of conservation of momentum , 0 = mv sin30 - M V sin 30 0 = 5 * v *0.5 - 2 V * 0.5 from this 2.5 v - V = 0 V = 2.5 v substitue this in eq ( 1) we get 5 v + 2 ( 2.5 v ) =346.41 v= 34.641 m / s So, the speed of the2 g particle after the collision is V = 2.5 v = 86.6 m /s 0 = 5 * v *0.5 - 2 V * 0.5 from this 2.5 v - V = 0 V = 2.5 v substitue this in eq ( 1) we get 5 v + 2 ( 2.5 v ) =346.41 v= 34.641 m / s So, the speed of the2 g particle after the collision is V = 2.5 v = 86.6 m /sRelated Questions
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