A collides with Sphere B which in turn goes on to collide off a wall and “Sphere bounce back.” The key in this situation is to look at the different time intervals and describe each individual time in terms of distance divided by velocity. A common error here is to assume that when a velocity is negative, the time will be also. But time is a scalar and therefore cannot be negative. Therefore if you get a negative time, simply ignore the minus sign. 1999 (a) A smooth sphere moves on a horizontal table. It strikes an identical sphere at rest on the table. The latter is at a distance from a vertical cushion. The impact is along the line of centres and normal to the cushion. y The next collision between the spheres takes place at a distance from the cushion. d Prove that d = , where e is the coefficient of restitution for impacts between spheres and between a sphere and cushion. Interpret the result when e = 1. Oblique Collisions - Introduction Introduction To answer these questions we assume that as a result of the collision the of the velocities will i-component change (if the is along the line of centres at impact). i-axis We will look at exceptions later. If a value for is not given then both velocities afterwards will be in terms of e e. Begin by setting up the table as follows: A U Cos U Sin m1 V1 U Sin i j i j B 0 0 m2 V2 0 i j i j (this assumes sphere B was initially at rest this be the case) – won’t always The first part of these questions is always the same; find the final velocities of the two spheres immediately after impact Two laws as before, but this time we just look at the i-direction only The Principle of Conservation of Momentum (P.C.M.) P.C.M.: M1 U1 + M2 U2 = M1 V1 + M 2V2 of Restitution (N.L.R.) Newton’s Law N.L.R.: V2 V1 = -e (U 2– U1) {along line joining the centres} –
“Sphere A collides with Sphere B which in turn goes on to collide off a wall and bounce back.” The key in this situation is to look at the different time intervals and describe each individual time in terms of distance divided by velocity. A common error here is to assume that when a velocity is negative, the time will be also. But time is a scalar and therefore cannot be negative. Therefore if you get a negative time, simply ignore the minus sign. 1999 (a) A smooth sphere moves on a horizontal table. It strikes an identical sphere at rest on the table. The latter is at a distance y from a vertical cushion. The impact is along the line of centres and normal to the cushion. The next collision between the spheres takes place at a distance d from the cushion. Prove that d = , where e is the coefficient of restitution for impacts between spheres and between a sphere and cushion. Interpret the result when e = 1. Oblique Collisions - Introduction Introduction To answer these questions we assume that as a result of the collision the i-component of the velocities will change (if the i-axis is along the line of centres at impact). We will look at exceptions later. If a value for e is not given then both velocities afterwards will be in terms of e. Begin by setting up the table as follows: A U Cos i U Sin j m1 V1 i U Sin j B 0 i 0 j m2 V2 i 0 j (this assumes sphere B was initially at rest – this won’t always be the case) The first part of these questions is always the same; find the final velocities of the two spheres immediately after impact Two laws as before, but this time we just look at the i-direction only The Principle of Conservation of Momentum (P.C.M.) P.C.M.: M1 U1 + M2 U2 = M1 V1 + M 2V2 Newton’s Law of Restitution (N.L.R.) N.L.R.: V2 – V1 = -e (U 2– U1) {along line joining the centres}
q5-impacts-and-collisions-revision.docx