Example A 45 kg swimmer runs with a horizontal velocity of 5 1 m s off of a boat dock into a stationary 12 kg rubber raft Find the velocity that the swimmer and raft would have after impact if there were no friction and resistance due to the water Solution Given m1 45 kg m2 12 kg v i1 5 1 ms x Find v f 1 v f 2 v i2 0 Consider motion of boy and raft just before and just after impact Boy and raft define the system Neglect friction and air resistance no external forces which act in the direction of motion Therefore we can use the Conservation of Linear Momentum Pi Pf pi1 pi 2 p f 1 p f 2 m1 v i1 m2 v i 2 m1 v f 1 m2 v f 2 v i1 Since the boy moves with the raft after the impact f1 f2 f v v v m1v i1 m1 m2 v f m1 45 vf v i1 5 1 m1 m2 45 12 4 0 ms m v f 4 0 s x v f 1 v f 2 What if we have the case where vf1 vf2 We then have two unknowns So we need another equation Collisions Revisited Return to the boy and the raft conservation of momentum problem But let s assume that the boy misses the raft Then the final velocities of the boy and raft are not equal vf1 vf2 We then have two unknowns with the conservation of momentum equation in one dimension given by m1v i1 m1v f 1 m2 v f 2 We need another equation We can use Conservation of Mechanical Energy but applied to the system No change in y so only K Ei E f 2 2 2 1 1 1 2 m1 v i 1 2 m1 v f 1 2 m2 v f 2 From original conservation of momentum equation solve for vf2 Then substitute into conservation of energy equation vf2 1 2 m1 v i1 v f 1 m2 2 1 2 2 m 2 m1 v m1 v m2 v i1 v f 1 m 2 i1 1 2 2 f1 1 2 2 1 m 2 m1 v v v i1 v f 1 m2 Here is a trick 1 2 m1 v i1 v f 1 v i1 v f 1 2 1 m1 2 v i1 v f 1 v i1 v f 1 m2 m1 v i1 v f 1 v i1 v f 1 m2 m2 v i1 m2 v f 1 m1 v i1 m1 v f 1 v f 1 m1 m2 v i1 m1 m2 m1 m2 v i1 v f 1 Eq 9 12 m1 m2 1 2 2 i1 2 f1 1 2 m1 vf2 v i1 v f 1 m2 m1 m1 m2 vf2 v i1 v i1 m2 m1 m2 m1 m1 m2 m1 m1 m2 v i1 m2 m1 m2 m1m1 m1m2 m1m1 m1m2 v i1 m2 m1 m2 Eq 9 12 2m1m2 2m1 v i1 v i1 v f 2 m1 m2 m2 m1 m2 Use numerical data from boy and raft example 45 12 m v f1 5 1 2 95 s 45 12 2 45 m vf2 5 1 8 1 s 45 12 p f 1 m1 v f 1 45 2 95 133 kg ms p f 2 m2 v f 2 12 8 1 97 kg ms m pi1 m1 v i1 45 5 1 230 kg s Momentum is conserved In almost all of the 2 body problems we will consider the total momentum will be conserved The total mechanical energy may or may not be conserved Two kinds of collisions 1 Elastic energy conserved special case example boy misses raft 2 Inelastic energy not conserved general example boy lands on raft completely inelastic Collisions in 2D Start with Conservation of Linear Momentum vector equation Similar to Newton s 2 nd Law problems break into x and y components P P Pix Pfx i f Piy Pfy Example Problem Three guns are aimed at the center of a circle They are mounted on the circle 120 apart They fire in a timed sequence such that the three bullets collide at the center and mash into a stationary lump Two of the bullets have identical masses of 4 50 g each and speeds of v1 and v2 The third bullet has a mass of 2 50 g and a speed of 575 m s Find the unknown speeds Solution Given m1 m2 4 50 g m3 2 50 g vi3 575 m s vf1 vf2 vf3 0 Find vi1 and vi2 Method If we neglect air resistance then there are no external forces in the horizontal x y plane gravity acts in the vertical direction we can use Conservation of Linear Momentum y Pi Pf m1 v i1 Piy Pfy 120 60 m1 v i1 sin 60 x m2 v i 2 sin 60 0 120 m3 v i 3 60 v i1 v i 2 Pix Pfx m2 v i 2 m1 v i1 cos 60 m2 v i 2 cos 60 m3 v i 3 0 m1 v i1 12 m1 v i1 12 m3 v i 3 m1 v i1 m3 v i 3 m3 2 5 v i1 vi3 575 319 ms v i 2 m1 4 5 Center of Mass The average location for the total mass of a collection of particles system Consider two particles located along a straight line x x xcm m1 x1 m2 x2 m1 m2 If m1 m2 m xcm 1 m1 cm m2 cm x2 mx1 mx2 m x1 x2 x1 x2 m m 2m 2 x Also can define the center of mass for multiple particles N number of particles xcm m1 x1 m2 x2 m3 x3 mN x N m1 m2 m3 mN N N m x m x i i xcm i 1 N i i m i i 1 M N M mi i 1 i 1 Where M is the total mass of the system Can also define the center of mass along other coordinate axes ycm and zcm Can also define the velocity of the center of mass N v cm m1v1 m2 v 2 or v cm m1 m2 N m v i i i M but mi v i P total momentum i P Mv cm If linear momentum is conserved then P0 Pf Mv 0 cm Mv f cm v 0 cm v f cm The velocity of the center of mass is constant It is the same before and after a collision Return to earlier Example m1 45 kg m2 12 kg v01 5 1 m s v02 0 We found vf 4 0 m s Now compute the velocity of the center of mass m1v 01 m2 v 02 45 5 1 12 0 v 0 cm 4 0 ms m1 m2 45 12 m1v f 1 m2 v f 2 45 4 0 12 4 0 v f cm 4 0 ms m1 m2 45 12 m1 x1 m2 …
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