Impulse and Momentum Review … This is what’s on the test

Basic Relations, momentum Momentum: a vector that is mv Find the momentum of a 40kg bike that is going 10m/s South If the momentum of a go cart is 2500kgm/s east when it is going 10m/s east, what is its mass? If the momentum of a 100kg go cart is 2500kgm/s east, find its velocity

Basic Relations, momentum Find the momentum of a 40kg bike that is going 10m/s South = 400kgm/s South If the momentum of a go cart is 2500kgm/s east when it is going 10m/s east, what is its mass? 250kg If the momentum of a 100kg go cart is 2500kgm/s east, find its velocity 25m/s East

Basic Relations, momentum If the momentum of a go cart is 2500kgm/s east, what would its momentum be if it doubled its speed? If the momentum of a go cart is 2500kgm/s east, what would the momentum of another go cart be if its mass is 25% greater but it has the same velocity?

Basic Relations, momentum If the momentum of a go cart is 2500kgm/s east, what would its momentum be if it doubled its speed? Double, 5000kgm/s east If the momentum of a go cart is 2500kgm/s east, what would the momentum of another go cart be if its mass is 25% greater but it has the same velocity? 25% greater, 3250kgm/s

Basic Relations, impulse Impulse = Ft Find the impulse of a 4000N average force to the west that acts for 50s. Find the impulse of a 4000N average force to the west that acts for half the time as above. Will the change in momentum be the same, greater, or less for the 50s impulse?

Basic Relations, impulse Impulse = Ft Find the impulse of a 4000N average force to the west that acts for 50s. = 200,000Ns west Find the impulse of a 4000N average force to the west that acts for half the time as above. = 100,000Ns west Will the change in momentum be the same, greater, or less for the 50s impulse? Greater (2x greater)

Basic Relations, impulse is the change in momentum Impulse = Ft = change in momentum If (and only if) the momentum change is the same, force will decrease if time increases. An object has 400kgm/s of momentum to the north when a 10N force to the north is applied for 2s. Find the change in momentum and the final momentum. An object has 400kgm/s of momentum to the north when a 10N force to the south is applied for 2s. Find the change in momentum and the final momentum.

Basic Relations, impulse is the change in momentum An object has 400kgm/s of momentum to the north when a 10N force to the north is applied for 2s. Find the change in momentum and the final momentum. = 20N*s north; 420kg*m/s north An object has 400kgm/s of momentum to the north when a 10N force to the south is applied for 2s. Find the change in momentum and the final momentum. = 20N*s south; 380kg*m/s north

Basic Relations, impulse is the change in momentum An object has 400kgm/s of momentum to the north when a 10N force to the east is applied for 20s. Find the change in momentum and the final momentum. An object has 400kgm/s of momentum to the north. Find the impulse required to stop it.

Basic Relations, impulse is the change in momentum An object has 400kgm/s of momentum to the north when a 10N force to the east is applied for 20s. Find the change in momentum and the final momentum. Pythag = 44.7m/s tan-1 (40/20) = 63 deg North of E

Basic Relations, impulse is the change in momentum An object has 400kgm/s of momentum to the north. Find the impulse required to stop it.  mv = - 400kgm/s = -400N*s = 400N*s south

Impulse and Momentum at an angle An object has 400kgm/s of momentum 30deg east of south. Find the components in the north and east. An object has 400kgm/s of momentum 30deg east of south when a 20N force is applied to it for 20s at 30deg east of south. Find its final momentum and velocity.

Impulse and Momentum at an angle An object has 400kgm/s of momentum 30deg east of south. Find the components in the north and east. 400 sin(30deg) = 200kgm/s east 400 cos(30deg) = 346.4kgm/s east Vel = (400kgm/s)/mass 30deg east of south.

Impulse and Momentum at an angle An object has 400kgm/s of momentum 30deg east of south. Find the components in the north and east. 400 sin(30deg) = 200kgm/s east 400 cos(30deg) = 346.4kgm/s east

Impulse and Momentum at an angle An object has 400kgm/s of momentum 30deg east of south. If it strikes another object that is at rest, what must the total momentum be after the collision in the north and east directions? What condition are we assuming for this to be true?

Impulse and Momentum at an angle Since momentum must be conserved (if there are no outside forces), it’s the same as before 400 sin(30deg) = 200kgm/s east 400 cos(30deg) = 346.4kgm/s east

Conservation of Momentum The total linear momentum of an isolated system remains constant. An isolated system is one for which the vector sum of the external forces acting on the system is zero. All of the momentum before an event must be the same after the event. m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i

Conservation of Momentum If a 80kg guy is rolling on his skates very fast towards his girlfriend at 8m/s while she is rolling towards him at 2m/s. Find their final velocity if they hold on to one another after the collision assuming that she is only half his mass. m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i 8okg*v 1f + 40kg*v 2f = 80kg*8m/s+ 40kg*(-2m/s) Since they move together, v 1f = v 2f = v f

Conservation of Momentum 8okg*v 1f + 40kg*v 2f = 80kg*8m/s+ 40kg*(-2m/s) Since they move together, v 1f = v 2f = v f 8okg*v f + 40kg*v f = 80kg*8m/s+ 40kg*(-2m/s) 8okg*v 1f + 40kg*v 2f = 560kg*m/s 120kg(v 1f ) = 560kg*m/s 4.67m/s in the boyfriend’s initial direction (which was made the pos direction above)

Conservation of Momentum If he pushed her fast enough so that he stops, how fast would the girlfriend have to be going? m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i 8okg*0m/s+ 40kg*v 2f = 80kg*8m/s+ 40kg*(-2m/s) 0+ 40kg*v 2f = 560kg*m/s v 2f = (560kg*m/s)/40kg*v 2f = 14 m/s in the orig direction of the boyfriend

Conservation of Momentum If a superball is 0.4kg and going 25m/s east before striking a wall and returning at 25m/s, find the force that the wall exerted if the time of contact was 0.20 sec.  mv = F*t = 0.4kg(-25m/s – 25m/s) = F*t = 0.4kg(-50m/s) = -20kg*m/s = F*t -20kg*m/s = F*(0.20s) F = -100N

Conservation of Momentum The drawing shows a collision between two pucks on an air hockey table. Puck A has a mass of 0.25 kg and is moving along the x-axis with a velocity of 5.5 m/s. It makes a collision with puck B, which has a mass of 0.5 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a) puck A and (b) puck B.

p initial, total = 5.5(.25) = 1.375kg*m/s to the right. After collision in the y direction mv A sin(65) - mv B sin(37) = 0 After collision in the x direction mv A cos(65)+ mv B cos(37) = 1.375kg*m/s right. After collision in the y direction.25v A (.9063)+.5v B (-.60) = 0; v B = 0.753v A. After collision in the x direction.25v A (.4226)+.5v B (.7986) = 1.375kg*m/s r.25v A (.4226)+.5(0.753v A )(.7986) = 1.375kg*m/s r v A = 1.375kg*m/s rt v A = 3.4m/s; Now plug back into v B = 0.753v A. = 0.753(3.4m/s) v B = 0.753v A. = 0.753(3.4m/s) = 2.55m/s

Impulse again If two objects collide, Newton’s 3 rd law tells us that the forces must be equal but opposite directions. Common sense tells us that if A is in contact with B for some time, t, then B is also in contact with A for time, t. Therefore, the impulse between two interacting bodies will also be equal but opposite (if there are no outside forces)

Test Taking Skills Tip 1: Take everything one step at a time and include units for each step. Tip 2: Watch the signs/directions Tip 3: When adding vectors, don’t forget to add head-to-tail Tip 4: If you put the correct answer, don’t change it!! Tip 5: Ask yourself if the answer makes sense.