B5-QRT29: COLLIDING BALL SYSTEMS—MOMENTUM DIRECTION BEFORE...

B5-QRT29: COLLIDING BALL SYSTEMS—MOMENTUM DIRECTION BEFORE AND AFTER COLLIDING In the figures below, two balls traveling in different directions are about to collide. The balls have the same size and shape, but they have different masses and are traveling at different velocities as shown. A 3 m/s 2 kg 4 m/s 2 kg B 4 m/s 2 kg 2 kg 3 m/s C 2 kg 4 m/s 2 kg 4 m/s D 3 m/s 2 kg 3 m/s 3 kg For the questions below, use the directions indicated by the arrows in the direction rosette, or use J for no direction, K for into the page, or L for out of the page. (a) Identify the closest directional match for the direction of the momentum of the two-ball systems before they collide. Explain your reasoning. (b) Identify the closest directional match for the direction of the momentum of the two-ball systems after they collide if the balls stick together. Explain your reasoning.

Transcript

00:01 This problem two balls collide and we have to identify the closest directional match for direction of momentum of the two ball system before they collide.

00:13 So for a, the momentum is px, x component of momentum, 2 kg times 3 meter per second, 6 kg meter per second.

00:51 And the second ball is moving along positive y axis so y component momentum is 2 kg times 4 meter per second which is 8 so x component is 6 and y component is 8 so the resultant momentum will be at an angle theta with x axis so the closest direction is for collision a the closed direction is now we consider b so for b x component of momentum is 2 kg times 3 meter per second 6 kg meter per second and y component is downward because velocity is downward 2kg times minus 4 meter per second so this is minus 8 kg meter per second so here x component is positive but y component is negative 6 -8 so the resultant will be the resultant will be in the fourth quadrant so the closest directional match is p for b now for collision c the momentum is 2 kg times 4 meter per cent along positive y axis and 2 kg minus 4 meter per second along negative y x so the net momentum is 0 so for c the momentum has no direction so it is j for collision d the net momentum is also 0 because we have momentum velocity times mass along positive x -axis minus 3 meter per second is along negative x -axis sorry here the momentum will not be 0 3 times 2 3 times 3 kg so this is 6 minus 9 equal to minus 3 kg meter per second so this is along negative x axis and y component is 0 so the direction for d will be along negative x axis which is s so for a the direction is for b the direction is p for c the direction is j and for d the direction is s now we have to identify the closest directional match for direction of momentum of the two -bots system after the collide if the balls stick together so from conservation of momentum the momentum momentum after collision is equal to the momentum before collision...

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