Two billiard balls of equal mass undergo a perfectly head-on collision. If one's ball initial speed was 2.00 m/s and the other's was 3.00 m/s in the opposite direction, what will be their speeds after the collision?
2007-11-14 05:06:52 · 4 answers · asked by Rasheena E 1 in Science & Mathematics ➔ Physics
Momentum is always conserved.
Let m be the mass of each billiard ball.
v1 will be the final velocity of the first ball.
v2 will be the final velocity of the second ball.
The initial momentum is (m)(+2.00 m/s) + (m)(-3.00 m/s).
The final momentum is m v1 + m v2.
Since this is a perfectly elastic collision, kinetic energy is also conserved.
Initial KE is 1/2 m (2.00)^2 + 1/2 m (3.00)^2
Final KE is 1/2 m (v1)^2 + 1/2 m (v2)^2
Initial P = Final P
Initial KE = Final KE
You now have two equations with two unknowns (v1 and v2).
Solve them using whatever method of solving linear equations you like best.
2007-11-14 05:19:19 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
You have to write 3 equations: 1) Conservation of kinetic energy before and after the collision, because it's an elastic collision 2) Conservation of linear momentum in the initial direction (let's call it x) 3) Conservation of linear momentum in the y direction Since the oxygen is initially at rest and the helium moves in the y direction after the collision, the equations are not difficult in this case Then you need to solve the system to find the velocities Do you think you can do it?
2016-05-23 03:41:29 · answer #2 · answered by ? 3 · 0⤊ 0⤋
I believe you should be looking to the conservation of momentum equation. Momentum being defined as the product of the objects mass and velocity (not speed, i.e., it's a vector). Therefore, the sum of each balls' momentum before the collision is equal to the sum of the two balls' momentum after collision. Note that you may have to make an assumption to solve the equation, like the two balls will have equal velocity after the collision (they roll away together). Hope that helps. The answer I believe is 0.5 m/s for each one.
2007-11-14 05:15:47 · answer #3 · answered by M 1 · 0⤊ 0⤋
V1f= [ (m1-m2)/(m1+m2) ]*V1i + [ (2*m2)/(m1+m2) ]*V2i
V2f= [ (2*m2)/(m1+m2) ]*V1i + [ (m1-m2)/(m1+m2) ]*V2i
V1i= 2 m/s
V2i= -3m/s
I think thats the equations for Vf for both 1 and 2 of elastic collision equations
2007-11-14 05:24:18 · answer #4 · answered by Anonymous · 0⤊ 0⤋