- #1
perfectionist17
- 4
- 0
Problem:
"A ball with the mass of .440 kg moving east (+x direction) with a speed of 3.70 m/s collides head-on with a .220 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each object after the collision?
Relevant equations.
Conservation of momentum:
m* vi of b1 + m* vi of b2= m* vf of b1 + m* vi of b2
and conservation of kinetic energy...
1/2mvf^2= 1/2 mvi^2
Attempt at Solution:
.440(3.70) + .220(0)= .440(vf1) + .220(vf2)
1.63= .44(vf1) + .220(Vf2)
and then...
1/2 (.440)(3.7^2)= 1/2 (.440)(Vf^2) + 1/2(.220)(vf^2)
3.01= " "
**** Where do I go from here? Also, do I have to use the component form (as in, momentum of initial x component= momentum of final x component), etc? What about angles? I'm so confused because I feel like I should be using angles if its 2-D collisions, but none are given in the problem and I'm not sure if they'll go off a 90 degree angle or not?
"A ball with the mass of .440 kg moving east (+x direction) with a speed of 3.70 m/s collides head-on with a .220 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each object after the collision?
Relevant equations.
Conservation of momentum:
m* vi of b1 + m* vi of b2= m* vf of b1 + m* vi of b2
and conservation of kinetic energy...
1/2mvf^2= 1/2 mvi^2
Attempt at Solution:
.440(3.70) + .220(0)= .440(vf1) + .220(vf2)
1.63= .44(vf1) + .220(Vf2)
and then...
1/2 (.440)(3.7^2)= 1/2 (.440)(Vf^2) + 1/2(.220)(vf^2)
3.01= " "
**** Where do I go from here? Also, do I have to use the component form (as in, momentum of initial x component= momentum of final x component), etc? What about angles? I'm so confused because I feel like I should be using angles if its 2-D collisions, but none are given in the problem and I'm not sure if they'll go off a 90 degree angle or not?