geez -- posting it three times will not get anyone to help any sooner. It just annoys those who have to read the same dang thing three times!
Check your first post, "Chris."
Can someone please solve this? I have stuck for hours on this and i need this before tomorrow's morning class! Please and thank you.
Two equal mass hockey pucks collide in a perfectly elastic collision. The first
puck is stationary, and the second puck collides with the first while
travelling 15 m/s [E]. The first puck travels in the direction of [E4°S], while the second puck travels with an unknown velocity, but with a direction of [E18°N]. Determine the final velocities of both pucks
2 answers
"perfectly elastic" means that momentum and K.E. are conserved
let m be the masses of the pucks
initial K.E. = 1/2 * m * 15^2
initial momentum is all in the E direction ... m * v = 15 m
... so the N and S components of the final momenta are equal (and opposite)
let v1 equal the final velocity of the 1st puck , and v2 the velocity of the 2nd
K.E. ... v1^2 + v2^2 = v^2 = 225
momentum ... N/S ... v1 sin(4º) = v2 sin(18º)
... E/W ... v1 cos(4º) + v2 cos(18º) = 15
two equations and two unknowns ... basic algebra ... good luck
let m be the masses of the pucks
initial K.E. = 1/2 * m * 15^2
initial momentum is all in the E direction ... m * v = 15 m
... so the N and S components of the final momenta are equal (and opposite)
let v1 equal the final velocity of the 1st puck , and v2 the velocity of the 2nd
K.E. ... v1^2 + v2^2 = v^2 = 225
momentum ... N/S ... v1 sin(4º) = v2 sin(18º)
... E/W ... v1 cos(4º) + v2 cos(18º) = 15
two equations and two unknowns ... basic algebra ... good luck
4 answers