Elastic collision in 2D?

Question

marble 1 has an initial velocity of 15.5 m/s and collides elastically with marbles 2 and 3 which are at rest and touching one another. Marble 1 is twice the mass but equal size to the other two marbles. If marble 1 is directed right at the contact point of the other two marbles, find the velocity of (a) marble 2, (b) marble 3 and (c) marble 1.
I will try to draw the figure but assume marbel2 and 3 are touching in figure.
________Marbel2 (o)
Marbel 1(o)----------->
________Marbel3 (o)

Answer 1

If everything were lined up exactly inline and marble 1 hits 2 which hits 3, then both marbles 2 and 3 would move together at the same rate as marble 1. (1~~>2~~>3) You have probably seen that demonstrated with the newton's pendulum toy that has 5 chrome balls hanging by strings. When you pull 1 of the balls back and let it swing into the other 4, 1 ball swings off the other end. 2 balls cause 2 to swing off the other end.

In your experiment, the 1st marble will hit the other 2 and release all of it's energy. They will then go at the same rate as the first marble.

V1 = 0
V2 = 15.5 in a direction 60 degrees off to left.
V3 = 15.5 in a direction 60 degrees off to the right.

Answer 2

In an elastic collision the two kinetic power and momentum are conserved. kinetic power in the previous = kinetic power afterwards v = velocity of first merchandise in the previous collision = 5 m/s east u = velocity of first merchandise after collision w = velocity of 2nd merchandise after collision m = mass of first merchandise = 4 kg n = mass of 2nd merchandise = 6 kg (a million/2)mv^2 = (a million/2)mu^2 + (a million/2)nw^2 mv^2 = mu^2 + nw^2 4(5^2) = 4u^2 + 6w^2 a hundred = 4u^2 + 6w^2 50 = 2u^2 + 3w^2 Momemtum ought to be conserved. on the grounds that there is not any circulation interior the N-S direction, this factor to the momentum ought to be 0. Plus the factor in E-W ought to equivalent that for the time of the previous the collision. Momentum in the previous = mv After collision the 1st merchandise is shifting South of East at 30 stages. p(N-S) = m*u*cos(30) p(E-W) = m*u*sin(30) For the 2nd merchandise if it strikes off at an attitude A: q(N-S) = n*w*cos(A) a(E-W) = n*w*sin(A) The N-S aspects ought to be equivalent so: m*u*cos(30) = n*w*cos(A) wcos(A) = (m*u)cos(30)/(n) = (4)(u)[SQRT(3)/2] / (6) wcos(A) = (u/3)SQRT(3) The momentum in the previous ought to equivalent the sum of the E-W after: mv = m*u*sin(30) + n*w*sin(A) 4(5) = 4u(a million/2) + 6wsin(A) 20 = 2u + 6wsin(A) 10 = u + 3wsin(A) this supplies us 3 equations: 50 = 2u^2 + 3w^2................. w^2 = [50 - 2u^2]/3 wcos(A) = (u/3)SQRT(3) .... cos(A) = (u/w)SQRT(3)/3 10 = u + 3wsin(A) ................ sin(A) = (10 - u)/(3w) on the grounds that we could choose for to locate u, we are able to apply the 2nd 2 and the certainty sin^2 + cos^2 =a million to get an equation in only u and w. we are able to combine this with equation a million which only makes use of u and w to do away with w and get a answer for u. sin^2(A) + cos^2(A) = a million [(10 - u)/(3w)]^2 + [(u/w)SQRT(3)/3]^2 = a million a hundred - 20u + u^2 + 3u^2 = 9w^2 From the conservation of power equation: 50 = 2u^2 + 3w^2 or 9w^2 = a hundred and fifty - 6u^2 a hundred - 20u + u^2 + 3u^2 = a hundred and fifty - 6u^2 10u^2 - 20u - 50 = 0 u^2 - 2u - 5 = 0 u = [2 +/- SQRT(4 + 20] / 2 u = a million +/- SQRT(6) u = 3.forty 5 m/s

Answer 3

The velocity of marble~
(a) 1 is 3.875 m/s
(b) 2 is 5.8125 m/s
(c) 3 is 5.8125 m/s
Hope it's correct

Answer 4

Posts: 29 2D Collision

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A rocket of mass M = 30kg is traveling with velocity V = 1500m/s horizontally. When it explodes into two chunks of mass m1 = 20kg and m2 = 10kg. The mass observed to be moving at an angle of (theta1) 20 above the horizontal with a speed of 1000m/s just after the explosion