A 2.08-kg particle has a velocity (2.05 i 2.94 j) m/s, and a 2.92-kg particle ha

Question

A 2.08-kg particle has a velocity (2.05 i 2.94 j) m/s, and a 2.92-kg particle has a velocity (1.01 i 6.03 j) m/s. (a) Find the velocity of the center of mass 25.3 How is the velocity of the center of mass related to the total momentum of the system? i 5.1 Remember that you can work with x and y components completely independently. j m/s (b) Find the total momentum of the system 7.4 X i 5.1 X j kg m/s 14 0/4 points I Previous Answers SerPSE9 9.P.055 My Notes C+D Ask Your Teacher A ball of mass 0.201 kg with a velocity of 1.47 i m/s meets a ball of mass 0.293 kg with a velocity of -0.408 i m/s in a head-on, elastic collision. (a) Find their velocities after the collision Vif 780i X m/s m/s (b) Find the velocity of their center of mass before and after the collision m/s cm, before m/s cm, after

Explanation / Answer

While solving this problem we need to keep everything as vectors.

We are going to work on obtaining the total momentum for the system first, then work backwards to get velocity of the center of mass.

We know that the Momentum (p) is simply mass (m) * velocity (v), so work it in pieces. (p = m v)

mass,m=2.08 kg

p1 = 2.08 kg * (2.05i - 2.94 j) m/s = (4.264i - 6.1152j) kg·m/s

p2 = 2.92 kg * (1.01i + 6.03j) m/s = (2.94i + 17.6076 j) kg·m/s

Total momentum is just the sum of the momenta which is (7.589 i + 11.49 j) kg·m/s.

For velocity of the center of mass, our formula is correct but it simply is just a rephrasing of the p = m v formula (But now, p is the total momentum of the system, m is total mass, etc.)

Solving for v: v = p / m = [(7.589 i + 11.49 j) kg·m/s] / (5 kg) = (1.51 i + 2.298 j) m/s.

14.

f v1 =2 and v2 = -3 then v1 = 5 + v2

It appears to v1 that v2 is approaching at 5 m/s as if v1 is not moving.

m1v1             + m2v2                      = m1fv1f           +        m2fv2f

0.201*1.47        + 0.293*(-.408)   = 0.201 * v1f    +        0.293v2f

        v2f - v1f = 1.878 m/s            v2f = 1.878 m/s + v1f

0.201*1.47        + 0.293*(-.408)   = 0.201 * v1f    + 0.293(1.878 m/s + v1f)

0.295         +        - 0.11954              = 0.201v1f        + 0.55 + 0.293v1f

0.295          - 0.12                    = 0.494v1f        + 0.55

v1f      = -1.317 m/s                   

v2f      = 1.50 m/s  

(b)

vCM = (m1fv1f + m2fv2f)/ m

vCM = 0.201*1.47 + 0.293*(-0.408) / 0.494 = +0.356 m/s

This is vCM before the collision, since momentum is conserved, this is also the vCM after the collision.

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