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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

An unstable nucleus of muss 1.7 × 10−26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One panicle, having a mass of m1 = 5.0 × 10−27 kg, moves in the positive y-direction with speed v1 = 6.0 × 106 m/s. Another particle, of mass m2 = 8.4 × 10−27 kg, moves in the positive x-direction with speed v2 = 4.0 × 105 m/s. Find the magnitude and direction of the velocity of the third particle.

To determine
The magnitude and direction of the third particle.

Explanation

Given Info:

The mass of the nucleus is 1.7×1026kg the mass of first particle is 5.0×1027kg , mass of second particle is 8.4×1027kg , speed of the first particle is 6.0×106ms1 , speed of the second particle is 4.0×106ms1 .

Explanation

Formula to calculate mass of the third particle is,

m3=M(m1+m2)

  • M is the mass of the nucleus
  • m1 is the mass of the first particle
  • m2 is the mass of the second particle

Substitute 1.7×1026kg for M, 5.0×1027kg for m1 , and 8.4×1027kg for m2 to calculate m3 .

m3=1.7×1026kg(5.0×1027kg+8.4×1027kg)=3.6×1027kg

Apply conservation of momentum in y direction.

m1v1y+m2v2y+m3v3y=0

Since m2 has no component along y direction therefore v2y is zero.

Use 0m/s for v3y in the above equation and rewrite in terms of v3y .

m3v3y=m1v1yv3y=m1v1ym3

Substitute 5.0×1027kg for m1 , 6.0×106ms1 for v1y and 3.6×1027kg for m3 in the above equation to calculate v3y .

v3y=(5.0×1027kg)(6.0×106ms1)(3.6×1027kg)=3×10203.6×1027=8.33×106ms1

Apply conservation of momentum in x direction

m1v1x+m2v2x+m3v3x=0

Since m1 has no component along x direction therefore v1x is zero.

Use 0m/s for v3x in the above equation to calculate v3x

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