An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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