An unstable nucleus of mass 17.0*10^-27 kg, initially at rest at the origin of a

Question

An unstable nucleus of mass 17.0*10^-27 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m1 = 5.0*10^-27 kg moves in the positive y-direction with speed v1 = 6.00*10^6 m/s. Another particle, of mass m2 = 8.40*10^-27 kg moves in the positive x-direction with speed v2 = 3.60*10^6 m/s. Find the velocity of the third particle. I ALREADY SOLVED THIS PART. I need to know: If the initial speed of nuclear mass was 30m/s when the disintegration occurred, what would the velocity of the third particle be? See the work I already did for the first part.

Explanation / Answer

draw this out as momentum vectors the sum of the given two is exactly opposite the third vector (sum of momenta is 0) now you can put the appropriate value of m*v on the length of each vector and use trig to solve for m3v3 then find m3 from the total mass and solve for v3 a quick 'cheat' is that m1v1 is on the Y axis and m2v2 is on the X axis so the x,y components of their sum is (m2v2 , m1v1) so the x,y components of the third vector which is the equilibrant is ( -m2v2 , -m1v1) divide each by m3 to get the x and y components of v3. I like 'easy' so I use a vector capable calculator to get away from all the annoying math. I got v3 = 9.76x10^6 m/s at 220 degrees taking x as 0 and y as 90 this gave me x comp = -7.48x10^6 y comp = -6.27x10^6

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