number 9 please The nucleus of a particular atom has a mass of 1.7 Times 10^-26
ID: 1361331 • Letter: N
Question
number 9 please
The nucleus of a particular atom has a mass of 1.7 Times 10^-26 kg and may be assumed to be at rest/ It decays into three particles, with the first having a mass of 5.0 Times^-27 kg and a velocity of 6.0 Times 10^6 m/s in the direction of the positive y-axis, and the second having a mass of 8.4 Times 10^-27 kg and a velocity of 4.0 Times 10^6 m/s in the direction of the positive x-axis. Find the magnitude and direction of the velocity of the third particle (which has a mass of 3.6 Times 10^-27 kg, found by requiring that the total mass is conserved). Gordon Summer is playing billiards^8 and hits the cue ball with a velocity of 3.50 It strikes the 8-ball and is deflected at an angle of 40.0 degree relative to its initial and with a final speed of 2.68 m/s. If the collision is elastic, find the angel of the after the collision and the speed of 8-ball. If you require it, the mass of each 0.165 kg. A diagram is provided.Explanation / Answer
let,
mass m1=5*10^-27 kg and velocity v1=6*10^6 m/sec
mass m2=8.4*10^-27 kg and velocity v2=4*10^6 m/sec
mass m3=3.6*10^-27 kg and velocity v3=(v3x)i+(v3y)j
total mass M=(m1+m2+m3) =1.7*10^-26 kg
by using conservation momentum,
first particle is moving along +ve y axis,
momentum along y-axis,
m1*v1+m3*v3y=0
(5*10^-27)*(6*10^6)+3.6*10^-27*v3y=0
==> v3y=-8.3*10^6 m/sec
velocity y-component of third particle is,
v3y=-8.3*10^6 m/sec
and
second particle is moving along +ve x axis,
momentum along x-axis,
m2*v2+m3*v3x=0
(8.4*10^-27)*(4*10^6)+3.6*10^-27*v3x=0
==> v3x=-9.3*10^6 m/sec
velocity x-component of third particle is
v3x=-9.3*10^6 m/sec
magnitude of velocity of third particle is,
V3=sqrt((-9.3*10^6)^2+(-8.3*10^6)^2)
v3=12.46*10^6 m/sec
and
tan(theta)=v3y/v3x
tan(theta)=(-9.3*10^6)/(-8.3*10^6)
===>
theta=48.25 degrees,
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