Physics, 18.09.2019 03:00 tinalmath

An unstable atomic nucleus of mass 1.58×10−26 kg initially at rest disintegrates into three particles. one of the particles, of mass 5.04×10−27 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. assume mass is conserved in this process. a. find the x-component of the velocity of the third particle. b. find the y-component of the velocity of the third particle. c. find the total kinetic energy increase in the process.

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The x-component of the velocity of the third particle is $- 1.50 \ 10 ^{7} \frac{m}{s}$ The y-component of the velocity of the third particle is $- 1.34 \ 10 ^{-20} \frac{\ m}{s}$ The increase in kinetic energy is $E= 6.1586 \ 10 ^ {-13} Joules$

Explanation:

We can take conservation of linear momentum to find the velocities:

The initial momentum of the nucleus will be:

$\vec{P}_0=0$

as is at rest.

After the decay, the first particle has a momentum

$\vec{P}_1 = m_1 \ \vec{V}_1 = 5.04 \ 10^{-27} kg \ * ( 0 \ , \ 6.00 \ 10 ^{6} \frac{m}{s})$

$\vec{P}_1 = ( 0 \ , \ 3.024 \ 10 ^{-20} \frac{kg \ m}{s})$

the second one has a momentum

$\vec{P}_2 = m_2 \ \vec{V}_2 = 8.50 \ 10^{-27} kg \ * ( 4.00 \ 10 ^{6} \frac{m}{s} \ , \ 0)$

$\vec{P}_2 = ( 3.4 \ 10 ^{-20} \frac{kg \ m}{s} \ , \ 0)$

By conservation of linear momentum we have:

$\vec{P}_0= \vec{P}_1 + \vec{P}_2 + \vec{P}_3$

$\vec{P}_3= -\vec{P}_1 - \vec{P}_2$

$\vec{P}_3= ( - 3.4 \ 10 ^{-20} \frac{kg \ m}{s} \ , \ - 3.024 \ 10 ^{-20} \frac{kg \ m}{s})$

for the third particle, we know that mass is conserved:

m_0 = m_1 + m_2 + m_3

m_3 = m_0 - m_1 - m_2

$m_3 = 1.58 \ 10^{-26} kg - 5.04 \ 10^{-27} kg - 8.50 \ 10^{-27} kg$

$m_3 = 2.26 \ 10^{-27} kg$

The velocity will be:

$\vec{v}_3 = \frac{1}{m_3} \vec{P}_3$

$\vec{v}_3 = \frac{1}{ 2.26 \ 10^{-27} kg} ( - 3.4 \ 10 ^{-20} \frac{kg \ m}{s} \ , \ - 3.024 \ 10 ^{-20} \frac{kg \ m}{s})$

$\vec{v}_3 = ( - 1.50 \ 10 ^{7} \frac{m}{s} \ , \ - 1.34 \ 10 ^{-7} \frac{\ m}{s})$

The kinetic energy is given by

$E= \frac{1}{2} m_1 v_1^2 + m_2 v_2^2 + m_3 v_3^2$

$E= \frac{1}{2} 5.04 \ 10^{-27} kg \ (6.00 \ 10 ^{6} \frac{m}{s}) ^2 \\ \\ + \frac{1}{2} 8.50 \ 10^{-27} kg \ (4.00 \ 10 ^{6} \frac{m}{s}) ^2 \\ \\ + \frac{1}{2} 2.26 \ 10^{-27} kg \ ((- 1.50 \ 10 ^{7} \frac{m}{s}) ^2+(- 1.34 \ 10 ^{-7} \frac{\ m}{s})^2)$

$E= 9.072 \ 10 ^ {-14} Joules + 6.8 \ 10 ^ {-14} Joules + 2.5425 \ 10^{-13} Joules + 2.029 \ 10^{-13} Joules$

$E= 6.1586 \ 10 ^ {-13} Joules$

And, as the initial kinetic energy is zero, this must be the increase in energy.

Answer from: xem74

(a) - 4.9 x 10^6 m/s

(b) - 4.38 x 10^6 m/s

Explanation:

M = 1.58 x 10^-26 kg

U = 0

m1 = 5.04 x 10^-27 kg

v1 = 6 x 10^6 m/s along Y axis

m2 = 8.5 x 10^-27 kg

v2 = 4 x 10^6 m/s along X axis

Let m3 be the mass of third particle which is moving with velocity v having X axis component is vx and y axis component is vy.

So, m3 = M - m1 - m2 = 1.58 x 10^-26 - 5.04 x 10^-27 - 8.5 x 10^-27

m3 = 6.9 x 10^-27 kg

(a) Use the conservation of momentum along X axis

M x U = m1 x 0 + m2 x v2 + m3 x vx

0 = 0 + 8.5 x 10^-27 x 4 x 10^6 + 6.9 x 10^-27 x vx

vx = - 4.9 x 10^6 m/s

(b) Use the conservation of momentum along Y axis

M x U = m1 x v1 + m2 x 0 + m3 x vy

0 = 5.04 x 10^-27 x 6 x 10^6 + 0 + 6.9 x 10^-27 x vy

vy = - 4.38 x 10^6 m/s

$v_{3}=\sqrt{4.9}^{2}+4.38^{2}}\times 10^{6}=6.57 \times 10^{6}m/s$

The formula for the kinetic energy is $\frac{1}{2}mv^{2}$

Total kinetic energy = $\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}+\frac{1}{2}m_{3}v_{3}^{2}$

$K = \frac{1}{2}\times 5.04\times 10^{-27}\times \left ( 6 \times 10^{6} \right )^{2}+\frac{1}{2}\times 8.5\times 10^{-27}\times \left ( 4 \times 10^{6} \right )^{2}+\frac{1}{2}\times 6.9\times 10^{-27}\times \left ( 6.57 \times 10^{6} \right )^{2}$

K = 17.36 x 10^-14 J

Answer from: Quest

This is what i found on google “what happens to the bond angle when you add or remove an electron domain? the bond angle decreases as you add domains 5. can you force the atoms into new configurations by pushing atoms around? note: we can think of a bond or a lone pair of electrons as a “domain” of electrons.” i hope this you

Answer from: Quest

mistakes are ok

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An unstable atomic nucleus of mass 1.58×10−26 kg initially at rest disintegrates into three particle...

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