Question

. 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.

Answer 1

Answer:

• 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 2

Answer:

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

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

(c) 17.36 x 10^-14 J

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

(c) The resultant velocity of third mass, $v_{3}=\sqrt{v_{x}^{2}+v_{y}^{2}}$

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

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