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

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

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

(c) 17.36 x 10^-14 J

Step-by-step 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)=√(4.9)^(2)+4.38^(2)}* 10^(6)=6.57 * 10^(6)m/s`

The formula for the kinetic energy is
`(1)/(2)mv^(2)`

Total kinetic energy =
`(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)+(1)/(2)m_(3)v_(3)^(2)`

`K = (1)/(2)* 5.04* 10^(-27)* \left ( 6 * 10^(6) \right )^(2)+(1)/(2)* 8.5* 10^(-27)* \left ( 4 * 10^(6) \right )^(2)+(1)/(2)* 6.9* 10^(-27)* \left ( 6.57 * 10^(6) \right )^(2)`

K = 17.36 x 10^-14 J

Answer 2

• The x-component of the velocity of the third particle is
  `- 1.50 \ 10 ^(7) (m)/(s)`

• The y-component of the velocity of the third particle is
  `- 1.34 \ 10 ^(-20) (\ m)/(s)`

• The increase in kinetic energy is
  `E= 6.1586 \ 10 ^ {-13} Joules`

Step-by-step 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) (m)/(s))`

`\vec{P}_1 = ( 0  \ , \ 3.024 \ 10 ^(-20) (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) (m)/(s) \ , \ 0)`

`\vec{P}_2 = (  3.4 \ 10 ^(-20) (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) (kg \ m)/(s) \ , \ - 3.024 \ 10 ^(-20) (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 = (1)/(m_3) \vec{P}_3`

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

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

The kinetic energy is given by

`E= (1)/(2) m_1 v_1^2 + m_2 v_2^2 + m_3 v_3^2`

`E= (1)/(2) 5.04 \ 10^(-27) kg \ (6.00 \ 10 ^(6) (m)/(s)) ^2 \\ \\ + (1)/(2) 8.50 \ 10^(-27) kg \ (4.00 \ 10 ^(6) (m)/(s)) ^2 \\ \\ + (1)/(2) 2.26 \ 10^(-27) kg \ ((- 1.50 \ 10 ^(7) (m)/(s)) ^2+(- 1.34 \ 10 ^(-7) (\ 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.