Chapter 9: Q71P (page 253)

In Fig. 9-21, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle $θ_{1} = 64.0 °$ and the oxygen nucleus recoils with speed localid="1661254577203" $1.20 \times 10^{5} m / s$ and at angle $θ_{1} = 51.0 °$ .In atomic mass units, the mass of the alpha particle is 4.00 uand the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle?

Short Answer

Expert verified

Final speed of alpha particle 1 is $4.15 \times 10^{5} m / s$ .
Initial speed of particle 1 is $4.84 \times 10^{5} m / s$ .

Step by step solution

Step 1: Given Data

Mass of particle 2 is, $m_{2} = 16 a m u$

Mass of particle 1 is, $m_{2} = 4 a m u$

Velocity of particle 2 is, $v_{2} = 1.2 \times 10^{5} m / s$

Velocity of particle 1 is, $v_{1} = 4.15 \times 10^{5} m / s$

Scattering angle $θ_{2} = 51 °$

Scattering angle $θ_{1} = 64 °$

Determining the concept

Consider particle 1 as alpha particle and particle 2 as oxygen. When they scatter, they may conserve energy as well as momentum. By using the linear momentum equation, find the solution for the initial and final velocities.According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formula is as follow:

Initial momentum = Final momentum

(a) Determining the final speed of alpha particle

Conservation of linear momentum can be expressed as,

$m_{1} v_{1 i n i t i a l} = m_{1} v_{1 f i n a l} \cos θ_{1} + m_{2} v_{2 f i n a l} \cos θ_{2} . . . . . . . . . . . (1) 0 = m_{1} v_{1 f i n a l} \sin θ_{1} - m_{2} v_{2 f i n a l} \sin θ_{2} . . . . . . . . . . . (2)$

From equation (2), to calculatethefinal speed of alpha particle,

$v_{1 f i n a l} = \frac{m_{2} v_{2 f i n a l} \sin θ_{2}}{m_{1} \sin θ_{1}}$

From equation (1), to calculate the initial speed of alpha particle,

$v_{1 i n i t i a l} = \frac{m_{1} v_{1 f i n a l} \cos θ_{1} + m_{2} v_{2 f i n a l} \cos θ_{2}}{m_{1}}$

Calculating final speed of a alpha particle,

$v_{1 f i n a l} = \frac{m_{2} v_{2 f i n a l} \sin θ_{2}}{m_{1} \sin θ_{1}} v_{1 f i n a l} = \frac{(16) (1.2 \times 10^{5}) \sin 51 °}{(4) \sin 64 °} v_{1 f i n a l} = 4.15 \times 10^{5} m / s$

Hence, final speed of alpha particle 1 is $4.15 \times 10^{5} m / s$ .

(b) Determining the initial speed of alpha particle

Calculating initial speed of a alpha particle,

$v_{1 i n i t i a l} = \frac{m_{1} v_{1 f i n a l} \cos θ_{1} + m_{2} v_{2 f i n a l} \cos θ_{2}}{m_{1}} v_{1 i n i t i a l} = \frac{(4) \times (4.15 \times 10^{5}) \cos 64 ° + (16) (1.2 \times 10^{5}) \cos 51 °}{4} v_{1 i n i t i a l} = 4.84 \times 10^{5} m / s$