Q71P In Fig. 9-21, projectile particl... [FREE SOLUTION]

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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 $胃_{1} = 64.0 掳$ and the oxygen nucleus recoils with speed localid="1661254577203" $1.20 脳 10^{5} m / s$ and at angle $胃_{1} = 51.0 掳$ .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 $4.15 脳 10^{5} m / s$ .
Initial speed of particle 1 is $4.84 脳 10^{5} m / s$ .

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, $v_{2} = 1.2 脳 10^{5} m / s$

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

Scattering angle $胃_{2} = 51 掳$

Scattering angle $胃_{1} = 64 掳$

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,

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

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

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

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

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

Calculating final speed of a alpha particle,

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

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

(b) Determining the initial speed of alpha particle

Calculating initial speed of a alpha particle,

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