An alpha-particle with mass 4u (u is the atomic mass unit) travels from the nega

Question

An alpha-particle with mass 4u (u is the atomic mass unit) travels from the negative x-direction and collides with an oxygen nucleus with mass 16u that is placed at the origin of the x-y coordinate system. After the collision the alpha-particle scatters along direction -64o (below the x-axis) and the oxygen nucleus scatters along direction 51o (above the x-axis), and travels with speed 1.2x105 m/s.

(A) Determine the initial and final speeds of the alpha-particle.

(B) How much of the kinetic energy was lost in the collision?

Explanation / Answer

Mass of alpha particle Ma = 4u

Mass of oxygen =Mo= 16u

initial spped of oxygen = Voi = 0

final speed of oxygen = Vof= 1.2*105m/s

inital speed of alpha particle = Vai = ?

final spped of alpha particle = Vaf = ?

so we need to apply law of conservation of momentum in x and y direction seperately

i.e momentum before and after colision is conserved

along x axis

Ma * Vaix + Mo * Voix = Ma * Vafx + Mo * Vofx

here subscript a-alpha particle

i initial

f final

x - along x axis

y - along y axis

so,

4* Vai + 16 * 0 = 4 * Vaf*Cos64 + 16 * 1.2*105*Cos51

along y axis

inital component of alpha particle and oxygen nucleus =0

so

4*0 + 16*0 = -4*Vaf*sin64 + 16 * 1.2*105 *sin51

so we have 2 equation and 2 unknowns (Vai + Vaf)

solving by method of substitution,

we get

Vai = inital speed of alpha particle = 6.61 * 105 m/s

Vaf = final speed of alpha particle = 4.15 * 105 m/s

b) kinetic energy lost =

inital kinetic energy - final kinetic energy

= 1/2*4*(6.61*105)2 - [1/2*4*(4.15*105)2 + 1/2*16*(1.2*105)2] = 41.42 Joules

36.125 32.160199999999996

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