An atomic nucleus of mass m travelling with speed v collides elastically with a

Question

An atomic nucleus of mass

m

travelling with speed v collides elastically with a target particle of

mass 3.0

m

(initially at rest) and is scattered at 45

o

.

(a).

What are the final speeds of the two particles?

Advice: eliminate the tar

get particle’s recoil angle by manipulating the equations of momentum

conservation so that can use the identity sin

2

+ cos

2

= 1.

(b).

At what angle does the target particle move after the collision?

(c).

What fraction of the initial kinetic energy i

s transferred to the target particle?

3. An atomic nucleus of mass m travelling with speed v collides elastically with a target particle of mass 3.0m (initially at rest) and is scattered at 45°. (a). What are the final speeds of the two particles? Advice: eliminate the target particle's recoil angle by manipulating the equations of momentum conservation so that can use the identity sin+ cos2 = 1. (b). At what angle does the target particle move after the collision? (c). What fraction of the initial kinetic energy is transferred to the target particle?

Explanation / Answer

(a). low of Conserve of momentum
In original direction: p = mv = 2mUcosA             
where U is the post-collision velocity of the 2m particle and A is measured relative to the original direction of the m particle.Then v = 2UcosA
In transverse direction: p = 0 = 2mUsinA - mV
where V is the post-collision velocity of the m particle, now moving at 90º to its original direction.
Then V = 2UsinA
Divide second by first:
V / v = 2UsinA / 2UcosA = tanA
V = v*tanA

(b). One object (m) is moving off at 900,

      the original momentum as to still be their

    The 2m object is moving at 45 degree .

(C). For an elastic collision, we can also conserve energy:
½mv² = ½mV² + ½(2m)U² m/2 cancels, leaving
v² = V² + 2U² substitute for V and v
(2UcosA)² = (2UsinA)² + 2U²
4U²cos²A = 4U²sin²A + 2U² 2U² cancels, leaving
2cos²A = 2sin²A + 1
I know of only two ways to solve this; iteration and wolfram.
wolfram gives A = /6 = 30º
Then V = vtan30 = 0.577v = v / 3 = v3 / 3
and U = v / 2cos30 = v / (2*3/2) = v / 3 = 0.577v
They have equal speeds.
Since they have the same speeds and the target particle has twice the mass, then 2/3 of the original KE was transferred to the target particle.

      

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