Question

A neutron collides elastically with a helium nucleus (at rest
initially) whose mass is four times that of the neutron. The helium
nucleus is observed to move off at an angle *θ*′2=45∘. The
neutron's initial speed is 4.1×10^{5} m/s

Determine the speeds of the two particles, *v*′n and
*v*′He, after the collision.

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Answer #1

**mn = m**

**mHe = 4*mn = 4 m**

**vn = 4.1*10^5 m/s**

**total momentum is conserved**

**along the horizontal ( x axis)**

**mn*vnx + mHe*vHex = mn*v'n*cosalpha +
mHe*v'He*costheta**

**vn + 0 = v'n*cosalpha + 4*v'He*cos45**

**v'n*cosalpha = vn - 4*v'He*cos45 -
.........(1)**

**along the vertical ( y axis)**

**mn*vny + mHe*vHey = mn*v'n*sinalpha -
mHe*v'He*sintheta**

**0 + 0 = v'n*sinalpha - 4*v'He*sin45**

**v'n*sinalpha = 4*v'He*sin45
................(2)**

**(1)^2 + (2)^2**

**(v'n)^2 = (vn)^2 - 2*vn*4*v'He*cos45 + 16*(v'He)^2
..............(3)**

**from energy conservation**

**total kinetic energy is constant**

**(1/2)*mn*vn^2 = (1/2)*mn*v'n^2 +
(1/2)*mHe*v'He^2**

**(vn)^2 = (v'n)^2 + 4*(v'He)^2.............(4)**

**substituting 3 and 4**

**(v'n)^2 = (v'n)^2 + 4*(v'He)^2 - 2*vn*4*v'He*cos45 +
16*(v'He)^2**

**- 2*vn*4*v'He*cos45 + 20*(v'He)^2 = 0**

**v'He = (8/17)*vn*cos45 =
(8/20)*4.1*10^5*cos45**

**v'He = 1.2*10^5 m/s**

**using v'He in eq 4**

**v'n = 3.4*10^5 m/s**

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