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 x2=45degree. The neutron's initial speed is 3.6×10^5 m/s.

a) Determine the angle of the neutron, x1, after the collision.
b) Determine the speeds of the two particles, v'n and v'He, after the collision.

Answer 1

The angle of the neutron after the collision is 45 degrees. The speed of the neutron after the collision is 2.74 × 105 m/s and the speed of the helium nucleus after the collision is 3.05 × 105 m/s.

Step-by-step explanation:

In an elastic collision between a neutron and a helium nucleus, the neutron's angle after the collision is the same as the helium nucleus's angle before the collision. Since the helium nucleus moves off at an angle of 45 degrees, the neutron will also move off at an angle of 45 degrees.

The speeds of the particles after the collision can be found using the conservation of momentum and kinetic energy. By applying these principles, we find that the speed of the neutron after the collision is 2.74 × 10⁵ m/s and the speed of the helium nucleus after the collision is 3.05 × 10⁵m/s.

Answer 2

Answer: (a)theta = 76°

(b). See explanation

Step-by-step explanation:

Since the collision is elastic, we say that there is conservation of momentum and energy.

(a). P(x);

P(ix) = p(fx)

m(n)V(i) = m(¶) V(¶) cos 45° + m(n) V(n) cos theta.

P(y);

P(iy)= p(fy)

0= m(¶} V(¶) sin 45° - m(n) V(n) sin theta

m(¶) V (¶) sin 45° = m(n) V(n) sin theta

Kinetic energy; k(i)= k(f)

1/2 m(n) V(i)^2 = 1/2 m(¶) V(¶)^2 + 1/2 m(n) V(n)^2.

V^2(i) = 4 V(¶)^2 + V(n) ^2 ----(1)

V(i)= 4V(¶) cos 45° + V(n) cos theta ------------------(2).

4V(¶) sin45° = V(n) sin theta ------(3).

Solving the three equations, we have;

Tan theta= 4

Theta= 76°

(b).using the momentum conservation principle that is

Momentum before collision= momentum after collision.

M(1) V(1)+ M(2) V(2) = M'(1) V'(1) + M'(2) V'(2)

4xV(1) + xV(2) = 4x V(1) + xV(2)

Factorising x out;

x[4V(1) + V(2)]= x[4V(1) + V(2)]

Multiply both sides by 1/x

4V(1) + V(2) = 4V(1) + V(2)

4V(1) + V(2)/ 4 = V(1) + V(2)