Aneutron 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.

(a)theta = 76°

(b). See explanation

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)

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach