Two spheres, A and B, with masses m and 2m respectively, are involved in a head-on collision. Sphere A has an initial velocity u, and sphere B is initially at rest. The coefficient of restitution is 1

$$m_A u_A + m_B u_B = m_A v_A + m_B v_B$$

Substitute the given values:

$$m(u) + 2m(0) = m v_A + 2m v_B$$ $$mu = m v_A + 2m v_B$$

Divide by $m$:

$$u = v_A + 2v_B \quad (1)$$

$$e = \frac{v_B - v_A}{u_A - u_B}$$

Substitute the given values:

$$\frac{1}{2} = \frac{v_B - v_A}{u - 0}$$ $$\frac{u}{2} = v_B - v_A \quad (2)$$

From (2), $v_A = v_B - \frac{u}{2}$. Substitute this into (1):

$$u = \left(v_B - \frac{u}{2}\right) + 2v_B$$ $$u = 3v_B - \frac{u}{2}$$ $$u + \frac{u}{2} = 3v_B$$ $$\frac{3u}{2} = 3v_B$$ $$v_B = \frac{u}{2}$$

Now substitute $v_B$ back into (2) to find $v_A$:

$$\frac{u}{2} = \frac{u}{2} - v_A$$ $$v_A = 0$$

The speeds of the spheres after impact are:

• Speed of sphere A: $v_A = 0$
• Speed of sphere B: $v_B = \frac{u}{2}$

$$J = m_B v_B - m_B u_B$$

Substitute the values:

$$J = (2m)\left(\frac{u}{2}\right) - (2m)(0)$$ $$J = mu - 0$$ $$J = mu$$

The magnitude of the impulse exerted by A on B is $\boxed{mu}$.

$$KE_{initial} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2$$ $$KE_{initial} = \frac{1}{2} m (u)^2 + \frac{1}{2} (2m) (0)^2$$ $$KE_{initial} = \frac{1}{2} mu^2$$

$$KE_{final} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2$$ $$KE_{final} = \frac{1}{2} m (0)^2 + \frac{1}{2} (2m) \left(\frac{u}{2}\right)^2$$ $$KE_{final} = 0 + \frac{1}{2} (2m) \left(\frac{u^2}{4}\right)$$ $$KE_{final} = m \frac{u^2}{4} = \frac{1}{4} mu^2$$

$$Loss = KE_{initial} - KE_{final}$$ $$Loss = \frac{1}{2} mu^2 - \frac{1}{4} mu^2$$ $$Loss = \left(\frac{1}{2} - \frac{1}{4}\right) mu^2$$ $$Loss = \frac{1}{4} mu^2$$

The loss of kinetic energy due to the impact is $\boxed{\frac{1}{4} mu^2}$.