The $$Q$$-value of a nuclear reaction and kinetic energy of the projectile particle, $$K_p$$ are related as

$$x + p \rightarrow \gamma + b$$

Here,

p = projectile particle

x = target nucleus  (remains stationary so its kinetic energy will be zero)

$$\gamma$$ and $$b$$ = products of the reaction

For a nuclear reaction,

$$Q = \text{Final K.E.} - \text{Initial K.E.}$$

So,

Q = $$K_{\gamma} + K_b - K_p$$

Rearranging,

$$Q + K_p = K_{\gamma} + K_b$$

Since kinetic energies of the products are always positive,

$$K_{\gamma} + K_b > 0$$

Hence,

$$\boxed{Q + K_p > 0}$$

Therefore, for the reaction to occur, the sum of the projectile kinetic energy and the Q-value must be positive.