1. **Problem statement:** Two blocks K and L slide towards each other on a frictionless surface and collide. The contact time during collision is 0.084 s. After collision, block L moves back with momentum 0.12 kgms⁻¹. Find the magnitude of the average force exerted on block L by block K during the collision. 2. **Relevant formula:** The average force $F_{avg}$ exerted during a collision is related to the change in momentum $\Delta p$ and the contact time $\Delta t$ by: $$F_{avg} = \frac{\Delta p}{\Delta t}$$ 3. **Understanding the problem:** The momentum of block L after collision is 0.12 kgms⁻¹ in the opposite direction. The initial momentum of block L before collision is given as 0.46 kgms⁻¹ (towards K). The change in momentum is the difference between final and initial momentum vectors. 4. **Calculate change in momentum:** Since block L reverses direction, the change in momentum magnitude is: $$\Delta p = |p_{final} - p_{initial}| = | -0.12 - 0.46| = 0.58 \text{ kgms}^{-1}$$ 5. **Calculate average force:** Using the contact time $\Delta t = 0.084$ s, $$F_{avg} = \frac{0.58}{0.084} \approx 6.9 \text{ N}$$ 6. **Interpretation:** The average force exerted on block L by block K during the collision is approximately 6.9 N in the direction opposite to L's initial motion. **Final answer:** $$\boxed{6.9 \text{ N}}$$