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Question: Two trolleys of mass 1 kg and 0.8 kg on an air track collide head on. The 1 kg vehicle is initially travelling at 2 ms-1 towards the 0.8 kg vehicle which is travelling at 3 ms-1. If after the collision the 1 kg vehicle returns along its original path at 0.5 ms-1, what is the velocity of the 0.8 kg vehicle.

We can apply the principle of conservation of momentum to solve this problem. The momentum before the collision should be equal to the momentum after the collision. So we set the equation: Pre-collision momentum = Post-collision momentum Let's denote the velocity of the 0.8 kg vehicle after the collision as 'v'. Before the collision, the momentum of the 1 kg vehicle is 1 kg * 2 m/s = 2 kg*m/s (to the right, let's say), and the momentum of the 0.8 kg vehicle is 0.8 kg * 3 m/s = 2.4 kg*m/s (to the left). So the total momentum before the collision is 2 - 2.4 = -0.4 kg*m/s. (The negative sign indicates that the momentum is to the left.) After the collision, the momentum of the 1 kg vehicle is 1 kg * -0.5 m/s = -0.5 kg*m/s (as it is moving to the left). The momentum of the 0.8 kg vehicle is then 0.8 kg * v. So the total momentum after the collision is -0.5 + 0.8v = -0.4. Setting -0.5 + 0.8v = -0.4, we solve for v and get v = 0.125 m/s. Therefore, the velocity of the 0.8 kg vehicle after collision would be 0.125 m/s to the right.

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