A car is travelling on a straight level road at a uniform speed of 26 m s⁻¹ when the driver notices a tractor 91.2 m ahead - Leaving Cert Applied Maths - Question 1 - 2020

To find the height where the two masses collide:

1. Define the motion equations for both masses:

   • For the 60-gram mass:

   y_{1} = 15t - rac{1}{2} imes 9.8t^2

2. For the 40-gram mass (starts 0.5 seconds later):

   y_{2} = 22.65(t - 0.5) - rac{1}{2} imes 9.8(t - 0.5)^2

3. Set the heights equal for collision:

   $15t - 4.9t^2 = 22.65(t - 0.5) - 4.9(t - 0.5)^2$

4. Resolve and simplify the equation to find t:

   • This will yield an approximate collision time at t = 1s after solving.

5. Substitute t back to find height:

   $h = 15 imes 1 - 4.9 imes (1^2) = 10.1 ext{ m}$

With both masses combining:

1. Calculate the combined mass:

   $m_{total} = 0.06 + 0.04 = 0.1 ext{ kg}$

2. Using conservation of momentum to find initial velocity of combined mass:

   v_{combined} = rac{(0.06 imes 15) + (0.04 imes 22.65)}{0.1}

   After calculating:

   $v_{combined} = 17.75 ext{ m/s}$

3. Determine maximum height reached using Kinematic equations:

   $v^2 = u^2 + 2as$

   Setting final velocity v = 0 at the max height:

   $0 = (17.75)^2 - 2 imes 9.8 imes s$

4. Solve for s:

   s = rac{(17.75)^2}{2 imes 9.8} ext{ m}

   After solving, we can find the total max height as:

   $ext{Greatest height} = h + s ext{ m}$