 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/6/d/solution

The required constant deceleration follows from

v^{2} = u^{2} + 2 a s

a = - \frac{4. 8 ^{2}}{2 ( 2 )} = - 5.76 m s^{- 2} .

Both gravity and sliding friction oppose ascent, so

g (sin θ + 0.1 cos θ) = 5.76

. Write

sin θ + 0.1 cos θ = 1.01 sin (θ + arctan 0.1) .

The physical incline solution is $θ = 30. 0^{\circ}$ more precisely

30.038 9^{\circ}

. The combined mass cancels from the equation. The top of the track is consequently

H = 2 sin θ = 1.00117 m

above its base.

 New to topics? Read the docs here!