Q1.
Using the values of time measured, find the ratio of acceleration for these two cases. Do you find that for the same force, when you increased the mass of the cart, the acceleration decreased?
Answer
Yes. The heavier cart takes longer over the same distance, which means its acceleration is smaller.
Both runs start from rest and cover the same distance s
s = ½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22
Doubling the cart's mass makes it slower, so T2 > T1
→ T12 ÷ T22 < 1 → a2 < a1
s = ½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22
Doubling the cart's mass makes it slower, so T2 > T1
→ T12 ÷ T22 < 1 → a2 < a1
Worked illustration: T1 = 1.0 s, T2 = 1.4 s
a2 ÷ a1 = (1.0 s)2 ÷ (1.4 s)2 = 1.00 ÷ 1.96 ≈ 0.51 ≈ ½
Doubling the mass roughly halves the acceleration
a2 ÷ a1 = (1.0 s)2 ÷ (1.4 s)2 = 1.00 ÷ 1.96 ≈ 0.51 ≈ ½
Doubling the mass roughly halves the acceleration
What it means: for a given magnitude of force, the acceleration produced is inversely related to the mass of the object. Put together with Activity 6.3 (a ∝ F for fixed m), you get Newton's second law: a = F ÷ m, or F = ma.
Tip: the measured ratio is usually a little off exactly ½ — friction at the wheels and timing error from the video both contribute.