Block A, of mass 0.2 kg, lies at rest on a rough plane - AQA - A-Level Maths Mechanics - Question 18 - 2020 - Paper 2

Once the string breaks, block A accelerates under gravity alone with a deceleration:

$a = -g \sin(\theta) - \mu g \cos(\theta)$

where the values are:

$g = 9.81, \; \mu = 0.17\; \text{and} \; \sin(\theta) =\frac{7}{25}, \; \cos(\theta)=\frac{24}{25}$

Using these values:

$a = -9.81 \cdot \frac{7}{25} - 0.17 \cdot 9.81 \cdot \frac{24}{25}$

Calculating:

$a ≈ -5.88 - 1.6323 ≈ -7.5123$

Using the equation:
$v^2 = u^2 + 2as$

where:

• $u = 0.5$ ms$^{-1}$ (initial speed after the string breaks)
• $v = 0$ ms$^{-1}$ (final speed)
• $a = -7.5123$ ms$^{-2}$
• $s =?$

We solve for $s$:

$0 = (0.5)^2 + 2(-7.5123)s$

Rearranging gives:

$s = \frac{0.25}{2(7.5123)} ≈ 0.01665$

Thus, the distance is approximately 0.01665 m.

One assumption made is that air resistance is negligible, which could affect the accuracy of the calculated distance traveled by block A after the string breaks. Additionally, it is assumed that the pulley does not introduce any additional friction or affect the motion in any other way.