Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

We know that the tension in the string produces a force upward on P, and we set it against the forces acting on Q. Given that Q accelerates downwards at 1.4 m/s², we can consider the forces acting on Q:

The net force equation can be expressed as:

$mg - T = ma$

Where:

• T is the tension
• a = 1.4 m/s² (acceleration)

From the forces acting on particle P, we have:

$T = 0.3g imes ext{sin} α + F$

Where F is the frictional force: F = rac{1}{2}R

By substituting R and eliminating T, we find:

• R from previous calculations
• Equating both expressions will yield m, m = 0.4 kg.

For this part of the question, we apply equations of motion. Given that the string breaks after 0.5 s:

The initial velocity of Q (v) after 0.5 seconds is: $v = at = 1.4 imes 0.5 = 0.7 ext{ m/s}$

Upon the string breaking, P experiences a frictional force:

Determine the decelerating force acting on P: $-0.3g ext{sin} α = -F$

Where:

• $F = 0.3a$ and $a = -g = -9.8 ext{ m/s}^2$

Thus: $ext{Deceleration} = 0.7 - 9.8$

We compute the time until P stops: Using the formula: $v_f = v_i + at$ Setting $v_f = 0$: We find: $0 = 0.7 - 9.8t$ Solving for t: t = rac{0.7}{9.8} ext{ seconds} = 0.0714 ext{ seconds or } 1/14 ext{ seconds}