Four masses on a horizontal, frictionless surface are linked together by strings P, Q, and R - Scottish Highers Physics - Question 5 - 2019

To determine the tension in the strings P, Q, and R, we first need to analyze the forces acting on each mass.

1. Identify the total mass: The total mass of the system is the sum of all individual masses: $m_{total} = 30 ext{ kg} + 20 ext{ kg} + 10 ext{ kg} + 40 ext{ kg} = 100 ext{ kg}$

2. Apply Newton's Second Law: The total force applied (let's denote it as F) can be calculated using: $F = m_{total} imes a$ where $a$ is the acceleration of the system.

3. Consider the tensions: The tension in string P must support the mass of the first block (30 kg) and the total mass behind it (70 kg). Hence, the tension in P is highest.

4. Analyze individual tensions:

   • Tension in string P (T_P) has to support the entire system behind it.
   • Tension in string Q (T_Q) supports the 20 kg and 10 kg masses behind it, thus will be less than T_P.
   • Tension in string R (T_R) supports only the 40 kg mass at the end.

5. Compare the tensions: Based on these calculations, the tensions can be summarized as:

   • T_P > T_Q > T_R

Thus, the conclusion is that the tension is greatest in string P and least in string R, making the correct answer B: greatest in P and least in R.