Dynamics: Question 6

Worked solution

Part (a): Acceleration of the system

Treat both trolleys as a single system of total mass $m = 2.0+3.0=5.0\text{ kg}$. The only external horizontal force on this system is the applied $15\text{ N}$ force. The string tension is an internal force between the two trolleys, so it does not appear when the trolleys are treated together.

$a = \frac{F}{m} = \frac{15}{5.0} = 3.0\text{ m s}^{-2}$

Part (b): Tension in the string

Consider trolley $Q$ on its own. Since the track is frictionless, the tension $T$ in the string is the only horizontal force acting on trolley $Q$, pulling it forward with the same acceleration as the whole system. Applying Newton’s second law to $Q$ alone:

$T = m_Q a = 3.0\times 3.0 = 9.0\text{ N}$

Check using trolley $P$ alone: two horizontal forces act on $P$, the applied $15\text{ N}$ forward, and the tension $T$ acting backward (the string pulls back on $P$ as it pulls $Q$ forward):

$15 - T = m_P a \implies 15-T = 2.0\times3.0 = 6.0 \implies T = 9.0\text{ N}$

This agrees with the value found from trolley $Q$, confirming the tension.

Part (c): Motion after the string breaks

Once the string breaks, trolley $Q$ has no applied force acting on it, and the track is frictionless, so no horizontal force acts on trolley $Q$ at all: the resultant force is $0\text{ N}$.

By Newton’s first law, a resultant force of zero means trolley $Q$ continues to move at whatever constant velocity it had at the instant the string broke, in a straight line, it neither speeds up nor slows down.

Final answers

• (a) $a = \boxed{3.0}\text{ m s}^{-2}$
• (b) $T = \boxed{9.0}\text{ N}$
• (c) Resultant force on $Q$ $= \boxed{0}\text{ N}$; it continues at constant velocity in a straight line