Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod. The system of the particles and the rod is at rest on a horizontal plane. At time t = 0, a constant force F of magnitude 4 N is applied to Q in the direction PQ, as shown in Figure 3. The system moves under the action of this force until t = 6 s. During the motion, the resistance to the motion of P has constant magnitude 1 N and the resistance to the motion of Q has constant magnitude 2 N. Find a) the acceleration of the particles as the system moves under the action of F, b) the speed of the particles at t = 6 s, c) the tension in the rod as the system moves under the action of F. At t = 6 s, F is removed and the system decelerates to rest. The resistances to motion are unchanged. d) Find the distance moved by P as the system decelerates, e) the thrust in the rod as the system decelerates.

A-Level Edexcel Maths Mechanics Forces: Two particles P and Q, of mass 0

Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

Question 7

Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod. The system of the particles and the rod is at rest on a horizont... show full transcript

Worked Solution & Example Answer:Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

Step 1

a) the acceleration of the particles as the system moves under the action of F

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Answer

Using Newton's Second Law (N2L), the net force acting on the system can be determined by accounting for the applied force and the resistance forces:

For the entire system:

$F_{net} = F - (Resistance_P + Resistance_Q) = 4 ext{ N} - (1 ext{ N} + 2 ext{ N}) = 1 ext{ N}$

Calculating the total mass:

$m_{total} = m_P + m_Q = 0.3 ext{ kg} + 0.5 ext{ kg} = 0.8 ext{ kg}$

Using Newton's second law:

$a = \frac{F_{net}}{m_{total}} = \frac{1 ext{ N}}{0.8 ext{ kg}} = 1.25 ext{ m/s}^2$

Step 2

b) the speed of the particles at t = 6 s

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Answer

To find the speed of the particles at t = 6 s, we use the kinematic equation:

$v = u + at$

where the initial speed $u = 0$ , thus:

$v = 0 + (1.25 ext{ m/s}^2)(6 ext{ s}) = 7.5 ext{ m/s}$

Step 3

c) the tension in the rod as the system moves under the action of F

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Answer

For particle P:

Using Newton's Second Law for P:

T - 1 ext{ N} = 0.3 ext{ kg} \cdot 1.25 ext{ m/s}^2\ T = 0.375 ext{ N} + 1 ext{ N} = 1.375 ext{ N}$$

Step 4

d) Find the distance moved by P as the system decelerates

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When F is removed, the system starts to decelerate due to the resistance. Using the equation:

$s = ut + \frac{1}{2} a t^2$

Re-arranging for deceleration:

$F_{net} = - (1 + 2) = -3 ext{ N}$

Total mass remains:

\implies a = \frac{-3 ext{ N}}{0.8 ext{ kg}} = -3.75 ext{ m/s}^2$$ Now substituting: $$s = 7.5 ext{ m/s} \cdot 6 ext{ s} + \frac{1}{2} (-3.75) \cdot (6 ext{ s})^2\ = 45 - 67.5\ = -22.5 ext{ m}$$

Step 5

e) the thrust in the rod as the system decelerates

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Answer

For Particle P:

Using N2L:

T = 0.3 \cdot (-3.75) - 1\ T = -1.125 ext{ N}$$ Since thrust can't be negative, we take the absolute value: Thrust = 0.125 N

Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

Worked Solution & Example Answer:Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

a) the acceleration of the particles as the system moves under the action of F

b) the speed of the particles at t = 6 s

c) the tension in the rod as the system moves under the action of F

d) Find the distance moved by P as the system decelerates

e) the thrust in the rod as the system decelerates

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