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
To find the acceleration of the system, we apply Newton's second law (N2L). The net force acting on the system can be calculated as follows:
Net Force = Applied Force - Resistance on P - Resistance on Q
Net Force = 4 N - 1 N - 2 N = 1 N
Total mass of the system = mass of P + mass of Q = 0.3 kg + 0.5 kg = 0.8 kg
Using N2L:
a = \frac{F_{net}}{m_{total}} = \frac{1}{0.8} = 1.25 , \text{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 after 6 seconds, we can use the formula:
v = u + at
v = 0 + 1.25 , \text{m/s}^2 \times 6 , \text{s} = 7.5 , \text{m/s}
Step 3
(c) the tension in the rod as the system moves under the action of F.
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Answer
Using N2L for particle P:
T - 1 N = m_P \cdot a
T - 1 = 0.3 imes 1.25
T = 0.375 , \text{N}
Step 4
(d) the distance moved by P as the system decelerates,
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Answer
During deceleration, the system experiences the following motion:
Using the kinematic equation:
s = ut + \frac{1}{2} a t^2
s = 0 + \frac{1}{2} \times (-1.25) \times (6^2) = - 22.5 , ext{m}
Since we require the distance, we take the absolute value: 22.5 m.
Step 5
(e) the thrust in the rod as the system decelerates.
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Answer
For particle P during deceleration, we have:
T - 1 N = 0.3 kg \cdot a \quad (a = -1.25 , \text{m/s}^2)
T - 1 = 0.3 \times (-1.25)
T = 1 - 0.375 = 0.625 , \text{N}
