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Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2010 - Paper 1
Question 8
Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small... show full transcript
Worked Solution & Example Answer:Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2010 - Paper 1
Step 1
Find the tension in the string immediately after the particles are released.
Answer
To find the tension in the string, we can apply Newton's second law to both particles. For particle A (mass = 0.4 kg), the forces acting on it are the tension T and its weight.
For particle A:
T - 0.4g &= 0.4a\ \ \ T &= 0.4g + 0.4a \ \ egin{align*} ext{For particle B (mass = 0.3 kg):}\ T - 0.3g &= -0.3a\ \ T &= 0.3g - 0.3a\ ext{Since both particles experience the same tension, we can rearrange the equations for } a: ext{From A: } T = 0.4g + 0.4a\\ ext{From B: } T = 0.3g - 0.3a\\ \ \ 0.4g + 0.4a = 0.3g - 0.3a \ \ \ 0.7a = 0.3g - 0.4g\ \ 0.7a = -0.1g \ \ \ a = - rac{0.1g}{0.7} = - rac{10}{7} m/s^2 \ \ \ a = -1.4 m/s^2\ \ \ ext{Substituting this value of } a ext{ in the tension for A:} \ T = 0.4(9.8) + 0.4(-1.4)\\ T = 3.36 ext{ N} ext{Hence, the tension in the string immediately after the particles are released is } 3.36 ext{ N.} \Step 2
Step 3
Find the further time that elapses until B hits the floor.
Answer
When the particles have been moving for 0.5 s, we can use the kinematic equation:
s = ut + rac{1}{2}at^2
Here, the initial velocity, (since they are released from rest), and we have
a = 1.4 m/s² (as found earlier).
We can calculate the distance traveled by B in 0.5 s:
= 0.5 imes 1.4 imes 0.25 = 0.175 ext{ m}$$ The height of B from the floor initially is 1 m, therefore: Height remaining = 1 m - 0.175 m = 0.825 m Using the same kinematic equation to find the additional time, we need to find: $$s = ut + rac{1}{2}at^2$$ Here, we know: $$0.825 = 0 + rac{1}{2} imes 9.8 imes t^2\\ 0.825 = 4.9 t^2\\ t^2 = rac{0.825}{4.9} \\ t^2 = 0.168367 \\ t = ext{approximately } 0.41 ext{ seconds}$$ Thus, the further time that elapses until B hits the floor after the initial 0.5 s is approximately 0.41 s.