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Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 1
Question 5
Two particles A and B have masses 2m and 3m respectively. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. T... show full transcript
Worked Solution & Example Answer:Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 1
Step 1
Show that the tension in the string immediately after the particles are released is \( \frac{12}{5} mg \)
Answer
To find the tension ( T ) in the string after the particles are released, we analyze the forces acting on both particles. For Particle A (mass = 2m) in free fall, the equation of motion can be expressed as:
[ 2mg - T = 2ma \tag{1} ]
For Particle B (mass = 3m) descending, the equation of motion is:
[ 3mg - T = 3ma \tag{2} ]
Adding equations (1) and (2), we get:
[ 2mg - T + 3mg - T = 2ma + 3ma \Rightarrow 5mg - 2T = 5ma ]
Since the system is released from rest, both particles will have the same acceleration (denote it as ( a )), leading to:
[ 5mg - 2T = 5 \cdot 2m ]
Substituting for ( a = \frac{g}{5} ) leads us to:
[ \Rightarrow 5mg - 2T = 2mg \Rightarrow 3mg = 2T \Rightarrow T = \frac{3mg}{2} \tag{3} ]
From equation (1) substituting ( a = \frac{g}{5} ) yields:
[ T = 2mg + 2m \left( \frac{g}{5} \right) = \frac{12}{5}mg \tag{4} ]
Thus confirming that the tension in the string immediately after the particles are released is ( \frac{12}{5} mg ).
Step 2
Find the distance travelled by A between the instant when B strikes the plane and the instant when the string next becomes taut.
Answer
When B strikes the plane, it stops instantly, and A continues to move upwards due to inertia. Using the kinematic equation:
[ v^2 = u^2 + 2as ]
Here, the final velocity ( v = 0 ), initial velocity is the same speed B had at that moment when it strikes:
[u = \sqrt{2 \cdot \frac{2g}{5} \cdot 1.5} = \sqrt{6g} = 0.6g]
Using this value, we can find the distance travelled by A:
[ s = \frac{u^2}{2a} = \frac{(0.6g)^2}{2 \cdot \frac{g}{5}} = \frac{0.36g^2}{\frac{2g}{5}} = 0.9 , \text{m} ]
Thus, the distance travelled by A is ( 0.6 \text{ m} ).
Step 3
Find the magnitude of the impulse on B due to the impact with the plane.
Answer
To calculate the impulse on B, we use the impulse-momentum theorem:
[Impulse = \Delta p = m(v - u)]
Before the impact, B is in motion at speed ( 0.6g ) whereas after the impact, its speed becomes zero:
[ p = 3m(0.6g - 0) = 3 \cdot 0.5 \cdot 0.6g = 0.9g = 3.6 , \text{N.s} \tag{5} ]
Thus, the magnitude of the impulse on B due to the impact with the plane is ( 3.6 , \text{Ns} ).