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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
Let's consider the forces acting on the particles A and B immediately after they are released. For particle A (mass = 2m) going upward:
For particle B (mass = 3m) going downward:
Since A moves upward and B moves downward, we set their accelerations equal as ( a_A = a_B = a ).
Now, substituting ( a ) into both equations:
- Rearranging gives us: ( T = 2mg + 2ma ) from the first equation.
- From the second equation: ( T = 3mg - 3ma )
We can set the two equations for T equal to each other:
Combining like terms results in:
Substituting back into one of the tension equations:
Thus, ( T = \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 and comes to rest, the acceleration experienced by A will be equal to gravity minus the effect of the string.
Therefore, the motion of A becomes one of free fall:
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Using the kinematic equation: [ v^2 = u^2 + 2as ] Since A starts from rest, ( u = 0 ), and the distance fallen before becoming taut is ( 1.5 , m ): [ v^2 = 0 + 2 \times \left(\frac{g}{5}\right) \times 1.5 ]
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Thus, we find: [ v^2 = 0.6g ]
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To determine the distance traveled by A after impacting, we need to account for this velocity and the time it takes: When calculated, this yields: [ \text{total distance} = 2 \times 0.3 = 0.6 , m ]
Step 3
Find the magnitude of the impulse on B due to the impact with the plane.
Answer
The impulse (I) experienced by B during the impact can be calculated using the impulse-momentum theorem:
[ I = m(v - u) ]
- Before impact, B has a downward velocity ( v = 0.6g ).
- After impact, B comes to rest, hence ( u = 0 ):
Substituting these values: [ I = 3m(0 - 0.6g) = -3m(0.6g) ] [ |I| = 3(0.5)(0.6g) = 3 \times 0.5 \times 0.6g = 3.6 \text{ Ns} ].
Thus, the magnitude of the impulse is 3.6 Ns.