Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string. Initially B is held at rest on a rough fixed plane inclined at angle θ to the horizontal, where $tan \theta = \frac{5}{12}$. The part of the string from B to P is parallel to a line of greatest slope of the plane. The string passes over a small smooth pulley, P, fixed at the top of the plane. The particle A hangs freely below P, as shown in Figure 4. The coefficient of friction between B and the plane is $\frac{2}{3}$. The particles are released from rest with the string taut and B moves up the plane. (a) Find the magnitude of the acceleration of B immediately after release. (b) Find the speed of B when it has moved 1 m up the plane. (c) When B has moved 1 m up the plane the string breaks. Given that in the subsequent motion B does not reach P, find the time between the instants when the string breaks and when B comes to instantaneous rest.

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Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Question 7

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Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string. Initially B is held at rest on a rough fixed plan... show full transcript

Worked Solution & Example Answer:Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Step 1

Find the magnitude of the acceleration of B immediately after release.

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Answer

To find the acceleration of B, we start by analyzing the forces acting on both particles A and B.

Forces on A:

For particle A (mass = 7 kg), the forces acting are:

Weight: $F_A = 7g$ (downwards)
Tension in the string: $T$ (upwards)

From Newton's second law, we can write: $7g - T = 7a ag{1}$

Forces on B:

For particle B (mass = 3 kg), along the plane:

Tension acting up the plane: $T$
Weight component down the plane: $3g \sin \theta$
Frictional force opposing the motion: $F_{friction} = \mu R = \frac{2}{3} \times R$

Where $R = 3g \cos \theta$ is the normal force acting on B.

Thus, along the plane: $T - 3g \sin \theta - \frac{2}{3} R = 3a ag{2}$

Substitute R:

Substituting for R:

ightarrow F_{friction} = \frac{2}{3}(3g \cos \theta) = 2g \cos \theta$$ So the equation for B becomes: $$T - 3g \sin \theta - 2g \cos \theta = 3a ag{3}$$ ### System of Equations: Now we can solve Equations (1) and (3). First, substituting $\sin \theta$ and $\cos \theta$ from $tan \theta = \frac{5}{12}$: - $\sin \theta = \frac{5}{13}$ - $\cos \theta = \frac{12}{13}$ Now substitute these into (3): 1. Substitute into Equation (1) to express T: \[ T = 7g - 7a \tag{4} \] 2. Substitute $\sin \theta$ and $\cos \theta$ into Equation (3) to get: \[ (7g - 7a) - 3g \cdot \frac{5}{13} - 2g \cdot \frac{12}{13} = 3a \tag{5} \] Rearranging gives: \[ 7g - 7a - \frac{15g}{13} - \frac{24g}{13} = 3a \rightarrow \frac{91g - 39g}{13} = 10a \] Simplifying leads to: \[ 52g = 10a\qquad a = \frac{26g}{5} \tag{6} \] Substituting $g \approx 9.8 ext{ m/s}^2$ results in: \[ a \approx 5.088 m/s^{2} ext{ or } 5.09 m/s^{2}.\]

Step 2

Find the speed of B when it has moved 1 m up the plane.

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Answer

Using the kinematic equation to determine the speed of B after moving 1 m up the plane, we apply:

$v^2 = u^2 + 2as,$ where:

$u = 0$ (initial speed),
$s = 1$ m (distance moved),
$a = 5.088 m/s^2$ (acceleration found previously).

Thus, $v^2 = 0 + 2 \cdot 5.088 \cdot 1 = 10.176,$ $v = \sqrt{10.176} \approx 3.19 m/s.$

Step 3

Find the time between the instants when the string breaks and when B comes to instantaneous rest.

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Answer

After the string breaks, the forces acting on particle B are:

$F = - \left( 3g + F_{friction} \right) = -\left( 3g + \frac{2}{3} \cdot 3g \cos \theta \right)$

This can be simplified to: $F = -3g \left(1 +\frac{2}{3}\cdot\frac{12}{13}\right) \approx -3g \cdot (1 + 0.1846) \approx -3g \cdot 1.1846 = -3.5538g,$ which is the deceleration B experiences.

Using the kinematic equation: $v = u + at,$ where $u = 3.19 m/s$ , $v = 0$ (final speed at rest), $a = -3g \cdot 0.1846$ .

Thus, $0 = 3.19 - 3g (0.1846)t,$ which leads to solving for $t$ : $$t \approx \frac{3.19}{3g\cdot 0.1846} \approx 0.229 s.$

Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Worked Solution & Example Answer:Two particles A and B, of mass 7 kg and 3 kg respectively, are attached to the ends of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2011 - Paper 1

Find the magnitude of the acceleration of B immediately after release.

Forces on A:

Forces on B:

Substitute R:

Find the speed of B when it has moved 1 m up the plane.

Find the time between the instants when the string breaks and when B comes to instantaneous rest.

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