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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 $\frac{12}{5}mg$.

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Answer

To find the tension $T$ in the string immediately after the system is released, we can apply Newton's second law to both masses. For mass B (3m), we have:

$3mg - T = 3ma$

For mass A (2m), considering its upward motion:

$T - 2mg = 2ma$

Since both masses accelerate at the same rate $a$ , we can set $a = \frac{g}{5}$ ( using the derived value for $a$ from the motion of B). Now substituting this into both equations:

From mass B: $3mg - T = 3m * \frac{g}{5}$
From mass A: $T - 2mg = 2m * \frac{g}{5}$

Rearranging these equations:

$T = 3mg - \frac{3mg}{5} = \frac{15mg - 3mg}{5} = \frac{12mg}{5}$

Thus, the tension in the string is: $T = \frac{12}{5}mg$

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.

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Answer

After B strikes the plane and comes to rest, A continues to move upward. To find the distance $s$ travelled by A, we use:

Using the formula for the motion:

t = \sqrt{\frac{2s}{g}}

Where $s = 1.5 m$ . Thus the time taken by A to travel this distance is:

t = \sqrt{\frac{2 * 1.5}{g}} = 0.6 s

In this time, A travels additional distance using the formula:

d = ut + \frac{1}{2} a t^2 where the speed u can be calculated as:

d = 2 * \frac{g}{5} * (0.6)^2 = 0.3 m

So the total distance A travels becomes: $\text{Total distance} = 0.6 + 0.3 = 0.6 m$

Step 3

Find the magnitude of the impulse on B due to the impact with the plane.

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Answer

Impulse is given by the change of momentum.

Using:

$\text{Impulse} = m(v - u)$ where

m = 3m (mass of B, with m = 0.5 kg)
v = 0 (final velocity post impact)
u = v (initial velocity at the moment just before impact)

The initial velocity can be calculated using:

$v = \sqrt{2as} = \sqrt{2 * \frac{g}{5} * 1.5}$

So the magnitude of the impulse is: $\text{Impulse} = 3 imes 0.5 imes (v - 0) = 3m * 0.6g = 3.6 \, \text{Ns}$

Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 1

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

Show that the tension in the string immediately after the particles are released is $\frac{12}{5}mg$.

Find the distance travelled by A between the instant when B strikes the plane and the instant when the string next becomes taut.

Find the magnitude of the impulse on B due to the impact with the plane.

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