8. Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1

For particle B, the forces acting are its weight (mg) downwards and the tension (T) upwards. Applying Newton's second law, we obtain:

$mg - T = ma.$

From the equation for B, we can express T as:

$T = mg - ma.$

Substituting T in the equation for A gives us:

$mg - ma - \mu(2mg) = 2ma.$

Rearranging leads to:

$mg - \mu(2mg) = 5ma.$

This simplifies to:

$a = \frac{mg(1 - 2μ)}{5m} = \frac{g(1 - 2μ)}{5}.$

As we know, A's acceleration in the context of the problem is therefore:

$a = \frac{g}{3}(1 - 2μ).$

Using the principle of conservation of energy, the energy lost by B falling a height h is equal to the kinetic energy gained by A.

The potential energy change is:

$PE = mgh.$

The kinetic energy gained by A is:

$KE = \frac{1}{2} (2m) v^2.$

Setting potential energy equal to the kinetic energy gives:

$mgh = \frac{1}{2} (2m) v^2.$

This simplifies to:

$gh = v^2.$

Thus, we can express the speed of A as:

$v = \sqrt{2gh(1 - 2μ)}.$

To find the distance the particle A travels after B hits the ground until it comes to rest, we can use the equations of motion. Once B hits the floor, the acceleration is now influenced only by the friction. Using:

$F = ma,$

where the only force acting on A is the friction which can be written as:

$F_{friction} = \mu (2mg).$

Then,

$\mu (2mg) = 2ma'$

Thus,

$a' = -\mu g.$

To find the stopping distance s when A stops from its speed v, we can use:

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

where u is the speed at that instant, a = -\mu g, and v = 0.

So,

$0 = v^2 - 2(\mu g)s,$

which rearranges to:

$s = \frac{v^2}{2\mu g}.$

Thus, the distance A travels after B hits the floor can be expressed in terms of h as:

$s = \frac{2gh(1 - 2μ)}{2\mu g} = \frac{h(1 - 2μ)}{μ}.$

If the coefficient of friction μ is ( \frac{1}{2} ), we substitute this value into our equations to examine the behavior of the system. This specific coefficient means that the friction force acting on particle A is equal to half of the weight of particle A.

In this scenario, when B hits the floor, A will still experience a retarding force due to friction. This can alter the time taken for A to come to rest compared to situations with either lower or higher coefficients of friction.

As the friction is substantial, but not enough to completely halt motion immediately, A may slide a greater distance before it comes to rest at P, thus extending the total distance travelled.