A particle P is projected vertically upwards from a point A with speed u m s⁻¹. The point A is 17.5 m above horizontal ground. The particle P moves freely under gravity until it reaches the ground with speed 28 m s⁻¹. (a) Show that u = 21 (b) At time t seconds after projection, P is 19 m above A. (c) The ground is soft and, after P reaches the ground, P sinks vertically downwards into the ground before coming to rest. The mass of P is 4 kg and the ground is assumed to exert a constant resistive force of magnitude 5000 N on P. (d) Find the vertical distance that P sinks into the ground before coming to rest.

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A particle P is projected vertically upwards from a point A with speed u m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

Question 5

A particle P is projected vertically upwards from a point A with speed u m s⁻¹. The point A is 17.5 m above horizontal ground. The particle P moves freely under grav... show full transcript

Worked Solution & Example Answer:A particle P is projected vertically upwards from a point A with speed u m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

Step 1

Show that u = 21

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Answer

To find u, we can use the kinematic equation:

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

where:

v = 28 m/s (final speed upon reaching the ground)
s = -17.5 m (the displacement to the ground is negative since it is downward)
a = -9.8 m/s² (acceleration due to gravity, negative because it acts downward)

Plugging these values into the equation:

$28^2 = u^2 + 2(-9.8)(-17.5)$

Calculating:

$784 = u^2 + 343$

Thus:

$u^2 = 784 - 343$ $u^2 = 441$ $u = 21$

Hence, we have shown that u = 21 m/s.

Step 2

Find the possible values of t.

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Answer

Using the equation for the height of the particle:

$s = ut + \frac{1}{2}at^2$

Substituting the values:

$19 = 21t - \frac{1}{2}(9.8)t^2$

This simplifies to:

$0 = -4.9t^2 + 21t - 19$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where a = -4.9, b = 21, and c = -19. Calculation gives:

$t = \frac{-21 \pm \sqrt{21^2 - 4(-4.9)(-19)}}{2(-4.9)}$

Solving the discriminant:

$\sqrt{441 - 372.4} = \sqrt{68.6} \approx 8.28$

Thus, the possible values for t are:

$t = \frac{21 \pm 8.28}{9.8}$

Calculating:

When using the positive root: $t \approx 2.99$
When using the negative root: $t \approx 1.30$

Step 3

Find the vertical distance that P sinks into the ground before coming to rest.

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Answer

When P reaches the ground, its speed is 28 m/s. To find how far it sinks before coming to rest, we can use the work-energy principle:

The kinetic energy (KE) when it hits the ground:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(4)(28^2) = 1568 \, J$

The work done against the resistive force (5,000 N) while sinking:

$W = F \cdot d$

Setting the work done equal to the kinetic energy:

$1568 = 5000d$

Solving for d gives:

$d = \frac{1568}{5000} \approx 0.316 \, m$

Thus, P sinks approximately 0.316 m into the ground before coming to rest.

A particle P is projected vertically upwards from a point A with speed u m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

Worked Solution & Example Answer:A particle P is projected vertically upwards from a point A with speed u m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

Show that u = 21

Find the possible values of t.

Find the vertical distance that P sinks into the ground before coming to rest.

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