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 At time t seconds after projection, P is 19 m above A. (b) Find the possible values of t. 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. (c) Find the vertical distance that P sinks into the ground before coming to rest.

Photo AI

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

96%

114 rated

Answer

To find the initial speed $u$ , we use the formula for vertical motion:

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

where:

$v = 28 ext{ m/s}$ (final speed when reaching the ground),
$a = -9.8 ext{ m/s}^2$ (acceleration due to gravity),
$s = 17.5 ext{ m}$ (height from which P is projected).

Substituting these values into the equation:

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

This simplifies to:

$784 = u^2 - 343$

Rearranging gives:

$u^2 = 784 + 343 = 1127$

Taking the square root:

$u = ext{approximately } 21 ext{ m/s}.$

Step 2

Find the possible values of t

99%

104 rated

Answer

Using the equation of motion:

$s = ut + rac{1}{2}at^2$

Here, we know:

$s = 19 ext{ m}$ (height at time t),
$u = 21 ext{ m/s}$ ,
$a = -9.8 ext{ m/s}^2$ .

Thus:

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

This simplifies to:

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

Using the quadratic formula:

$t = rac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} = rac{21 \\pm \\sqrt{(-21)^2 - 4(4.9)(-19)}}{2(4.9)}$

This provides two potential solutions for $t$ : 2.99 s and 1.30 s.

Step 3

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

96%

101 rated

Answer

Using the work-energy principle, we set the work done by the resistive force equal to the change in kinetic energy:

$Work = F_d$ Substituting the values:

$F = 5000 ext{ N},
$d =$ ext{Distance sunk}
$m = 4 ext{ kg}$ .

The initial kinetic energy when P hits the ground is:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(4)(28^2) = 1576 ext{ J}$

Setting work done equal to kinetic energy:

$F_d = 1576$

Substituting gives:

ightarrow d = \frac{1576}{5000} = 0.3152 ext{ m},$$ or approximately 0.32 m.

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

Join the A-Level students using SimpleStudy...