A particle P of mass 0.4 kg is moving on rough horizontal ground when it hits a fixed vertical plane wall - Edexcel - A-Level Maths Mechanics - Question 3 - 2016 - Paper 1

Using Newton's second law (F = ma), we find the deceleration a:

$0.49 = 0.4a \Rightarrow a = \frac{0.49}{0.4} \approx 1.225 \, \text{m/s}^2$.

We know the particle starts with an initial speed (u) of 4 m/s and comes to rest (v = 0) over a distance (s) of 5 m. Applying the equation:

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

Substituting the known values:

$0 = 4^2 + 2(-1.225)(5)$,

this confirms that the deceleration is valid.

Impulse can be calculated using the momentum change formula:

$I = m(v - u)$

Given that the momentum change from 4 m/s before hitting the wall to -3.5 m/s after bouncing back:

• Before hitting the wall: Initial momentum = ( 0.4 \times 4 = 1.6 , \text{kg m/s} )
• After bouncing back (moving in the opposite direction): Final momentum = ( 0.4 \times (-3.5) = -1.4 , \text{kg m/s} )

Thus,

$I = 0.4 \times (-3.5 - 4) = 3 \, \text{Ns}$.