QUESTION 5 (Start on a NEW page.) A 5 g steel marble is fired horizontally to the east from a 2 kg model canon that is stationary on a flat frictionless horizontal surface - NSC Physical Sciences - Question 5 - 2017 - Paper 1

5.2.1 Canon collides the barrier
To find the speed, we can use the formula:

ext{Speed} = rac{ ext{Distance}}{ ext{Time}}
Substituting the values:

ext{Speed} = rac{0.32 ext{ m}}{0.33 ext{ s}} \\ = 0.97 ext{ m·s}^{-1}

5.2.2 Steel marble is fired from the canon
Applying the conservation of momentum:

$m_{ ext{canon}} v_{ ext{canon}} + m_{ ext{marble}} v_{ ext{marble}} = 0$
Given,
$m_{ ext{marble}} = 0.005 ext{ kg}, m_{ ext{canon}} = 2 ext{ kg}, v_{ ext{canon}} = -0.97 ext{ m·s}^{-1}$

Solving for $v_{ ext{marble}}$:

$0.005 ext{ kg} imes v_{ ext{marble}} = -2 ext{ kg} imes (-0.97 ext{ m·s}^{-1})$

oh v_{ ext{marble}} = rac{2 imes 0.97}{0.005} = 388 ext{ m·s}^{-1}.

Impulse can be calculated as:

$I = ext{F}_{ ext{net}} imes ext{t}$
Where:

• $ext{F}_{ ext{net}}$ is the change in momentum.
• $ext{t}$ is the time the force acts.

We calculate the change in momentum for the cannon:

$I = m v_{ ext{final}} - m v_{ ext{initial}}$
Where:
v_{ ext{final}} = -0.4 m·s^-1
v_{ ext{initial}} = 0.97 m·s^-1

Calculating:

$I = 2 ext{ kg} imes (-0.4) - 2 ext{ kg} imes (0.97) = -0.8 - 1.94 = -2.74 ext{ kg·m·s}^{-1}$
The magnitude of the impulse is:

$2.74 ext{ kg·m·s}^{-1}$.