Ball P of mass 0,16 kg, moving east at a speed of 10 m/s, collides head-on with another ball Q of mass 0,2 kg, moving west at a speed of 15 m/s - NSC Physical Sciences - Question 4 - 2020 - Paper 1

To find the velocity of ball Q after the collision, we apply the principle of conservation of momentum:

$ext{Total initial momentum} = ext{Total final momentum}$

Before the collision:

• Momentum of ball P = $m_P v_{P_i} = (0.16 kg)(10 m/s) = 1.6 ext{ kg m/s east}$
• Momentum of ball Q = $m_Q v_{Q_i} = (0.2 kg)(-15 m/s) = -3 ext{ kg m/s west}$ (negative sign indicates west direction)

Total initial momentum: $1.6 - 3 = -1.4 ext{ kg m/s}$ (west)

After the collision, the final momentum of ball P (moving west) is:

• Momentum of ball P = $m_P v_{P_f} = (0.16 kg)(-5 m/s) = -0.8 ext{ kg m/s}$

Let $v_{Q_f}$ be the final velocity of ball Q after the collision:

• Momentum of ball Q = $m_Q v_{Q_f} = (0.2 kg)(v_{Q_f})$

The total final momentum can then be expressed as: $-0.8 + (0.2)(v_{Q_f}) = -1.4$

Solving for $v_{Q_f}$ gives: $(0.2)(v_{Q_f}) = -1.4 + 0.8$ $(0.2)(v_{Q_f}) = -0.6$ v_{Q_f} = rac{-0.6}{0.2} = -3 ext{ m/s}

Thus, the velocity of ball Q after the collision is $3 m/s$ west.

Impulse is defined as the change in momentum of an object. The formula for impulse ($J$) is:

$J = riangle p = p_f - p_i$

For ball P, the initial momentum $p_i$ and final momentum $p_f$ are:

• Initial momentum: $p_i = 0.16 kg imes 10 m/s = 1.6 ext{ kg m/s} ext{ east}$
• Final momentum: $p_f = 0.16 kg imes (-5 m/s) = -0.8 ext{ kg m/s} ext{ (west)}$

Now, calculating the impulse: $J = (-0.8) - (1.6) = -2.4 ext{ kg m/s}$

The magnitude of the impulse on ball P is therefore $2.4 ext{ N s}$ and it acts in the westward direction.