A ball X, of mass 10 kg, is moving eastwards with a velocity of 2 m·s⁻¹ - NSC Physical Sciences - Question 4 - 2021 - Paper 1

To find the velocity vY of ball Y before the collision, we can use the principle of conservation of momentum:

$m_1 v_{X_i} + m_2 v_{Y_i} = m_1 v_{X_f} + m_2 v_{Y_f}$

Here:
$m_1 = 10 ext{ kg}$ (mass of ball X)
$m_2 = 2 ext{ kg}$ (mass of ball Y)
$v_{X_i} = 2 ext{ m·s⁻¹}$ (initial velocity of ball X)
$v_{Y_i} = v_Y$ (initial velocity of ball Y)
$v_{X_f} = 0$ (final velocity of ball X after collision)
$v_{Y_f}$ can be determined from the kinetic energy of ball Y:
rac{1}{2} m_2 v_{Y_f}^2 = 36 ext{ J}
Solving for $v_{Y_f}$ gives:
v_{Y_f} = rac{36 imes 2}{2} = 6 ext{ m·s⁻¹} Substituting into the momentum equation:
$10 imes 2 + 2 v_Y = 10 imes 0 + 2 imes 6$
This simplifies to:
$20 + 2 v_Y = 12$
Rearranging gives:
$2 v_Y = 12 - 20 = -8$
v_Y = -4 ext{ m·s⁻¹}
Thus, the velocity of ball Y before the collision is -4 m·s⁻¹ (indicating it was moving in the opposite direction).

To calculate the force exerted by ball X on ball Y, we can use the impulse-momentum theorem:

$F_{net} imes riangle t = m(v_{Y_f} - v_{Y_i})$

Here:

• $F_{net}$ is the net force exerted
• $riangle t = 0.1 ext{s}$ (duration of the collision)
• $v_{Y_f} = 6 ext{ m·s⁻¹}$ (final velocity of ball Y)
• $v_{Y_i} = -4 ext{ m·s⁻¹}$ (initial velocity of ball Y)

Substituting the values into the equation gives:

F_{net} imes 0.1 &= 2 imes (6 - (-4)) \ F_{net} imes 0.1 &= 2 imes (6 + 4) \ F_{net} imes 0.1 &= 2 imes 10 \ F_{net} &= rac{20}{0.1} \ F_{net} &= 200 ext{ N} dots Thus, the magnitude of the force that ball X exerted on ball Y during the collision is 200 N.