Momentum — practice questions

First, let's set up the problem. Let the direction of P's initial motion be the positive direction.

Diagram and Initial Values: Before collision: Particle P: $m_P = 2$ kg, $u_P = +5$ m s⁻¹ Particle Q: $m_Q = 3$ kg, $u_Q = -4$ m s⁻¹

After collision, let their velocities be $v_P$ and $v_Q$.

1. Conservation of Momentum (CoM): Total momentum before = Total momentum after $m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q$ $(2)(5) + (3)(-4) = 2v_P + 3v_Q$ $10 - 12 = 2v_P + 3v_Q$ $-2 = 2v_P + 3v_Q$ (Equation 1)

2. Newton's Law of Restitution (NLR): Speed of Separation = $e \times$ Speed of Approach $v_Q - v_P = e(u_P - u_Q)$ $v_Q - v_P = 0.6(5 - (-4))$ $v_Q - v_P = 0.6(9)$ $v_Q - v_P = 5.4$ (Equation 2)

3. Solve Simultaneous Equations: From Equation 2, $v_Q = v_P + 5.4$. Substitute this into Equation 1: $-2 = 2v_P + 3(v_P + 5.4)$ $-2 = 2v_P + 3v_P + 16.2$ $-18.2 = 5v_P$ $v_P = -3.64$ m s⁻¹

Now find $v_Q$ using the expression from Equation 2: $v_Q = -3.64 + 5.4$ $v_Q = 1.76$ m s⁻¹

Conclusion: After the collision:

• Particle P has a speed of 3.64 m s⁻¹ and its direction of motion is reversed.
• Particle Q has a speed of 1.76 m s⁻¹ and its direction of motion is also reversed.

We need to find the ball's speed just before and just after it hits the floor. We can use kinematics, with $g = 9.8$ m s⁻².

Part (a): Find the coefficient of restitution, e

1. Speed before impact ($u$): The ball is dropped from rest ($u_{initial}=0$) from a height $h_1 = 5$ m. Using $v^2 = u^2 + 2as$: $u^2 = 0^2 + 2(9.8)(5)$ $u^2 = 98$ $u = \sqrt{98} = 7\sqrt{2}$ m s⁻¹ (This is the speed of approach)

2. Speed after impact ($v$): The ball rebounds to a height $h_2 = 3.2$ m. At its peak, its final velocity is 0. Using $v^2 = u^2 + 2as$ (taking upwards as positive, so $a = -9.8$): $0^2 = v^2 + 2(-9.8)(3.2)$ $0 = v^2 - 62.72$ $v^2 = 62.72$ $v = \sqrt{62.72} \approx 7.92$ m s⁻¹ (This is the speed of separation) Note: The exact value is $v = \sqrt{62.72} = \sqrt{3136/50} = 56/\sqrt{50} = 56/(5\sqrt{2}) = (28\sqrt{2})/5$.

3. Apply Newton's Law of Restitution: The floor is stationary, so its speed is 0 before and after. $e = \frac{\text{Speed of Separation}}{\text{Speed of Approach}} = \frac{v}{u}$ $e = \frac{\sqrt{62.72}}{\sqrt{98}} = \sqrt{\frac{62.72}{98}} = \sqrt{0.64}$ $e = 0.8$

Part (b): Find the loss in kinetic energy

KE before impact = $\frac{1}{2} m u^2 = \frac{1}{2} (0.5) (98) = 24.5$ J KE after impact = $\frac{1}{2} m v^2 = \frac{1}{2} (0.5) (62.72) = 15.68$ J

Loss in KE = KE before - KE after Loss in KE = $24.5 - 15.68 = 8.82$ J