Momentum — practice questions

First, define a positive direction. Let's take the initial direction of particle A as positive. \ \ 1. State initial velocities with correct signs: \ Velocity of A, $u_A = +4$ m s⁻¹ \ Velocity of B, $u_B = -2$ m s⁻¹ \ \ 2. Apply the Principle of Conservation of Momentum: \ Total momentum before = Total momentum after \ $(m_A u_A) + (m_B u_B) = (m_A + m_B) v_C$ \ \ 3. Substitute values and solve for $v_C$: \ $(0.5 \times 4) + (0.3 \times -2) = (0.5 + 0.3) v_C$ \ $2 - 0.6 = 0.8 v_C$ \ $1.4 = 0.8 v_C$ \ $v_C = \frac{1.4}{0.8} = 1.75$ m s⁻¹ \ \ Since the result is positive, the combined particle C moves at 1.75 m s⁻¹ in the original direction of particle A.

The shell is initially at rest, so the total initial momentum is 0. \ \ 1. Apply Conservation of Momentum: \ Total initial momentum = Total final momentum \ $0 = m_P v_P + m_Q v_Q$ \ \ 2. Substitute values and solve for $v_Q$: \ Let the direction of P be positive, so $v_P = +6$ m s⁻¹. \ $0 = (1 \times 6) + (2 \times v_Q)$ \ $0 = 6 + 2v_Q$ \ $2v_Q = -6$ \ $v_Q = -3$ m s⁻¹ \ The speed of Q is the magnitude of its velocity, which is 3 m s⁻¹. The negative sign indicates it moves in the opposite direction to P. \ \ 3. Calculate Kinetic Energy Generated: \ The KE generated is the total final KE, since the initial KE was zero. \ $KE_{total} = KE_P + KE_Q$ \ $KE_{total} = \frac{1}{2}m_P v_P^2 + \frac{1}{2}m_Q v_Q^2$ \ $KE_{total} = \frac{1}{2}(1)(6)^2 + \frac{1}{2}(2)(-3)^2$ \ $KE_{total} = \frac{1}{2}(36) + 1(9)$ \ $KE_{total} = 18 + 9 = 27$ J \ The explosion generated 27 Joules of kinetic energy.