Linear momentum and its conservation — practice questions

Identify initial conditions:

• Trolley 1: $m_1 = 2.0 \text{ kg}$, $u_1 = +3.0 \text{ m/s}$ (take initial direction as positive)
• Trolley 2: $m_2 = 4.0 \text{ kg}$, $u_2 = 0 \text{ m/s}$ (stationary)

State the Principle of Conservation of Momentum:

• Total momentum before = Total momentum after
• $m_1u_1 + m_2u_2 = (m_1+m_2)V_f$ (since they stick together, they have a common final velocity $V_f$)

Substitute values into the equation:

• $(2.0 \text{ kg})(3.0 \text{ m/s}) + (4.0 \text{ kg})(0 \text{ m/s}) = (2.0 \text{ kg} + 4.0 \text{ kg})V_f$
• $6.0 \text{ kg m/s} + 0 = (6.0 \text{ kg})V_f$

Solve for $V_f$:

• $6.0 = 6.0 V_f$
• $V_f = \frac{6.0}{6.0} = +1.0 \text{ m/s}$

The common final velocity of the trolleys is 1.0 m/s in the initial direction of the 2.0 kg trolley.

Define variables & initial conditions:

• Ball 1: $m_1 = 0.50 \text{ kg}$, $u_1 = +4.0 \text{ m/s}$
• Ball 2: $m_2 = 0.30 \text{ kg}$, $u_2 = 0 \text{ m/s}$
• Let the final velocities be $v_1$ and $v_2$.

Apply Conservation of Momentum:

• $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
• $(0.50)(4.0) + (0.30)(0) = 0.50v_1 + 0.30v_2$
• $2.0 = 0.50v_1 + 0.30v_2$ (Equation 1)

Apply property of elastic collisions (Relative Speeds):

• For elastic collisions, the relative speed of approach equals the relative speed of separation.
• $u_1 - u_2 = -(v_1 - v_2) = v_2 - v_1$
• $4.0 - 0 = v_2 - v_1 \implies v_2 = 4.0 + v_1$ (Equation 2)

Solve the simultaneous equations:

• Substitute $v_2$ from Equation 2 into Equation 1:
• $2.0 = 0.50v_1 + 0.30(4.0 + v_1)$
• $2.0 = 0.50v_1 + 1.2 + 0.30v_1$
• $2.0 - 1.2 = 0.80v_1$
• $0.8 = 0.80v_1$
• $v_1 = +1.0 \text{ m/s}$

Find $v_2$:

• Substitute $v_1 = 1.0$ m/s into Equation 2:
• $v_2 = 4.0 + 1.0 = +5.0 \text{ m/s}$

Final Answer: The 0.50 kg ball moves at 1.0 m/s and the 0.30 kg ball moves at 5.0 m/s, both in the original direction of the first ball.