Dynamics: Question 4

Worked solution

Part (a): Total momentum before the collision

Before the collision, puck $A$ moves entirely along the $x$-direction (east) and puck $B$ is at rest, so it contributes no momentum: $p_x = m_A u_A + m_B(0) = (0.30)(5.0) = 1.5\text{ kg m s}^{-1}$ $p_y = 0\text{ kg m s}^{-1}$

The total momentum before the collision is $1.5\text{ kg m s}^{-1}$ due east, with zero component in the north-south direction.

Part (b): Velocity of puck $B$ after the collision

Resolve puck $A$‘s final velocity into components. Puck $A$ moves at $3.0\text{ m s}^{-1}$ at $53.1°$ from east (north side), so: $v_{Ax} = 3.0\cos53.1° = 3.0 \times 0.600 = 1.8\text{ m s}^{-1}$ $v_{Ay} = 3.0\sin53.1° = 3.0 \times 0.800 = 2.4\text{ m s}^{-1}$

Find puck $A$‘s momentum components: $p_{Ax}' = m_A v_{Ax} = (0.30)(1.8) = 0.54\text{ kg m s}^{-1}$ $p_{Ay}' = m_A v_{Ay} = (0.30)(2.4) = 0.72\text{ kg m s}^{-1}$

Apply conservation of momentum separately in each direction. The total momentum in each direction must be unchanged by the collision: $p_{Bx}' = p_x - p_{Ax}' = 1.5 - 0.54 = 0.96\text{ kg m s}^{-1}$ $p_{By}' = p_y - p_{Ay}' = 0 - 0.72 = -0.72\text{ kg m s}^{-1}$

The negative sign shows puck $B$‘s $y$-momentum (and hence its motion) is on the south side of the original line of travel, the opposite side to puck $A$‘s deflection, as required for the $y$-components to cancel to zero overall.

Convert to velocity components using puck $B$‘s mass: $v_{Bx} = \frac{p_{Bx}'}{m_B} = \frac{0.96}{0.50} = 1.92\text{ m s}^{-1}$ $v_{By} = \frac{p_{By}'}{m_B} = \frac{-0.72}{0.50} = -1.44\text{ m s}^{-1}$

Combine into a magnitude and direction using Pythagoras’ theorem and trigonometry: $v_B = \sqrt{v_{Bx}^2+v_{By}^2} = \sqrt{1.92^2+1.44^2} = \sqrt{3.6864+2.0736} = \sqrt{5.76} = 2.4\text{ m s}^{-1}$ $\tan\theta = \frac{|v_{By}|}{v_{Bx}} = \frac{1.44}{1.92} = 0.75 \implies \theta = 36.9°$

So puck $B$ moves off at $2.4\text{ m s}^{-1}$, at $36.9°$ from due east, on the south side of the original line of travel.

Consistency check: adding the $x$-momenta after the collision, $0.54+0.96=1.50\text{ kg m s}^{-1}$, matches the $1.5\text{ kg m s}^{-1}$ found in part (a). Adding the $y$-momenta, $0.72+(-0.72)=0$, also matches the zero found in part (a). Confirming momentum is conserved in both directions.

Part (c): Elastic or inelastic?

Kinetic energy before the collision (only $A$ moving): $E_{k,\text{before}} = \tfrac{1}{2}m_A u_A^2 = \tfrac{1}{2}(0.30)(5.0)^2 = \tfrac{1}{2}(0.30)(25) = 3.75\text{ J}$

Kinetic energy after the collision (both pucks moving): $E_{k,\text{after}} = \tfrac{1}{2}m_A v_A^2 + \tfrac{1}{2}m_B v_B^2 = \tfrac{1}{2}(0.30)(3.0)^2 + \tfrac{1}{2}(0.50)(2.4)^2$ $= \tfrac{1}{2}(0.30)(9.0) + \tfrac{1}{2}(0.50)(5.76) = 1.35 + 1.44 = 2.79\text{ J}$

Since $E_{k,\text{after}} = 2.79\text{ J}$ is less than $E_{k,\text{before}} = 3.75\text{ J}$, kinetic energy is not conserved. The collision is inelastic (though the pucks do not stick together, some kinetic energy is lost to other forms of energy, such as sound and heat, during the impact).

Final answers

• (a) $p_x = \boxed{1.5}\text{ kg m s}^{-1}$ due east, $p_y = \boxed{0}\text{ kg m s}^{-1}$
• (b) Puck $B$‘s velocity $= \boxed{2.4}\text{ m s}^{-1}$, at $\boxed{36.9°}$ from due east, on the south side of the original line of travel
• (c) $E_{k,\text{before}} = \boxed{3.75}\text{ J}$, $E_{k,\text{after}} = \boxed{2.79}\text{ J}$. The collision is inelastic