Question 8.6: Collision in a horizontal plane Figure 8.14a shows two battl......
Collision in a horizontal plane
Figure 8.14a shows two battling robots on a frictionless surface. Robot A, with mass 20 \mathrm{~kg}, initially moves at 2.0 \mathrm{~m} / \mathrm{s} parallel to the x-axis. It collides with robot B, which has mass 12 \mathrm{~kg} and is initially at rest. After the collision, robot A moves at 1.0 \mathrm{~m} / \mathrm{s} in a direction that makes an angle \alpha=30^{\circ} with its initial direction (Figure 8.14b). What is the final velocity of robot B ?
IDENTIFY and SET UP There are no horizontal external forces, so the x – and y-components of the total momentum of the system are conserved. Hence the sum of the x-components of momentum before the collision (subscript 1) must equal the sum after the collision (subscript 2), and similarly for the sums of the y-components. Our target variable is \overrightarrow{\boldsymbol{v}}_{B 2}, the final velocity of robot B.
EXECUTE The momentum-conservation equations and their solutions for v_{B 2 x} and v_{B 2 y} are
\begin{aligned} m_{A} v_{A 1 x}+m_{B} v_{B 1 x} & =m_{A} v_{A 2 x}+m_{B} v_{B 2 x} \\ v_{B 2 x} & =\frac{m_{A} v_{A 1 x}+m_{B} v_{B 1 x}-m_{A} v_{A 2 x}}{m_{B}} \\ & =\frac{\left[\begin{array}{c} (20 \mathrm{~kg})(2.0 \mathrm{~m} / \mathrm{s})+(12 \mathrm{~kg})(0) \\ -(20 \mathrm{~kg})(1.0 \mathrm{~m} / \mathrm{s})\left(\cos 30^{\circ}\right) \end{array}\right]}{12 \mathrm{~kg}}=1.89 \mathrm{~m} / \mathrm{s} \\ m_{A} v_{A 1 y}+m_{B} v_{B 1 y} & =m_{A} v_{A 2 y}+m_{B} v_{B 2 y} \\ v_{B 2 y} & =\frac{m_{A} v_{A 1 y}+m_{B} v_{B 1 y}-m_{A} v_{A 2 y}}{m_{B}} \\ & =\frac{\left[\begin{array}{l} (20 \mathrm{~kg})(0)+(12 \mathrm{~kg})(0) \\ -(20 \mathrm{~kg})(1.0 \mathrm{~m} / \mathrm{s})\left(\sin 30^{\circ}\right) \end{array}\right]}{12 \mathrm{~kg}}=-0.83 \mathrm{~m} / \mathrm{s} \end{aligned}
Figure 8.14 \mathrm{~b} shows the motion of robot B after the collision. The magnitude of \overrightarrow{\boldsymbol{v}}_{B 2} is
v_{B 2}=\sqrt{(1.89 \mathrm{~m} / \mathrm{s})^{2}+(-0.83 \mathrm{~m} / \mathrm{s})^{2}}=2.1 \mathrm{~m} / \mathrm{s}
and the angle of its direction from the positive x-axis is
\beta=\arctan \frac{-0.83 \mathrm{~m} / \mathrm{s}}{1.89 \mathrm{~m} / \mathrm{s}}=-24^{\circ}
EVALUATE Let’s confirm that the components of total momentum before and after the collision are equal. Initially robot A has x-momentum m_{A} v_{A 1 x}=(20 \mathrm{~kg})(2.0 \mathrm{~m} / \mathrm{s})=40 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} and zero y-momentum; robot B has zero momentum. Afterward, the momentum components are m_{A} v_{A 2 x}=(20 \mathrm{~kg})(1.0 \mathrm{~m} / \mathrm{s})\left(\cos 30^{\circ}\right)=17 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} and m_{B} v_{B 2 x}= (12 \mathrm{~kg})(1.89 \mathrm{~m} / \mathrm{s})=23 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}; the total x-momentum is 40 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}, the same as before the collision. The final y-components are m_{A} v_{A 2 y}=(20 \mathrm{~kg})(1.0 \mathrm{~m} / \mathrm{s})\left(\sin 30^{\circ}\right)=10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} and m_{B} v_{B 2 y}= (12 \mathrm{~kg})(-0.83 \mathrm{~m} / \mathrm{s})=-10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}; the total y-component of momentum is zero, as before the collision.
KEYCONCEPT In problems that involve a two-dimensional collision, write separate conservation equations for the x-component and y-component of total momentum.