COLLISION IN HORIZONTAL PLANE
Figure 8.14a shows two battling robots on a frictionless surface. Robot A, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. It collides with robot B, which has mass 12 kg and is initially at rest. After the collision, robot A moves at 1.0 m/s in a direction that makes an angle α = 30° with its initial direction (Fig. 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} \text { and } v_{B 2 y} are
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 m/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 m/s
Figure 8.14b 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 kg . m/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 kg . m/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 kg . m/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.