A 600-kg car makes a 90° turn.Its speed before the turn is 21.0 m/s and after the turn it is 24.0 m/s.What is the magnitude of the change in the car's momentum during the turn?
A)19.1 × 1 $0 ^ { 3 }$ kg∙m/s
B)10.2 × 1 $0 ^ { 3 }$ kg∙m/s
C)9.55 × 1 $0 ^ { 3 }$ kg∙m/s
D)12.0 × 1 $0 ^ { 3 }$ kg∙m/s
E)22.2 × 1 $0 ^ { 3 }$ kg∙m/s

Question

Quizplus · Accepted Answer

The change in momentum ($\Delta p$) can be calculated using the vector difference between the final and initial momentum since momentum is a vector quantity. Given the 90° turn, the final and initial momentum vectors are perpendicular. The magnitude of the initial momentum ($p_i$) is $m \times v_i = 600 \, \text{kg} \times 21.0 \, \text{m/s} = 12600 \, \text{kg}\cdot\text{m/s}$, and the magnitude of the final momentum ($p_f$) is $m \times v_f = 600 \, \text{kg} \times 24.0 \, \text{m/s} = 14400 \, \text{kg}\cdot\text{m/s}$. Using the Pythagorean theorem for the vector magnitudes, $\Delta p = \sqrt{(p_f)^2 + (p_i)^2} = \sqrt{(14400)^2 + (12600)^2} = 19140 \, \text{kg}\cdot\text{m/s}$ or $19.1 \times 10^3 \, \text{kg}\cdot\text{m/s}$.

A 600-Kg Car Makes a 90° Turn 030 ^ { 3 }03

A 600-Kg Car Makes a 90° Turn $0 ^ { 3 }$